Primer on Quantitative Aptitude
1. Ten years ago, P was half of Q in age. If the ratio of their present ages is 3:4, what will be the total of their pre
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1. Ten years ago, P was half of Q in age. If the ratio of their present ages is 3:4, what will be the total of their present ages? A. 45
B. 40
C. 35
D. 30
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the present age of P and Q be 3x and 4x respectively.
Ten years ago, P was half of Q in age => (3x – 10) = ½ (4x – 10) => 6x – 20 = 4x – 10 => 2x = 10 => x = 5
total of their present ages = 3x + 4x = 7x = 7 × 5 = 35
2. Father is aged three times more than his son Sunil. After 8 years, he would be two and a half times of Sunil's age. After further 8 years, how many times would he be of Sunil's age? A. 4 times
B. 4 times
C. 2 times
D. 3 times
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Assume that Sunil's present age = x. Then father's present age = 3x + x = 4x
After 8 years, fathers age = 2 ½ times of Sunils' age => (4x+8) = 2 ½ × (x+8)
=> 4x + 8 = (5/2) ×(x + 8) => 8x + 16 = 5x+ 40 => 3x = 4016 = 24 => x = 24/3 = 8
After further 8 years, Sunil's age = x + 8+ 8 = 8+8+8 = 24 Father's age = 4x + 8 + 8 = 4 × 8 + 8 + 8 = 48 Father's age/Sunil's age = 48/24 = 2
3. A man's age is 125% of what it was 10 years ago, but 83 1/3 % of what it will be after ten 10 years. What is his present age? A. 70
B. 60
C. 50
D. 40
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the age before 10 years = x Then 125x/100 = x + 10 =>125x = 100x + 1000 => x = 1000/25 = 40
Present age = x + 10 = 40 +10 = 50
4. A man is 24 years older than his son. In two years, his age will be twice the age of his son. What is the present age of his son? A. 23 years
B. 22 years
C. 21 years
D. 20 years
Hide Answer | Notebook | Discuss
Here is the answer and explanation Answer : Option B Explanation : Let the present age of the son = x years Then present age the man = (x+24) years
Given that in 2 years, man's age will be twice the age of his son => (x+24) +2 = 2(x+2) => x = 22
5. Present ages of Kiran and Syam are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Syam's present age in years? A. 28
B. 27
C. 26
D. 24
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Ratio of the present age of Kiran and Syam = 5 : 4 => Let the present age of Kiran = 5x Present age of Syam = 4x
After 3 years, ratio of their ages = 11:9 => (5x + 3) : (4x + 3) = 11 : 9 => (5x+3) / (4x+3) = 11/9 => 9(5x + 3) = 11(4x + 3) => 45x + 27 = 44x + 33 => x = 3327 =6
Syam's present age = 4x = 4×6 = 24
6. The sum of ages of 5 children born at the intervals of 3 years each is 50 years. Find out the age of the youngest child? A. 6 years
B. 5 years
C. 4 years
D. 3 years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the age of the youngest child = x Then the ages of 5 children can be written as x, (x+3), (x+6), (x+9) and (x+12)
X + (x+3) + (x+6) + (x+9) + (x+12) = 50 => 5x + 30 =50 => 5x = 20 => x = 20/5 = 4
7. A is two years older than B who is twice as old as C. The total of the ages of A, B and C is 27. How old is B? A. 10
B. 9
C. 8
D. 7
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the age of C = x. Then Age of B = 2x Age of A = 2 + 2x
The total age of A,B and C =27 => (2+2x) + 2x + x = 27 => 5x = 25 => 25/5 = 5
B's age = 2x = 2×5 = 10
8. The Average age of a class of 22 students is 21 years. The average increased by 1 when the teacher's age also included. What is the age of the teacher? A. 48
B. 45
C. 43
D. 44
View Answer | Notebook | Discuss
9. A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 38 years now, what was the son's age five years back? A. 20 years
B. 18 years
C. 14 years
D. 22 years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the son's present age be x years. Then, (38 x) = x => 2x = 38 => x = 38/2 = 19
Son's age 5 years back = 195 = 14
10. Ayisha's age is 1/6th of her father's age. Ayisha 's father's age will be twice the age of Shankar's age after 10 years. If Shankar's eight birthdays was celebrated two years before, then what is Ayisha 's present age. A. 10 years
B. 12 years
C. 8 years
D. 5 years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Consider Ayisha's present age = x Then her father's age = 6x
Given that Ayisha 's father's age will be twice the age of Shankar's age after 10 years => Shankar's age after 10 years = ½(6x + 10) = 3x + 5
Also given that Shankar's eight birthdays was celebrated two years before => Shankar's age after 10 years = 8 + 12 = 20 => 3x + 5 = 20 => x = 15/3 = 5 => Ayisha 's present age = 5 years
11. The sum of the present ages of a son and his father is 60 years. Six years ago, father's age was five times the age of the son. After 6 years, what will be son's age? A. 23 years
B. 22 years
C. 21 years
D. 20 years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the present age of the son = x, then Present age of the father = 60x
Six years ago father's age was 5 times the age of the son
=> (60x)6 = 5(x6) => 84 = 6x => x = 84/6 = 14
Son's age after 6 years = x + 6 = 14 + 6 = 20
12. Kiarn is younger than Bineesh by 7 years and their ages are in the respective ratio of 7 : 9, how old is Kiran? A. 25
B. 24.5
C. 24
D. 23.5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the ages of Kiran and Bineesh are 7x and 9x respectively
7x = 9x7 => x = 7/2 = 3.5
Kiran's age = 7x = 7 × 3.5 = 24.5
13. Six years ago, the ratio of the ages of Vimal and Saroj was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Saroj's age at present? A. 18
B. 17
C. 16
D. 15
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Given that , six years ago, the ratio of the ages of Vimal and Saroj = 6 : 5
Hence we can assume that The age of Vimal six years ago = 6x The age of Saroj six years ago = 5x
After 4 years, the ratio of their ages = 11 : 10 => (6x+10) / (5x + 10) = 11/10 => 10(6x + 10) = 11(5x + 10) => 5x = 10 => x = 10/5 = 2
Saroj's present age = 5x + 6 = 5×2 +6 = 16
14. At present, the ratio between the ages of Shekhar and Shobha is 4 : 3. After 6 years, Shekhar's age will be 26 years. Find out the age of Shobha at present? A. 15 years
B. 14 years
C. 13 years
D. 12 years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : After 6 years, Shekhar's age will be 26years => Present age of Shekhar = 26 – 6 = 20
Let present age of Shobha = x Then 20/x = 4/3 => x = 20×3/4 = 15
15. My brother is 3 years elder to me. My father was 28 years of age when my sister was born while my
mother was 26 years of age when I was born. If my sister was 4 years of age when my brother was born, then what was the age my father when my brother was born? A. 35 years
B. 34 years
C. 33 years
D. 32 years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let my age = x Then My brother's age = x + 3 My mother's age = x + 26 My sister's age = (x + 3) + 4 = x + 7 My Father's age = (x + 7) + 28 = x + 35
=> age my father when my brother was born = x + 35 – (x + 3) = 32
16. The present ages of A,B and C are in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. What are their present ages (in years)? A. Insufficient data
B. 16, 30, 40
C. 16, 28 40
D. 16, 28, 36
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let's take the present age of A,B and C as 4x, 7x and 9x respectively Then (4x 8) + (7x – 8) + (9x – 8) = 56 => 20x = 80 => x = 4
Hence the present age of A, B and C are 4×4, 7×4 and 9×4 respectively ie, 16,28 and 36 respectively.
17. A person's present age is twofifth of the age of his mother. After 8 years, he will be onehalf of the age of his mother. What is the present age of the mother? A. 60
B. 50
C. 40
D. 30
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the present age of the person = x . Then present age of the mother = 5x/2
Given that , after 8 years, the person will be onehalf of the age of his mother. => (x + 8) = (1/2)(5x/2 + 8) => 2x + 16 = 5x/2 + 8 => x/2 = 8 => x = 16
Present age of the mother = 5x/2 = 5×16/2 = 40
18. A is as much younger than B and he is older than C. If the sum of the ages of B and C is 50 years, what is definitely the difference between B and A's age? A. Data inadequate
B. 3 years
C. 2 years
D. 5 years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Age of C Sobha's father's age Sobha's mother's age = (x + 38) – [(x4) + 36] = x + 38 –x +4 – 36 = 6
20. The age of father 10 years ago was thrice the age of his son. Ten years hence, father's age will be twice that of his son. What is the ratio of their present ages? A. 7 : 3
B. 3 : 7
C. 9 : 4
D. 4 : 9
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the age of the son before 10 years = x and age of the father before 10 years = 3x
Now we can write as (3x + 20) = 2(x + 20) => x = 20
=> Age of Father the son at present = x + 10 = 20 + 10 = 30 Age of the father at present = 3x + 10 = 3×20 + 10 = 70 Required ratio = 70 : 30 = 7 : 3
21. The ages of two persons differ by 16 years. 6 years ago, the elder one was 3 times as old as the younger one. What are their present ages of the elder person? A. 10
B. 20
C. 30
D. 40
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let's take the present age of the elder person = x and the present age of the younger person = x – 16
(x – 6) = 3 (x166) => x – 6 = 3x – 66 => 2x = 60 => x = 60/2 = 30
22. The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. What is father's present age? A. 30 years
B. 31 years
C. 32 yeas
D. 33 years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the present age the son = x. Then present age of the father = 3x + 3
Given that ,three years hence, father's age will be 10 years more than twice the age of the son => (3x+3+3) = 2(x + 3) +10 => x = 10
Father's present age = 3x + 3 = 3×10+ 3 = 33
23. Kamal was 4 times as old as his son 8 years ago. After 8 years, Kamal will be twice as old as his son. Find out the present age of Kamal. A. 40 years
B. 38 years
C. 42 years
D. 36 years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the age of the son before 8 years = x. Then age of Kamal before 8 years ago = 4x
After 8 years, Kamal will be twice as old as his son => 4x + 16 = 2(x + 16) => x = 8 Present age of Kamal = 4x + 8 = 4×8 +8 = 40
24. If 6 years are subtracted from the present age of Ajay and the remainder is divided by 18, then the present age of Rahul is obtained. If Rahul is 2 years younger to Denis whose age is 5 years, then what is Ajay 's present age? A. 50 years
B. 60 years
C. 55 years
D. 62 years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Present age of Denis = 5 years Present age of Rahul = 52 = 3
Let the present age of Ajay = x Then (x6)/18 = present age of Rahul = 3 => x 6 = 3×18 = 54 => x = 54 + 6= 60
25. The ratio of the age of a man and his wife is 4:3. At the time of marriage the ratio was 5:3 and After 4 years this ratio will become 9:7. How many years ago were they married? A. 8 years
B. 10 years
C. 11 years
D. 12 years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the present age of the man and his wife be 4x and 3x respectively.
After 4 years this ratio will become 9:7 => (4x + 4)/ (3x + 4) = 9/7 => 28x + 28 = 27x + 36 => x = 8
=> Present age of the man = 4x = 4×8 = 32 Present age of his wife = 3x = 3×8 = 24
Assume that they got married before t years. Then (32 – t) / (24 – t) = 5/3 => 96 – 3t = 120 – 5t => 2t = 24 => t = 24/2 = 12
26. The product of the ages of Syam and Sunil is 240. If twice the age of Sunil is more than Syam's age by 4 years, what is Sunil's age? A. 16
B. 14
C. 12
D. 10
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the age of Sunil = x and age of Syam = y. Then xy = 240 (1)
2x = y + 4 => y = 2x – 4
=> y = 2(x – 2) (2)
Substituting equation (2) in equation (1). We get x × 2(x2) = 240 => x (x2) = 240/2 => x (x 2) = 120 (3) We got a quadratic equation to solve. Always time is precious and objective tests measures not only how accurate you are but also how fast you are. We can either solve this quadratic equation in the traditional way. But more easy way is just substitute the values given in the choices in the quadratic equation (equation 3 ) and see which choice satisfy the equation. Here the option A is 10. If we substitute that value in the quadratic equation, x(x2) = 10 × 8 which is not equal to 120 Now try option B which is 12. If we substitute that value in the quadratic equation, x(x2) = 12 × 10 = 120. See, we got that x = 12. Hence Sunil's age = 12
(Or else, we cal solve the quadratic equation by factorization as, x(x 2) = 120 => x2 – 2x 120 = 0 =>(x12)(x+10) = 0 => x = 12 or 10. Since x is age and can not be negative, x = 12 Or
by
using
−−−−−−− −b ± √ b2 − 4ac x= = 2a
quadratic
formula
as
−−−−−−−−−−−−−−−−−−− 2 ± √ (−2) 2 − 4 × 1 × (−120) 2×1
−−− 2±√− 4−−−− + 480− 2 ± √ 484 2 ± 22 = = = = 12 or − 10 2 2 2 Since age is positive, x = 12 )
27. One year ago, the ratio of Sooraj's and Vimal's age was 6: 7 respectively. Four years hence, this ratio would become 7: 8. How old is Vimal? A. 32
B. 34
C. 36
D. 38
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let take the age of Sooraj and Vimal , 1 year ago as 6x and 7x respectively.
Given that, four years hence, this ratio would become 7: 8. => (6x + 5)/(7x + 5) = 7/8 => 48x + 40 = 49x + 35 => x = 5
Vimal's present age = 7x + 1 = 7×5 + 1 = 36
28. The total age of A and B is 12 years more than the total age of B and C. C is how many year younger than A? A. 10
B. 11
C. 12
D. 13
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Given that A+B = 12 + B + C => A – C = 12 + B – B = 12 => C is younger than A by 12 years
29. Sachin's age after 15 years will be 5 times his age 5 years back. Find out the present age of Sachin? A. 10 years
B. 11 years
C. 12 years
D. 13 years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the present age of Sachin = x Then (x+15) = 5(x5) => 4x = 40 => x = 10
30. Sandeep's age after six years will be threeseventh of his father's age. Ten years ago the ratio of their ages was 1 : 5. What is Sandeep's father's age at present? A. 30 years
B. 40 years
C. 50 years
D. 60 years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the age of Sandeep and his father before 10 years be x and 5x respectively.
Given that Sandeep's age after six years will be threeseventh of his father's age => x + 16 = (3/7)(5x + 16) => 7x + 112 = 15x + 48 => 8x = 64 => x = 8
Sandeep's father's present age = 5x + 10 = 5×8 + 10 = 50
Comments(244)
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daniel bishop 02 Feb 2015 12:37 AM My Brother is four years older than I am. The sum of our ages is 34. How old are we? Like | Dislike | Reply | Flag
Jay 11 Feb 2015 9:46 PM take the ages as x and (x+4) sum = 2x+4=34 =>x=15 ages are 15 and 19 Like | Dislike | Reply | Flag
amritava 29 Jan 2015 12:48 PM the mean of the ages of father and his son is 27 years.after 18 years,father will be twice as old as his son. what would be their present ages? Like | Dislike | Reply | Flag
Raj 09 Feb 2015 11:51 PM let age of father be x and his son be y (x+y)/2 = 27 x+y=54 (eq1) After 18 years,father will be twice as old as his son (x+18) = 2(y+18) x+18=2y+36 x2y=18 (eq2) Subtracting eq2 from eq1 3y = 36 y=12 x=54y=42 Present age of the father = 42 Present age of the son =12
I mportant Formulas - Area 1. P ythagorean Theorem (P ythagoras' theorem )
In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides c 2 = a 2 + b 2 where c is the length of the hypotenuse and a and b are the lengths of the other two sides 2.
Pi is a mathematical constant which is the ratio of a circle's circumference to its diameter. It is denoted by π
π ≈ 3.14 ≈ 3.
22 7
Geom etric Shapes and solids and I m portant Form ulas
Geom etric Shapes
Description
Form ulas
Rectangle l = Length b = Breadth d= Length of
Area = lb Perimeter = 2(l + b)
−−−−−− d = √ l2 + b2
diagonal Squar e a = Length of a side d= Length of diagonal
Area = a 2 =
1 2 d 2
Perimeter = 4a d = √ 2 a
Par allelogr am b and c are sides b = base
Area = bh Perimeter = 2(b + c)
h = height Rhombus a = length of each side
Area = bh (Formula 1 for area)
b = base
Area =
1 d1 d2 2
h = height
(Formula 2 for area)
d1, d2 are the
Perimeter = 4a
diagonal
1 bh 2 (Formula 1 for area) Area =
a , b and c are sides
−−−−−−−−−−−−−−−−−−− Area = √ S(S − a)(S − b)(S − c) a+b+c where S is the semiperimeter = 2 (Formula 2 for area - Heron's formula)
b = base
h = height
Perimeter = a + b + c
Tr iangle
Radius of incircle of a triangle of area A = A a+b+c where S is the semiperimeter = S 2
Area = Equilater al Tr iangle a = side
√3 2 a 4
Perimeter = 3a Radius of incircle of an equilateral triangle of side a =
a 2√ 3
Radius of circumcircle of an equilateral triangle of side a =
Tr apezium
Base a is parallele to base b
(Trapezoid in
Area =
American English)
1 (a + b)h 2
h = height d = 2r Cir cle r = radius d = diameter
Area = πr 2 =
1 2 πd 4
Circumference = 2πr = πd Circumference =π d
a √3
θ πr 2 (if angle measure is in degrees) 360
Area, A =
1 2 r θ(if angle measure is in radians) 2 θ πr(if angle measure is in degrees) 180 Sector of Cir cle
Arc Length, s =
r = radius θ = central angle
rθ(if angle measure is in radians)
Plese note that in the radian system for angular measurement, 2π radians = 360° 180° ⇒ 1 radian = π π ⇒ 1° = radians 180
Hence, Angle in Degrees = Angle in Radians ×
180° π
Angle in Radians = Angle in Degrees ×
π 180°
Ellipse Major axis length = 2a Minor axis length = 2b
Area = πab
−−−−−−− a 2 + b2 √ Perimeter ≈ 2π 2
Rectangular Solid l = length w = width
Total Surface Area = 2lw + 2wh + 2hl = 2(lw + wh + hl) Volume = lwh
h = height Cube
Total Surface Area = 6s2
s = edge
Volume = s3
Right Cir cular
Lateral Surface Area = (2 π r)h
Cylinder h = height r = radius of base
Total Surface Area = (2 π r)h + 2 (π r2) Voulme = (π r2)h
Pyr amid h = height B = area of the
Total Surface Area = B + Sum of the areas of the trianguar sides
Volume =
base
Right Cir cular Cone
−−−−−− Lateral Surface Area =πr√ r 2 + h 2 = π rs −−−−−− where s is the slant height = √ r 2 + h 2
h = height
r = radius of base
4.
−−−−−− Total Surface Area = πr√ r 2 + h 2 + πr 2 = πrs + πr 2
Spher e
d = 2r
r = radius
Surface Area = 4πr 2 = πd 2
d = diameter
Volume =
I m portant properties of Geom etric Shapes I. P roperties of Triangle i.
1 Bh 3
Sum of the angles of a triangle = 180°
4 3 1 3 πr = πd 3 6
ii.
Sum of any two sides of a triangle is greater than the third side.
iii.
The line joining the midpoint of a side of a triangle to the positive vertex is called the median
iv.
The median of a triangle divides the triangle into two triangles with equal areas
v. vi. vii.
viii. ix.
Centroid is the point where the three medians of a triangle meet. Centroid divides each median into segments with a 2:1 ratio Area of a triangle formed by joining the midpoints of the sides of a given triangle is one-fourth of the area of the given triangle. An equilateral triangle is a triangle in which all three sides are equal In an equilateral triangle, all three internal angles are congruent to each other
x.
In an equilateral triangle, all three internal angles are each 60°
xi.
An isosceles triangle is a triangle with (at least) two equal sides
xii.
In isosceles triangle, altitude from vertex bisects the base.
II. P roperties of Quadrilaterals A. Rectangle i. ii.
The diagonals of a rectangle are equal and bisect each other opposite sides of a rectangle are parallel
iii.
opposite sides of a rectangle are congruent
iv.
opposite angles of a rectangle are congruent
v.
All four angles of a rectangle are right angles
vi.
The diagonals of a rectangle are congruent
B. Square i.
All four sides of a square are congruent
ii.
Opposite sides of a square are parallel
iii.
The diagonals of a square are equal
iv.
The diagonals of a square bisect each other at right angles
v.
All angles of a square are 90 degrees.
C. P arallelogram i. ii.
The opposite sides of a parallelogram are equal in length. The opposite angles of a parallelogram are congruent (equal measure).
iii.
The diagonals of a parallelogram bisect each other.
iv.
Each diagonal of a parallelogram divides it into two triangles of the same area
D. Rhom bus i.
All the sides of a rhombus are congruent
ii.
Opposite sides of a rhombus are parallel.
iii.
The diagonals of a rhombus are unequal and bisect each other at right angles
iv.
Opposite internal angles of a rhombus are congruent (equal in size)
v.
Any two consecutive internal angles of a rhombus are supplementary; i.e. the sum of their angles = 180° (equal in size)
Other properties of quadrilaterals i. ii.
The sum of the interior angles of a quadrilateral is 360 degrees A square and a rhombus on the same base will have equal areas.
iii.
A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
iv.
Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area.
v.
Each diagonal of a parallelogram divides it into two triangles of the same area
III. Sum of I nterior Angles of a polygon i.
The sum of the interior angles of a polygon = 180(n - 2) degrees where n = number of sides Example 1 : Number of sides of a triangle = 3. Hence, sum of the interior angles of a triangle = 180(3 - 2) = 180 × 1 = 180 ° Example 2 : Number of sides of a quadrilateral = 4. Hence, sum of the interior angles of any quadrilateral = 180(4 - 2) = 180 × 2 = 360 °
1. An error 2% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square? A. 4.04 %
B. 2.02 %
C. 4 %
D. 2 %
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Error = 2% while measuring the side of a square. Let the correct value of the side of the square = 100
Then the measured value = 100 ×
(100 + 2) = 102 (∵ error 2% in excess) 100
Correct Value of the area of the square = 100 × 100 = 10000 Calculated Value of the area of the square = 102 × 102 = 10404 Error = 10404 - 10000 = 404
Percentage Error =
Error 404 × 100 = × 100 = 4.04% 10000 Actual Value
2. A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. The area of the lawn is 2109 sq. m. what is the width of the road? A. 5 m
B. 4 m
C. 2 m
D. 3 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Please refer the diagram given above. Area of the park = 60 × 40 = 2400 m2 Given that area of the lawn = 2109 m2 ∴ Area of the cross roads = 2400 - 2109 = 291 m2 Assume that the width of the cross roads = x Then total area of the cross roads = Area of road 1 + area of road 2 - (Common Area of the cross roads) = 60x + 40x - x2 (Let's look in detail how we got the total area of the cross roads as 60x + 40x - x2 As shown in the diagram, area of the road 1 = 60x. This has the areas of the parts 1,2 and 3 given in the diagram Area of the road 2 = 40x. This has the parts 4, 5 and 6 You can see that there is an area which is intersecting (i.e. part 2 and part 5)
and the intersection area = x2. Since 60x + 40x covers the intersecting area (x2) two times ( part 2 and part 5) ,we need to subtract the intersecting area of (x2) once time to get the total area. . Hence total area of the cross roads = 60x + 40x - x2) Now, we have Total areas of cross roads = 60x + 40x - x2 But area of the cross roads = 291 m2 Hence 60x + 40x - x2 = 291 => 100x - x2 = 291 => x2 - 100x + 291 = 0 => (x - 97)(x - 3) = 0 => x = 3 (x can not be 97 as the park is only 60 m long and 40 m wide)
3. A towel, when bleached, lost 20% of its length and 10% of its breadth. What is the percentage of decrease in area? A. 30 %
B. 28 %
C. 32 %
D. 26 %
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : --------------------------------------------------------Solution 1 --------------------------------------------------------Let original length = 100 and original breadth = 100 Then original area = 100 × 100 = 10000
Lost 20% of length ⇒ New length = Original length ×
(100 − 20) 80 = 100 × = 80 100 100
Lost 10% of breadth ⇒ New breadth= Original breadth × New area = 80 × 90 = 7200
(100 − 10) 90 = 100 × = 90 100 100
Decrease in area = Original Area - New Area = 10000 - 7200 = 2800
Percentage of decrease in area =
Decrease in Area 2800 × 100 = × 100 = 28% 10000 Original Area
--------------------------------------------------------Solution 2 --------------------------------------------------------Let original length = l and original breadth = b Then original area = lb
Lost 20% of length ⇒ New length = Original length ×
(100 − 20) 80 80l =l× = 100 100 100
Lost 10% of breadth ⇒ New breadth= Original breadth ×
New area =
(100 − 10) 90 90b =b× = 100 100 100
80l 90b 7200lb 72lb × = = 100 100 10000 100
Decrease in area = Original Area - New Area = lb −
Percentage of decrease in area = ( =
28lb ) 100 lb
× 100 =
72lb 28lb = 100 100
Decrease in Area × 100 Original Area
28lb × 100 = 28% 100lb
4. If the length of a rectangle is halved and its breadth is tripled, what is the percentage change in its area? A. 25 % Increase
B. 25 % Decrease
C. 50 % Decrease
D. 50 % Increase
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : --------------------------------------------------------Solution 1 --------------------------------------------------------Let original length = 100 and original breadth = 100 Then original area = 100 × 100 = 10000
Length of the rectangle is halved Original length 100 ⇒ New length = = = 50 2 2 breadth is tripled ⇒ New breadth= Original breadth × 3 = 100 × 3 = 300 New area = 50 × 300 = 15000
Increase in area = New Area - Original Area = 15000 - 10000= 5000
Percentage of Increase in area =
Increase in Area 5000 × 100 = × 100 = 50% 10000 Original Area
--------------------------------------------------------Solution 2 --------------------------------------------------------Let original length = l and original breadth = b Then original area = lb
Length of the rectangle is halved Original length l ⇒ New length = = 2 2 breadth is tripled ⇒ New breadth = Original breadth × 3 = 3b New area =
l 3lb × 3b = 2 2
Increase in area = New Area - Original Area = Percentage of Increase in area = ( =
lb ) 2 lb
× 100 =
3lb lb − lb = 2 2
Increase in Area × 100 Original Area
lb × 100 = 50% 2lb
5. A person walked diagonally across a square plot. Approximately, what was the percent saved by not walking along the edges? A. 35%
B. 30 %
C. 20 %
D. 25%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : --------------------------------------------------------Solution 1 ---------------------------------------------------------
Consider a square plot as shown above and let the length of
each side = 1
−−−−−− Then length of the diagonal = √ (1 + 1) = √ 2
Distance travelled if walked along the edges = BC + CD = 1 + 1 = 2
Distance travelled if walked diagonally = BD = √ 2 = 1.41
Distance Saved = 2 - 1.41 = .59
Percent distance saved =
.59 × 100 = .59 × 50 ≈ 30% 2
--------------------------------------------------------Solution 2 ---------------------------------------------------------
Consider a square plot as shown above and let the length of each side = x
−−−−−− −− Then length of the diagonal = √ (x + x) = √ 2x
Distance travelled if walked along the edges = BC + CD = x + x = 2x
−− Distance travelled if walked diagonally = BD = √ 2x = 1.41x
Distance Saved = 2x - 1.41x = .59x
Percent distance saved =
.59x × 100 = .59 × 50 ≈ 30% 2x
6. A rectangular field has to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required? A. 95
B. 92
C. 88
D. 82
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Given that area of the field = 680 sq. feet => lb = 680 sq. feet Length(l) = 20 feet => 20 × b = 680
⇒b=
680 = 34 feet 20
Required length of the fencing = l + 2b = 20 + (2 × 34) = 88 feet
7. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet? A. 126 sq. ft.
B. 64 sq. ft.
C. 100 sq. ft.
D. 102 sq. ft.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let l = 9 ft. Then l + 2b = 37 => 2b = 37 - l = 37 - 9 = 28 => b = 28/2 = 14 ft.
Area = lb = 9 × 14 = 126 sq. ft.
8. The area of a rectangle plot is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the plot? A. 14 metres
B. 20 metres
C. 18 metres
D. 12 metres
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : lb = 460 m2 ------(Equation 1) Let the breadth = b
Then length, l = b ×
(100 + 15) 115b = ------(Equation 2) 100 100
From Equation 1 and Equation 2,
115b × b = 460 100 b2 =
46000 = 400 115
−− = 20 m ⇒b=√− 400
9. A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares? A. 400
B. 365
C. 385
D. 315
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the areas of the parts be x hectares and (700 - x) hectares.
Difference of the areas of the two parts = x - (700 - x) = 2x - 700 one-fifth of the Average of the two areas = =
1 700 350 × = = 70 5 2 5
1 [x + (700 − x)] 5 2
Given that difference of the areas of the two parts = one-fifth of the Average of the two areas => 2x - 700 = 70 => 2x = 770
⇒x=
770 = 385 2
Hence, Area of smaller part = (700 - x) = (700 – 385) = 315 hectares.
10. The length of a room is 5.5 m and width is 3.75 m. What is the cost of paying the floor by slabs at the rate of Rs. 800 per sq. metre. A. Rs.12000
B. Rs.19500
C. Rs.18000
D. Rs.16500.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Area = 5.5 × 3.75 sq. metre. Cost for 1 sq. metre. = Rs. 800 Hence total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = Rs. 16500
11. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. What is the length of the rectangle? A. 18 cm
B. 16 cm
C. 40 cm
D. 20 cm
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option C Explanation : Let breadth = x cm Then length = 2x cm Area = lb = x × 2x = 2x2 New length = (2x - 5) New breadth = (x + 5) New Area = lb = (2x - 5)(x + 5) But given that new area = initial area + 75 sq.cm. => (2x - 5)(x + 5) = 2x2 + 75 => 2x2 + 10x - 5x - 25 = 2x2 + 75 => 5x - 25 = 75 => 5x = 75 + 25 = 100 => x = 100/5 = 20 cm Length = 2x = 2 × 20 = 40cm
12. If a square and a rhombus stand on the same base, then what is the ratio of the areas of the square and the rhombus? A. equal to ½
B. equal to ¾
C. greater than 1
D. equal to 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : A square and a rhombus on the same base will have equal areas. Hence ratio of the areas of the square and the rhom bus w ill be equal to 1 since they stand on the sam e base ================================================================ Note : Please find the proof of the formula given below which you may like to go through
Let ABCD be the square and ABEF be the rhombus Consider the right-angled triangles ADF and BCE We know that AD = BC (∵ sides of a square) AF = BE (∵ sides of a rhombus) ∴ DF = CE [∵ DF2 = AF2 - AD2 and CE2 = BE2 - BC2] Hence Δ ADF = Δ BCE => Δ ADF + Trapezium ABCF= Δ BCE + Trapezium ABCF => Area of square ABCD = Area of rhombus ABEF
13. The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m, find out the area of the field. A. 37500 m2
B. 30500 m2
C. 32500 m2
D. 40000 m2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Given that breadth of a rectangular field is 60% of its length
⇒b=
60l 3l = 100 5
perimeter of the field = 800 m => 2 (l + b) = 800
⇒ 2 (l + ⇒l+
3l ) = 800 5
3l = 400 5
⇒
8l = 400 5
⇒
l = 50 5
⇒ l = 5 × 50 = 250 m
b =
3l 3 × 250 = = 2 × 50 = 150 m 5 5
Area = lb = 250 × 150 = 37500 m 2
14. A room 5m 44cm long and 3m 74cm broad needs to be paved with square tiles. What will be the least number of square tiles required to cover the floor? A. 176
B. 124
C. 224
D. 186
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : l = 5 m 44 cm = 544 cm b = 3 m 74 cm = 374 cm Area = 544 × 374 cm2 Now we need to find out HCF(Highest Common Factor) of 544 and 374. Let's find out the HCF using long division method for quicker results) 374) 544 (1 374 170) 374 (2 340
34) 170 (5 170 0
Hence, HCF of 544 and 374 = 34
Hence, side length of largest square tile we can take = 34 cm Area of each square tile = 34 × 34 cm2
Number of tiles required =
544 × 374 = 16 × 11 = 176 34 × 34
15. The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per metre is Rs. 5300, what is the length of the plot in metres? A. 60 m
B. 100 m
C. 75 m
D. 50 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Length of the plot is 20 metres more than its breadth. Hence, let's take the length as l metres and breadth as (l - 20) metres Length of the fence = perimeter = 2(length + breadth)= 2[ l + (l - 20) ] = 2(2l - 20) metres Cost per meter = Rs. 26.50 Total cost = 2(2l - 20) × 26.50 Total cost is given as Rs. 5300 => 2(2l - 20) × 26.50 = 5300 => (2l - 20) × 26.50 = 2650 => (l - 10) × 26.50 = 1325 => (l - 10) = 1325/26.50 = 50 => l = 50 + 10 = 60 metres
16. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then what is the area of the park (in sq. m)? A. 142000
B. 112800
C. 142500
D. 153600
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : l : b = 3 : 2 ------------------------------------------(Equation 1) Perimeter of the rectangular park = Distance travelled by the man at the speed of 12 km/hr in 8 minutes = speed × time = 12 × 8/60 (∵ 8 minute = 8/60 hour) = 8/5 km = 8/5 ×1000 m = 1600 m Perimeter = 2(l + b) => 2(l + b) = 1600 => l + b = 1600/2 = 800 m ---------------------------(Equation 2)
From (Equation 1) and (Equation 2) l = 800 × 3/5 = 480 m b = 800 × 2/5 = 320 m (Or b = 800 - 480 = 320m) Area = lb = 480 × 320 = 153600 m2
17. What is the percentage increase in the area of a rectangle, if each of its sides is increased by 20%? A. 45%
B. 44%
C. 40%
D. 42%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : --------------------------------------------------------Solution 1 ---------------------------------------------------------
Let original length = 100 and original breadth = 100 Then original area = 100 × 100 = 10000
Increase in 20% of length ⇒ New length = Original length ×
(100 + 20) 120 = 100 × = 120 100 100
Increase in 20% of breadth ⇒ New breadth= Original breadth ×
(100 + 20) 120 = 100 × = 120 100 100
New area = 120 × 120 = 14400
Increase in area = New Area - Original Area = 14400 - 10000 = 4400
Percentage increase in area =
Increase in Area 4400 × 100 = × 100 = 44% 10000 Original Area
--------------------------------------------------------Solution 2 --------------------------------------------------------Let original length = l and original breadth = b Then original area = lb
Increase in 20% of length ⇒ New length = Original length ×
(100 + 20) 120 120l =l× = 100 100 100
Increase in 20% of breadth ⇒ New breadth= Original breadth ×
New area =
(100 + 20) 120 120b =b× = 100 100 100
120l 120b 14400lb 144lb × = = 100 100 10000 100
Increase in area = New Area - Original Area =
Percentage of increase in area = ( =
44lb ) 100 lb
× 100 =
144lb 44lb − lb = 100 100
Increase in Area × 100 Original Area
44lb × 100 = 44% 100lb
18. If the difference between the length and breadth of a rectangle is 23 m and its perimeter is 206 m, what is its area? A. 2800 m2
B. 2740 m2
C. 2520 m2
D. 2200 m2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : l - b = 23 ...................(Equation 1) perimeter = 2(l + b) = 206 => l + b = 103.............(Equation 2) (Equation 1) + (Equation 2) => 2l = 23 + 103 = 126 => l = 126/2 = 63 metre
Substituting this value of l in (Equation 1), we get 63 - b = 23 => b = 63 - 23 = 40 metre Area = lb = 63 × 40 = 2520 m2
19. The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle? A. 16 cm
B. 18 cm
C. 14 cm
D. 20 cm
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Given that
2(l + b) =5 b
=> 2l + 2b = 5b => 2l = 3b
=> b =
2l 3
Also given that area = 216 cm2 => lb = 216 cm2
Substituting the value of b, we get, l × ⇒ l2 =
2l = 216 3
3 × 216 = 3 × 108 = (3 × 3) × 36 2
⇒ l = 3 × 6 = 18 cm
20. What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad? A. 814
B. 802
C. 836
D. 900
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : l = 15 m 17 cm = 1517 cm b = 9 m 2 cm = 902 cm Area = 1517 × 902 cm2 Now we need to find out HCF(Highest Common Factor) of 1517 and 902. Let's find out the HCF using long division method for quicker results)
902) 1517 (1 902 615) 902 (1 615 287) 615 (2 574 41) 287 (7 287 0
Hence, HCF of 1517 and 902 = 41
Hence, side length of largest square tile we can take = 41 cm Area of each square tile = 41 × 41 cm2
Number of tiles required =
1517 × 902 = 37 × 22 = 407 × 2 = 814 41 × 41
21. The diagonal of the floor of a rectangular room is 7
1 1 feet. The shorter side of the room is 4 feet. 2 2
What is the area of the room? A. 27 square feet
B. 22 square feet
C. 24 square feet
D. 20 square feet
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
1 15 feet = feet 2 2 1 9 Breadth, b = 4 feet = feet 2 2 Diagonal, d = 7
In the right-angled triangle PQR, 15 2 9 2 2 l =( ) −( ) 2 2 =
225 81 144 − = 4 4 4
−−−− 144 12 l=√ = feet = 6 feet 4 2
Area = lb = 6 ×
9 = 27 feet 2 2
22. The diagonal of a rectangle is √ rectangle?
−− 41 cm and its area is 20 sq. cm. What is the perimeter of the
A. 16 cm
B. 10 cm
C. 12 cm
D. 18 cm
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
For a rectangle, d 2 = l2 + b2 where l = length , b = breadth and d = diagonal of the of the rectangle
−− d = √ 41 cm d 2 = l2 + b2 −− ⇒ l2 + b2 = (√ 41 )
Area = lb = 20 cm
2
2
= 41........(Equation 1)
............(Equation 2)
Solving (Equation 1) and (Equation 2)
(a + b) 2 = a 2 + 2ab + b2
using the above formula, we have (l + b) 2 = l2 + 2lb + b2 = (l2 + b2 ) + 2lb = 41 + (2 × 20) = 81 ⇒ (l + b) = √ −− 81 = 9 cm perimeter = 2(l + b) = 2 × 9 = 18 cm
23. A tank is 25 m long, 12 m wide and 6 m deep. What is the cost of plastering of its walls and bottom at the rate of 75 paise per sq. m? A. Rs. 558
B. Rs. 502
C. Rs. 516
D. Rs. 612
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Consider a rectangular solid of length l, w idth w and height h. Then 1. Total Surface area of a rectangular solid, S = 2lw + 2lh + 2w h = 2(lw + lh + w h) 2. Volum e of a rectangular solid, V = lw h
In this case, l = 25 m, w = 12 m, h = 6 m and all surface needs to be plastered except the top Hence total area needs to be plastered = Total Surface Area - Area of the Top face = (2lw + 2lh + 2wh) - lw
= lw + 2lh + 2wh = (25 × 12) + (2 × 25 × 6) + (2 × 12 × 6) = 300 + 300 + 144 = 744 m2 Cost of plastering = 744 × 75 = 55800 paise = Rs.558
24. It is decided to construct a 2 metre broad pathway around a rectangular plot on the inside. If the area of the plots is 96 sq.m. and the rate of construction is Rs. 50 per square metre., what will be the total cost of the construction? A. Rs.3500
B. Rs. 4200
C. Insufficient Data
D. Rs. 4400
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Let length and width of the rectangular plot be l and b respectively Total Area of the rectangular plot = 96 sq.m. Width of the pathway = 2 m Length of the remaining area in the plot = (l - 4) breadth of the remaining area in the plot = (b - 4) Area of the remaining area in the plot = (l - 4)(b - 4) Area of the pathway = Total Area of the rectangular plot - remaining area in the plot
= 96 - [(l - 4)(b - 4)] = 96 - [lb - 4l - 4b + 16] = 96 - [96 - 4l - 4b + 16] = 96 - 96 + 4l + 4b - 16] = 4l + 4b - 16 = 4(l + b) - 16 We do not know the values of l and b and hence total area of the rectangular plot can not be found out. So we can not find out total cost of the construction.
25. The area of a parallelogram is 72 cm2 and its altitude is twice the corresponding base. What is the length of the base? A. 6 cm
B. 7 cm
C. 8 cm
D. 12 cm
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Area of a parallelogram , A = bh w here b is the base and h is the height of the parallelogram
Let the base = x cm. Then the height = 2x cm (∵ altitude is twice the base) Area = x × 2x = 2x2 But the area is given as 72 cm2 => 2x2 = 72
=> x2 = 36 => x = 6 cm
26. Two diagonals of a rhombus are 72 cm and 30 cm respectively. What is its perimeter? A. 136 cm
B. 156 cm
C. 144 cm
D. 121 cm
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Remember the following two properties of a rhombus which will be useful in solving this question 1. The sides of a rhombus are congruent. 2. The diagonals of a rhombus are unequal and bisect each other at right angles. Let the diagonals be PR and SQ such that PR = 72 cm and SQ = 30 cm
PO = OR =
72 = 36 cm 2
SO = OQ =
30 = 15 cm 2
−−−−−−−− −−−−−−− −−− = 39 cm PQ = QR = RS = SP = √ 36 2 + 15 2 = √ − 1296 + 225− = √ − 1521 perimeter = 4 × 39 =156 cm
27. The base of a parallelogram is (p + 4), altitude to the base is (p - 3) and the area is (p2 - 4), find out its actual area. A. 40 sq. units
B. 54 sq. units
C. 36 sq. units
D. 60 sq. units
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Area of a parallelogram , A = bh w here b is the base and h is the height of the parallelogram
Hence, we have p2 - 4 = (p + 4)(p - 3) => p2 - 4 = p2 - 3p + 4p - 12 => -4 = p - 12 => p = 12 - 4 = 8 Hence, actual area = (p2 - 4) = 82 - 4 = 64 - 4 = 60 sq. units
28. A circle is inscribed in an equilateral triangle of side 24 cm, touching its sides. What is the area of the remaining portion of the triangle? A. 144√
3 − 48π
cm2
B. 121√
3 − 36π
cm2
C. 144√
3 − 36π
cm2
D. 121√
3 − 48π
cm2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
√3 2 a 4 where a is length of one side of the equilateral triangle
Area of an equilateral triangle =
Area of the equilateral Δ ABC =
√3 2 √3 2 a = 24 = 144√ 3 cm 2 .............(1) 4 4
1 bh 2 where b is the base and h is the height of the triangle
Area of a triangle =
Let r = radius of the inscribed circle. Then Area of Δ ABC = Area of Δ OBC + Area of Δ OCA + area of Δ OAB = (½ × r × BC) + (½ × r × CA) + (½ × r × AB) = ½ × r × (BC + CA + AB) = ½ x r x (24 + 24 + 24) = ½ x r x 72 = 36r cm2 ------------------------------------------ (2)
From (1) and (2),
144√ 3 = 36r ⇒r=
144 √ 3 = 4√ 3 − − − − − − − − − − − −(3) 36
Area of a circle = πr 2 where = radius of the circle
From (3), the area of the inscribed circle = πr 2 = π(4√ 3 )
2
= 48π − − − − − − − − − − − −(4)
Hence , Area of the remaining portion of the triangle Area of Δ ABC – Area of inscribed circle 144√ 3 − 48π cm 2
29. A rectangular plot measuring 90 metres by 50 metres needs to be enclosed by wire fencing such that poles of the fence will be kept 5 metres apart. How many poles will be needed? A. 30
B. 44
C. 56
D. 60
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : P erim eter of a rectangle = 2(l + b) w here l is the length and b is the breadth of the rectangle
Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is
very important. Hence number of poles required = 280∕5 = 56
30. If the diagonals of a rhombus are 24 cm and 10 cm, what will be its perimeter A. 42 cm
B. 64 cm
C. 56 cm
D. 52 cm
View Answer | Notebook | Discuss
31. What will be the length of the longest rod which can be placed in a box of 80 cm length, 40 cm breadth and 60 cm height?
−−−−− 11600 cm −−−−− C. √ 10000 cm
A. √
−−−−− 14400 cm −−−−− D. √ 12040 cm
B. √
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
The longest road which can fit into the box will have one end at A and other end at G (or any other similar diagonal) Hence the length of the longest rod = AG Initially let's find out AC. Consider the right angled triangle ABC
AC2 = AB2 + BC2 = 402 + 802 = 1600 + 6400 = 8000
−−−− ⇒ AC = √ 8000 cm
Consider the right angled triangle ACG
AG2 = AC2 + CG2
−−− ) + 60 2 = 8000 + 3600 = 11600 = (√ − 8000 2
−−−− cm ⇒ AG = √ − 11600 −−−− cm ⇒ The length of the longest rod = √ − 11600
1. An error 2% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square? A. 4.04 %
B. 2.02 %
C. 4 %
D. 2 %
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Error = 2% while measuring the side of a square. Let the correct value of the side of the square = 100
Then the measured value = 100 ×
(100 + 2) = 102 (∵ error 2% in excess) 100
Correct Value of the area of the square = 100 × 100 = 10000 Calculated Value of the area of the square = 102 × 102 = 10404 Error = 10404 - 10000 = 404
Percentage Error =
Error Actual Value
× 100 =
404 × 100 = 4.04% 10000
2. A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. The area of the lawn is 2109 sq. m. what is the width of the road? A. 5 m
B. 4 m
C. 2 m
D. 3 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Please refer the diagram given above. Area of the park = 60 × 40 = 2400 m2 Given that area of the lawn = 2109 m2 ∴ Area of the cross roads = 2400 - 2109 = 291 m2 Assume that the width of the cross roads = x Then total area of the cross roads = Area of road 1 + area of road 2 - (Common Area of the cross roads) = 60x + 40x - x2 (Let's look in detail how we got the total area of the cross roads as 60x + 40x - x2 As shown in the diagram, area of the road 1 = 60x. This has the areas of the
parts 1,2 and 3 given in the diagram Area of the road 2 = 40x. This has the parts 4, 5 and 6 You can see that there is an area which is intersecting (i.e. part 2 and part 5) and the intersection area = x2. Since 60x + 40x covers the intersecting area (x2) two times ( part 2 and part 5) ,we need to subtract the intersecting area of (x2) once time to get the total area. . Hence total area of the cross roads = 60x + 40x - x2) Now, we have Total areas of cross roads = 60x + 40x - x2 But area of the cross roads = 291 m2 Hence 60x + 40x - x2 = 291 => 100x - x2 = 291 => x2 - 100x + 291 = 0 => (x - 97)(x - 3) = 0 => x = 3 (x can not be 97 as the park is only 60 m long and 40 m wide)
3. A towel, when bleached, lost 20% of its length and 10% of its breadth. What is the percentage of decrease in area? A. 30 %
B. 28 %
C. 32 %
D. 26 %
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : --------------------------------------------------------Solution 1 --------------------------------------------------------Let original length = 100 and original breadth = 100 Then original area = 100 × 100 = 10000
Lost 20% of length ⇒ New length = Original length ×
(100 − 20) 80 = 100 × = 80 100 100
Lost 10% of breadth ⇒ New breadth= Original breadth ×
(100 − 10) 90 = 100 × = 90 100 100
New area = 80 × 90 = 7200 Decrease in area = Original Area - New Area = 10000 - 7200 = 2800
Percentage of decrease in area =
Decrease in Area 2800 × 100 = × 100 = 28% 10000 Original Area
---------------------------------------------------------Solution 2--------------------------------------------------------Let original length = l and original breadth = b Then original area = lb
Lost 20% of length ⇒ New length = Original length ×
(100 − 20) 80 80l = l× = 100 100 100
Lost 10% of breadth ⇒ New breadth= Original breadth ×
New area =
(100 − 10) 90 90b = b× = 100 100 100
80l 90b 7200lb 72lb × = = 100 100 10000 100
Decrease in area = Original Area - New Area = lb −
Decrease in Area × 100 Original Area
Percentage of decrease in area = 28lb ) 100 = lb (
× 100 =
72lb 28lb = 100 100
28lb × 100 = 28% 100lb
4. If the length of a rectangle is halved and its breadth is tripled, what is the percentage change in its area? A. 25 % Increase
B. 25 % Decrease
C. 50 % Decrease
D. 50 % Increase
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : --------------------------------------------------------Solution 1 --------------------------------------------------------Let original length = 100 and original breadth = 100 Then original area = 100 × 100 = 10000
Length of the rectangle is halved Original length ⇒ New length = 2
=
100 = 50 2
breadth is tripled ⇒ New breadth= Original breadth × 3 = 100 × 3 = 300 New area = 50 × 300 = 15000 Increase in area = New Area - Original Area = 15000 - 10000= 5000
Percentage of Increase in area =
Increase in Area 5000 × 100 = × 100 = 50% 10000 Original Area
---------------------------------------------------------Solution 2--------------------------------------------------------Let original length = l and original breadth = b Then original area = lb
Length of the rectangle is halved Original length ⇒ New length = 2
=
l 2
breadth is tripled ⇒ New breadth = Original breadth × 3 = 3b New area =
l 3lb × 3b = 2 2
Increase in area = New Area - Original Area = Percentage of Increase in area = lb ) 2 = lb (
× 100 =
3lb lb − lb = 2 2
Increase in Area × 100 Original Area
lb × 100 = 50% 2lb
5. A person walked diagonally across a square plot. Approximately, what was the percent saved by not walking along the edges? A. 35%
B. 30 %
C. 20 %
D. 25%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : --------------------------------------------------------Solution 1 ---------------------------------------------------------
Consider a square plot as shown above and let the length of each side = 1
−−−−−− Then length of the diagonal = √ (1 + 1) = √ 2 Distance travelled if walked along the edges = BC + CD = 1 + 1 = 2
Distance travelled if walked diagonally = BD = √ 2 = 1.41 Distance Saved = 2 - 1.41 = .59
Percent distance saved =
.59 × 100 = .59 × 50 ≈ 30% 2
---------------------------------------------------------Solution 2---------------------------------------------------------
Consider a square plot as shown above and let the length of each side = x
−−−−−− −− Then length of the diagonal = √ (x + x) = √ 2x
Distance travelled if walked along the edges = BC + CD = x + x = 2x
−− Distance travelled if walked diagonally = BD = √ 2x = 1.41x Distance Saved = 2x - 1.41x = .59x
Percent distance saved =
.59x × 100 = .59 × 50 ≈ 30% 2x
6. A rectangular field has to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required? A. 95
B. 92
C. 88
D. 82
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Given that area of the field = 680 sq. feet => lb = 680 sq. feet Length(l) = 20 feet => 20 × b = 680
⇒b=
680 = 34 feet 20
Required length of the fencing = l + 2b = 20 + (2 × 34) = 88 feet
7. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet? A. 126 sq. ft.
B. 64 sq. ft.
C. 100 sq. ft.
D. 102 sq. ft.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let l = 9 ft. Then l + 2b = 37 => 2b = 37 - l = 37 - 9 = 28 => b = 28/2 = 14 ft. Area = lb = 9 × 14 = 126 sq. ft.
8. The area of a rectangle plot is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the plot? A. 14 metres
B. 20 metres
C. 18 metres
D. 12 metres
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : lb = 460 m2 ------(Equation 1) Let the breadth = b
Then length, l = b ×
(100 + 15) 115b = ------(Equation 2) 100 100
From Equation 1 and Equation 2,
115b × b = 460 100 b2 =
46000 = 400 115
−− = 20 m 400 ⇒b=√−
9. A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares? A. 400
B. 365
C. 385
D. 315
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the areas of the parts be x hectares and (700 - x) hectares.
Difference of the areas of the two parts = x - (700 - x) = 2x - 700 one-fifth of the Average of the two areas = =
1 700 350 × = = 70 5 2 5
1 [x + (700 − x)] 5 2
Given that difference of the areas of the two parts = one-fifth of the Average of the two areas => 2x - 700 = 70 => 2x = 770
⇒x=
770 = 385 2
Hence, Area of smaller part = (700 - x) = (700 – 385) = 315 hectares.
10. The length of a room is 5.5 m and width is 3.75 m. What is the cost of paying the floor by slabs at the rate of Rs. 800 per sq. metre. A. Rs.12000
B. Rs.19500
C. Rs.18000
D. Rs.16500.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Area = 5.5 × 3.75 sq. metre. Cost for 1 sq. metre. = Rs. 800 Hence total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = Rs. 16500
11. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. What is the length of the rectangle? A. 18 cm
B. 16 cm
C. 40 cm
D. 20 cm
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let breadth = x cm Then length = 2x cm Area = lb = x × 2x = 2x2 New length = (2x - 5) New breadth = (x + 5) New Area = lb = (2x - 5)(x + 5) But given that new area = initial area + 75 sq.cm. => (2x - 5)(x + 5) = 2x2 + 75 => 2x2 + 10x - 5x - 25 = 2x2 + 75 => 5x - 25 = 75 => 5x = 75 + 25 = 100 => x = 100/5 = 20 cm Length = 2x = 2 × 20 = 40cm
12. If a square and a rhombus stand on the same base, then what is the ratio of the areas of the square and the rhombus? A. equal to ½
B. equal to ¾
C. greater than 1
D. equal to 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : A square and a rhombus on the same base will have equal areas. Hence ratio of the areas of the square and the rhom bus w ill be equal to 1 since they stand on the sam e base ================================================================ Note : Please find the proof of the formula given below which you may like to go through
Let ABCD be the square and ABEF be the rhombus Consider the right-angled triangles ADF and BCE We know that AD = BC (∵ sides of a square) AF = BE (∵ sides of a rhombus) ∴ DF = CE [∵ DF 2 = AF 2 - AD2 and CE2 = BE2 - BC2] Hence Δ ADF = Δ BCE => Δ ADF + Trapezium ABCF= Δ BCE + Trapezium ABCF => Area of square ABCD = Area of rhombus ABEF
13. The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m, find out the area of the field. A. 37500 m2
B. 30500 m2
C. 32500 m2
D. 40000 m2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Given that breadth of a rectangular field is 60% of its length
⇒b=
60l 3l = 100 5
perimeter of the field = 800 m => 2 (l + b) = 800
⇒ 2 (l + ⇒ l+
3l ) = 800 5
3l = 400 5
⇒
8l = 400 5
⇒
l = 50 5
⇒ l = 5 × 50 = 250 m
b=
3l 3 × 250 = = 2 × 50 = 150 m 5 5
Area = lb = 250 × 150 = 37500 m 2
14. A room 5m 44cm long and 3m 74cm broad needs to be paved with square tiles. What will be the least number of square tiles required to cover the floor? A. 176
B. 124
C. 224
D. 186
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : l = 5 m 44 cm = 544 cm b = 3 m 74 cm = 374 cm Area = 544 × 374 cm2 Now we need to find out HCF(Highest Common Factor) of 544 and 374. Let's find out the HCF using long division method for quicker results) 374) 544 (1
374
170) 374 (2
340
34) 170 (5
170
0
Hence, HCF of 544 and 374 = 34 Hence, side length of largest square tile we can take = 34 cm Area of each square tile = 34 × 34 cm2
Number of tiles required =
544 × 374 = 16 × 11 = 176 34 × 34
15. The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per metre is Rs. 5300, what is the length of the plot in metres? A. 60 m
B. 100 m
C. 75 m
D. 50 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Length of the plot is 20 metres more than its breadth. Hence, let's take the length as l metres and breadth as (l - 20) metres Length of the fence = perimeter = 2(length + breadth)= 2[ l + (l - 20) ] = 2(2l - 20) metres Cost per meter = Rs. 26.50 Total cost = 2(2l - 20) × 26.50 Total cost is given as Rs. 5300 => 2(2l - 20) × 26.50 = 5300 => (2l - 20) × 26.50 = 2650 => (l - 10) × 26.50 = 1325 => (l - 10) = 1325/26.50 = 50
=> l = 50 + 10 = 60 metres
16. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then what is the area of the park (in sq. m)? A. 142000
B. 112800
C. 142500
D. 153600
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : l : b = 3 : (Equation 1)
2 ------------------------------------------
Perimeter of the rectangular park = Distance travelled by the man at the speed of 12 km/hr in 8 minutes = speed × time = 12 × 8/60 (∵ 8 minute = 8/60 hour) = 8/5 km = 8/5 ×1000 m = 1600 m Perimeter = 2(l + b) => 2(l + b) = 1600 => l + b = 1600/2 = 800 m --------------------------(Equation 2)
From (Equation 1) and (Equation 2) l = 800 × 3/5 = 480 m b = 800 × 2/5 = 320 m (Or b = 800 - 480 = 320m) Area = lb = 480 × 320 = 153600 m2
17. What is the percentage increase in the area of a rectangle, if each of its sides is increased by 20%? A. 45%
B. 44%
C. 40%
D. 42%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : --------------------------------------------------------Solution 1 --------------------------------------------------------Let original length = 100 and original breadth = 100 Then original area = 100 × 100 = 10000
Increase in 20% of length ⇒ New length = Original length ×
(100 + 20) 120 = 100 × = 120 100 100
Increase in 20% of breadth ⇒ New breadth= Original breadth ×
(100 + 20) 120 = 100 × = 120 100 100
New area = 120 × 120 = 14400 Increase in area = New Area - Original Area = 14400 - 10000 = 4400
Percentage increase in area =
Increase in Area 4400 × 100 = × 100 = 44% 10000 Original Area
---------------------------------------------------------Solution 2--------------------------------------------------------Let original length = l and original breadth = b Then original area = lb
Increase in 20% of length ⇒ New length = Original length ×
(100 + 20) 120 120l = l× = 100 100 100
Increase in 20% of breadth ⇒ New breadth= Original breadth ×
New area =
(100 + 20) 120 120b = b× = 100 100 100
120l 120b 14400lb 144lb × = = 100 100 10000 100
Increase in area = New Area - Original Area =
Percentage of increase in area = 44lb ) 100 = lb (
× 100 =
144lb 44lb − lb = 100 100
Increase in Area × 100 Original Area
44lb × 100 = 44% 100lb
18. If the difference between the length and breadth of a rectangle is 23 m and its perimeter is 206 m, what is its area? A. 2800 m2
B. 2740 m2
C. 2520 m2
D. 2200 m2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : l - b = 23 ...................(Equation 1) perimeter = 2(l + b) = 206 => l + b = 103.............(Equation 2) (Equation 1) + (Equation 2) => 2l = 23 + 103 = 126 => l = 126/2 = 63 metre Substituting this value of l in (Equation 1), we get 63 - b = 23 => b = 63 - 23 = 40 metre Area = lb = 63 × 40 = 2520 m2
19. The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle? A. 16 cm
B. 18 cm
C. 14 cm
D. 20 cm
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Given that
2(l + b) =5 b
=> 2l + 2b = 5b => 2l = 3b
=> b =
2l 3
Also given that area = 216 cm2 => lb = 216 cm2
Substituting the value of b, we get, l × ⇒ l2 =
2l = 216 3
3 × 216 = 3 × 108 = (3 × 3) × 36 2
⇒ l = 3 × 6 = 18 cm
20. What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad? A. 814
B. 802
C. 836
D. 900
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : l = 15 m 17 cm = 1517 cm b = 9 m 2 cm = 902 cm Area = 1517 × 902 cm2 Now we need to find out HCF(Highest Common Factor) of 1517 and 902. Let's find out the HCF using long division method for quicker results)
902) 1517 (1
902
615) 902 (1
615
287) 615 (2
574
41) 287 (7
Hence, HCF of 1517 and 902 = 41 Hence, side length of largest square tile we can take = 41 cm Area of each square tile = 41 × 41 cm2
Number of tiles required =
1517 × 902 = 37 × 22 = 407 × 2 = 814 41 × 41
21. The diagonal of the floor of a rectangular room is
7
1 1 feet. The shorter side of the room is 4 feet. What is the area of the room? 2 2
A. 27 square feet
B. 22 square feet
C. 24 square feet
D. 20 square feet
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
287
1 15 feet = feet 2 2 1 9 Breadth, b = 4 feet = feet 2 2 Diagonal, d = 7
In the right-angled triangle PQR, 15 2 9 2 l2 = ( ) − ( ) 2 2 =
225 81 144 − = 4 4 4
l=√
−144 −−− 12 = feet = 6 feet 4 2
Area = lb = 6 ×
9 = 27 feet 2 2
22. The diagonal of a rectangle is
√ −− 41 cm and its area is 20 sq. cm. What is the perimeter of the rectangle?
A. 16 cm
B. 10 cm
C. 12 cm
D. 18 cm
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
For a rectangle, d 2 = l 2 + b2 where l = length , b = breadth and d = diagonal of the of the rectangle
d = √ −− 41 cm d 2 = l 2 + b2 −− 2 ⇒ l 2 + b2 = (√ 41 ) = 41........(Equation 1)
Area = lb = 20 cm
2
............(Equation 2)
Solving (Equation 1) and (Equation 2)
(a + b) 2 = a 2 + 2ab + b2
using the above formula, we have (l + b) 2 = l 2 + 2lb + b2 = (l 2 + b2 ) + 2lb = 41 + (2 × 20) = 81 ⇒ (l + b) = √ −− 81 = 9 cm perimeter = 2(l + b) = 2 × 9 = 18 cm
23. A tank is 25 m long, 12 m wide and 6 m deep. What is the cost of plastering of its walls and bottom at the rate of 75 paise per sq. m? A. Rs. 558
B. Rs. 502
C. Rs. 516
D. Rs. 612
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Consider a rectangular solid of length l, w idth w and height h. Then 1. Total Surface area of a rectangular solid, S = 2lw + 2lh + 2w h = 2(lw + lh + w h) 2. Volum e of a rectangular solid, V = lw h
In this case, l = 25 m, w = 12 m, h = 6 m and all surface needs to be plastered except the top Hence total area needs to be plastered = Total Surface Area - Area of the Top face = (2lw + 2lh + 2wh) - lw = lw + 2lh + 2wh = (25 × 12) + (2 × 25 × 6) + (2 × 12 × 6) = 300 + 300 + 144 = 744 m2 Cost of plastering = 744 × 75 = 55800 paise = Rs.558
24. It is decided to construct a 2 metre broad pathway around a rectangular plot on the inside. If the area of the plots is 96 sq.m. and the rate of construction is Rs. 50 per square metre., what will be the total cost of the construction? A. Rs.3500
B. Rs. 4200
C. Insufficient Data
D. Rs. 4400
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Let length and width of the rectangular plot be l and b respectively Total Area of the rectangular plot = 96 sq.m. Width of the pathway = 2 m Length of the remaining area in the plot = (l - 4) breadth of the remaining area in the plot = (b - 4) Area of the remaining area in the plot = (l - 4)(b - 4) Area of the pathway = Total Area of the rectangular plot - remaining area in the plot = 96 - [(l - 4)(b - 4)] = 96 - [lb - 4l - 4b + 16] = 96 - [96 - 4l - 4b + 16] = 96 - 96 + 4l + 4b - 16] = 4l + 4b - 16 = 4(l + b) - 16 We do not know the values of l and b and hence total area of the rectangular plot can not be found out. So we can not find out total cost of the construction.
25. The area of a parallelogram is 72 cm2 and its altitude is twice the corresponding base. What is the length of the base? A. 6 cm
B. 7 cm
C. 8 cm
D. 12 cm
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Area of a parallelogram , A = bh w here b is the base and h parallelogram
is
the
height of
Let the base = x cm. Then the height = 2x cm (∵ altitude is twice the base) Area = x × 2x = 2x2 But the area is given as 72 cm2 => 2x2 = 72 => x2 = 36 => x = 6 cm
the
26. Two diagonals of a rhombus are 72 cm and 30 cm respectively. What is its perimeter? A. 136 cm
B. 156 cm
C. 144 cm
D. 121 cm
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Remember the following two properties of a rhombus which will be useful in solving this question 1. The sides of a rhombus are congruent. 2. The diagonals of a rhombus are unequal and bisect each other at right angles. Let the diagonals be PR and SQ such that PR = 72 cm and SQ = 30 cm
PO = OR =
72 = 36 cm 2
SO = OQ =
30 = 15 cm 2
−−−−−−−− −−−−−−− −−− = 39 cm 1296 + 225− = √ − 1521 PQ = QR = RS = SP = √ 362 + 152 = √ − perimeter = 4 × 39 =156 cm
27. The base of a parallelogram is (p + 4), altitude to the base is (p - 3) and the area is (p2 - 4), find out its actual area. A. 40 sq. units
B. 54 sq. units
C. 36 sq. units
D. 60 sq. units
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Area of a parallelogram , A = bh w here b is the base and h parallelogram
is
the
height of
the
Hence, we have p2 - 4 = (p + 4)(p - 3) => p2 - 4 = p2 - 3p + 4p - 12 => -4 = p - 12 => p = 12 - 4 = 8 Hence, actual area = (p2 - 4) = 82 - 4 = 64 - 4 = 60 sq. units
28. A circle is inscribed in an equilateral triangle of side 24 cm, touching its sides. What is the area of the remaining portion of the triangle? A.
144√ 3 − 48π cm2
B.
121√ 3 − 36π cm2
C.
144√ 3 − 36π cm2
D.
121√ 3 − 48π cm2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
√3 2 a 4 where a is length of one side of the equilateral triangle
Area of an equilateral triangle =
Area of the equilateral Δ ABC =
√3 2 √3 2 a = 24 = 144√ 3 cm2 .............(1) 4 4
1 bh 2 where b is the base and h is the height of the triangle
Area of a triangle =
Let r = radius of the inscribed circle. Then Area of Δ ABC = Area of Δ OBC + Area of Δ OCA + area of Δ OAB = (½ × r × BC) + (½ × r × CA) + (½ × r × AB) = ½ × r × (BC + CA + AB) = ½ x r x (24 + 24 + 24) = ½ x r x 72 = 36r cm2 ------------------------------------------ (2) From (1) and (2),
144√ 3 = 36r ⇒r=
144 √ 3 = 4 √ 3 − − − − − − − − − − − −(3) 36
Area of a circle = πr 2 where = radius of the circle
2
From (3), the area of the inscribed circle = πr 2 = π(4√ 3) = 48π − − − − − − − − − − − −(4)
Hence , Area of the remaining portion of the triangle Area of Δ ABC – Area of inscribed circle 144√ 3 − 48π cm 2
29. A rectangular plot measuring 90 metres by 50 metres needs to be enclosed by wire fencing such that poles of the fence will be kept 5 metres apart. How many poles will be needed? A. 30
B. 44
C. 56
D. 60
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option C Explanation : Perim eter of a rectangle = 2(l + b) w here l is the length and b is the breadth of the rectangle
Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is very important. Hence number of poles required = 280⁄5 = 56
30. If the diagonals of a rhombus are 24 cm and 10 cm, what will be its perimeter A. 42 cm
B. 64 cm
C. 56 cm
D. 52 cm
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Let the diagonals be PR and SQ such that PR = 24 cm and SQ = 10 cm
PO = OR =
24 = 12 cm 2
SO = OQ =
10 = 5 cm 2
−−−−−−− −−−−− −− = 13 cm 144 + 25− = √ − 169 PQ = QR = RS = SP = √ 122 + 52 = √ − perimeter = 4 × 13 = 52 cm
31. What will be the length of the longest rod which can be placed in a box of 80 cm length, 40 cm breadth and 60 cm height?
−−−− √− 11600 cm − −−−− C. √ 10000 cm A.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
−−−− √− 14400 cm − −−−− D. √ 12040 cm B.
The longest road which can fit into the box will have one end at A and other end at G (or any other similar diagonal) Hence the length of the longest rod = AG Initially let's find out AC. Consider the right angled triangle ABC
AC2 = AB2 + BC2 = 402 + 802 = 1600 + 6400 = 8000
−−−− ⇒ AC = √ 8000 cm Consider the right angled triangle ACG
AG2 = AC2 + CG2
−−− ) 2 + 602 = 8000 + 3600 = 11600 8000 = (√ − −−−− cm ⇒ AG = √ − 11600 −−−− cm ⇒ The length of the longest rod = √ − 11600
I mportant Formulas - Average 1.
Average Average =
2.
Sum of observations Number of observations
Average Speed If a car covers a certain distance at x kmph and an equal distance at y kmph. Then,the average speed of the whole journey =
Comments(26)
2xy x+y
kmph
Newest
teja 18 Nov 2014 8:49 PM there were 35 students in a hostel.due to the admission of 7 new students the expenses for the mess were increased by 42/- per day while the average expenditure per head diminished by Re 1 .what was the original expenditure of the mess. Like | Dislike | Reply | Flag
Raj 21 Nov 2014 1:00 AM Let the initial average expenditure per head be x Then the initial total expenditure= 35x After the admission of 7 new students, number of students will become 42 average expenditure per head = (x-1) New total expenditure = 42(x-1) Given that Expenses for the mess were increased by 42 42(x-1) - 35x = 42
7x = 84 x = 12 Original expenditure = 35x = Rs.420
1. In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs? A. 6.25
B. 5.5
C. 7.4
D. 5
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Runs scored in the first 10 overs = 10 × 3.2 = 32 Total runs = 282 remaining runs to be scored = 282 - 32 = 250 remaining overs = 40
Run rate needed =
250 = 6.25 40
2. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500? A. 4800
B. 4991
C. 5004
D. 5000
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the sale in the sixth month = x
Then
6435 + 6927 + 6855 + 7230 + 6562 + x 6
= 6500
=> 6435 + 6927 + 6855+ 7230 + 6562 + x = 6 × 6500 = 39000 => 34009 + x = 39000 => x = 39000 - 34009 = 4991
3. The average of 20 numbers is zero. Of them, How many of them may be greater than zero , at the most? A. 1
B. 20
C. 0
D. 19
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Average of 20 numbers = 0
=>
Sum of 20 numbers =0 20
=> Sum of 20 numbers = 0 Hence at the most, there can be 19 positive numbers. (Such that if the sum of these 19 positive numbers is x, 20th number will be -x)
4. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. Find out the average age of the team. A. 23 years
B. 20 years
C. 24 years
D. 21 years
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Number of members in the team = 11 Let the average age of of the team = x
=>
Sum of the ages of all the 11 members of the team 11
=x
=> Sum of the ages of all the 11 members of the team = 11x
Age of the captain = 26 Age of the wicket keeper = 26 + 3 = 29 Sum of the ages of 9 members of the team excluding captain and wicket keeper
= 11x - 26 - 29 = 11x - 55 Average age of 9 members of the team excluding captain and wicket keeper
=
11x − 55 9
Given that
11x − 55 = (x − 1) 9
=> 11x - 55 = 9(x - 1) => 11x - 55 = 9x - 9 => 2x = 46
=> x =
46 = 23 years 2
5. The average monthly income of A and B is Rs. 5050. The average monthly income of B and C is Rs. 6250 and the average monthly income of A and C is Rs. 5200. What is the monthly income of A? A. 2000
B. 3000
C. 4000
D. 5000
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the monthly income of A = a monthly income of B = b monthly income of C = a
a + b = 2 × 5050 − − − − − − − − − (Equation1) b + c = 2 × 6250 − − − − − − − − − (Equation2) a + c = 2 × 5200 − − − − − − − − − (Equation3) (Equation 1) + (Equation 3) - (Equation 2)
=> a + b + a + c − (b + c) = (2 × 5050) + (2 × 5200) − (2 × 6250) => 2a = 2(5050 + 5200 - 6250)
=> a = 4000 => Monthly income of A = 4000
6. A car owner buys diesel at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of diesel if he spends Rs. 4000 each year? A. Rs. 8
B. Rs. 7.98
C. Rs. 6.2
D. Rs. 8.1
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Total Cost = 4000 × 3 Total diesel used =
4000 4000 4000 + + 7.5 8 8.5
average cost per litre of diesel =
4000 × 3 4000 4000 4000 ( ) + + 7.5 8 8.5
3
= (
1 1 1 ) + + 7.5 8 8.5
It is important how you proceed from this stage. Remember time is very important here and if we solve this completely in the traditional way, it may take lot of time. Instead, we can find out the approximate value easily and select the right answer from the given choices 3
In this case answer = (
≈
3 1 1 1 ( + + ) 8 8 8
≈
1 1 1 ) + + 7.5 8 8.5 3 3 ( ) 8
≈8
Means we got that answer is approximately equal to 8. From the given choices, the answer
can be 8 or 7.98 or 8.1 . But which one from these?
It will be easy to figure out. Just see here the denominator was and we approximated it as
1 1 1 + + 7.5 8 8.5
3 . However 8
1 1 1 1 8 + .5 + 8 − .5 + = + = 7.5 8.5 8 − .5 8 + .5 (8 − .5) (8 + .5) =
16 (8 2 − .5 2 )
[because a 2 − b2 = (a − b) (a + b)]
=
16 (64 − .25)
ie,
1 1 16 + = 7.5 8.5 (64 − .25)
We know that =>
1 1 1 16 + = = 8 8 4 64
1 1 1 1 + > + 7.5 8.5 8 8
Early we had approximated the denominator as
3 8
However from the above mentioned equations, now you know that actually denominator is slightly greater than
3 8
It means that answer is slightly lower that 8. Hence we can pick the choice 7.98 as the answer
Try to remember the relations between numbers and which can help you to save a lot of time
which can be very precious in competitive exams
7. In Kiran's opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Kiran and he thinks that Kiran's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Kiran? A. 70 kg
B. 69 kg
C. 61 kg
D. 67 kg
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let Kiran's weight = x. Then According to Kiran, 65 20 = x - 65 => x = 20 + 65 = 85
16. There are two divisions A and B of a class, consisting of 36 and 44 students respectively. If the average weight of divisions A is 40 kg and that of division b is 35 kg. What is the average weight of the
whole class? A. 38.25
B. 37.25
C. 38.5
D. 37
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Total weight of students in division A = 36 × 40 Total weight of students in division B = 44 × 35 Total students = 36 + 44 = 80 (36 × 40) + (44 × 35) Average weight of the whole class = 80 (9 × 40) + (11 × 35) (9 × 8) + (11 × 7) 72 + 77 149 = = = = = 37.25 20 4 4 4
17. A batsman makes a score of 87 runs in the 17th inning and thus increases his averages by 3. What is his average after 17th inning? A. 39
B. 35
C. 42
D. 40.5
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the average after 17 innings = x Total runs scored in 17 innings = 17x then average after 16 innings = (x-3) Total runs scored in 16 innings = 16(x-3) We know that Total runs scored in 16 innings + 87 = Total runs scored in 17 innings => 16(x-3) + 87 = 17x => 16x - 48 + 87 = 17x => x = 39
18. A student needed to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14
and x. He found the mean to be 12. What is the value of x? A. 12
B. 5
C. 7
D. 9
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 + x 12 137 + x => = 12 12
= 12
=> 137 + x = 144 => x = 144 - 137 = 7
19. Arun obtained 76, 65, 82, 67 and 85 marks (out in 100) in English, Mathematics, Chemistry, Biology and Physics. What is his average mark? A. 53
B. 54
C. 72
D. 75
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Average mark =
76 + 65 + 82 + 67 + 85 5
=
375 = 75 5
20. Distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56km per hour. Find the average speed of the train during the whole journey? A. 69.0 km /hr
B. 69.2 km /hr
C. 67.2 km /hr
D. 67.0 km /hr
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : ------------------------------------------Solution 1 (Quick)
--------------------------------------------
If a car covers a certain distance at x kmph and an equal distance at y kmph. Then, 2xy the average speed of the whole journey = kmph. x+y
By using the same formula, we can find out the average speed quickly 2 × 84 × 56 2 × 84 × 56 2 × 21 × 56 average speed = = = 84 + 56 140 35 =
2 × 3 × 56 336 = = 67.2 5 5
------------------------------------------Solution 2 (Fundamentals) -------------------------------------------Though it is a good idea to solve the problems quickly using formulas, you should know the fundamentals too. Let's see how we can solve this problems using basics Train travels from A to B at 84 km per hour Let the distance between A and B = x
Total time taken for traveling from A to B = Train travels from B to A at 56 km per hour
distance x = speed 84
Total time taken for traveling from B to A =
distance x = speed 56
Total distance travailed = x + x = 2x Total time taken = Average speed =
=
=
2 1 1 + 84 56
=
x x + 84 56
Total distance traveled Total time taken
2x = x x + 84 56
2 × 84 × 56 2 × 84 × 56 2 × 21 × 56 2 × 3 × 56 = = = 56 + 84 140 35 5
336 = 67.2 5
21. The average age of boys in a class is 16 years and that of the girls is 15 years. What is the average age for the whole class? A. 15
B. 16
C. 15.5
D. Insufficient Data
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : We do not have the number of boys and girls. Hence we can not find out the answer.
22. The average age of 36 students in a group is 14 years. When teacher's age is included to it, the average increases by one. Find out the teacher's age in years? A. 51 years
B. 49 years
C. 53 years
D. 50 years
Hide Answer | Notebook | Notebook | Discuss
Here is the answer and explanation Answer : Option A Explanation : average age of 36 students in a group is 14
=> Sum of the ages of 36 students = 36 × 14 When teacher's age is included to it, the average increases by one => average = 15
=> Sum of the ages of 36 students and the teacher =37 × 15 Hence teachers age = 37 × 15 − 36 × 14 = 37 × 15 − 14 (37 − 1) = 37 × 15 − 37 × 14 + 14 = 37 (15 − 14) + 14 = 37 + 14 = 51
23. The average of five numbers id 27. If one number is excluded, the average becomes 25. What is the excluded number? A. 30
B. 40
C. 32.5
D. 35
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Sum of 5 numbers = 5 × 27 Sum of 4 numbers after excluding one number = 4 × 25 Excluded number = 5 × 27 − 4 × 25 = 135 − 100 = 35
24. The batting average for 40 innings of a cricket player is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. Find out the highest score of the player. A. 150
B. 174
C. 180
D. 166
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Total runs scored by the player in 40 innings = 40 × 50 Total runs scored by the player in 38 innings after excluding two innings 38 × 48 Sum of the scores of the excluded innings = 40 × 50 − 38 × 48 = 2000 − 1824 = 176
Given that the scores of the excluded innings differ by 172. Hence let's take the highest score as x + 172 and lowest score as x Now x + 172 + x = 176 => 2x = 4
=> x =
4 =2 2
highest score as x + 172 = 2 + 172 = 174
25. The average score of a cricketer for ten matches is 38.9 runs. If the average for the first six matches is 42, what is the average for the last four matches? A. 34.25
B. 36.4
C. 40.2
D. 32.25
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Total runs scored in 10 matches = 10 × 38.9 Total runs scored in first 6 matches = 6 × 42 Total runs scored in the last 4 matches = 10 × 38.9 − 6 × 42 Average of the runs scored in the last 4 matches = =
10 × 38.9 − 6 × 42 4
389 − 252 137 = = 34.25 4 4
26. The average of six numbers is x and the average of three of these is y. If the average of the remaining three is z, then
A. None of these
B. x = y + z
C. 2x = y + z
D. x = 2y + 2z
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Average of 6 numbers = x => Sum of 6 numbers = 6x Average of the 3 numbers = y => Sum of these 3 numbers = 3y Average of the remaining 3 numbers = z => Sum of the remaining 3 numbers = 3z Now we know that 6x = 3y + 3z => 2x = y + z
27. Suresh drives his car to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/hr. What is his average speed for the whole journey ? A. 32.5 km/hr.
B. 35 km/hr.
C. 37.5 km/hr
D. 40 km/hr
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : ------------------------------------------Solution 1 (Quick) --------------------------------------------
If a car covers a certain distance at x kmph and an equal distance at y kmph. Then, 2xy the average speed of the whole journey = kmph. x+y
By using the same formula, we can find out the average speed quickly 2 × 50 × 30 2 × 50 × 30 2 × 50 × 3 average speed = = = 50 + 30 80 8 =
50 × 3 25 × 3 75 = = = 37.5 4 2 2
------------------------------------------Solution 2 (Fundamentals) -------------------------------------------Though it is a good idea to solve the problems quickly using formulas, you should know the fundamentals too. Let's see how we can solve this problems using basics.
Total time taken for traveling one side =
Total time taken for return journey =
distance 150 = speed 50
distance 150 = speed 30
Total distance travailed = 150 + 150 = 2 × 150 Total time taken = Average speed =
=
2 1 1 + 50 30
=
150 150 + 50 30
Total distance traveled Total time taken
=
2 × 150 150 150 + 50 30
2 × 50 × 30 30 + 50
=
2 × 50 × 30 2 × 50 × 3 = 80 8
=
50 × 3 25 × 3 75 = = = 37.5 4 2 2
28. The average age of a husband and his wife was 23 years at the time of their marriage. After five years they have a one year old child. What is the average age of the family ? A. 21 years
B. 20 years
C. 18 years
D. 19 years
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Total of the age of husband and wife = 2 × 23 = 46 Total of the age of husband and wife after 5 years + Age of the 1 year old child = 46 + 5 + 5 + 1 = 57 Average age of the family =
57 = 19 3
29. In an examination, a student's average marks were 63. If he had obtained 20 more marks for his Geography and 2 more marks for his history, his average would have been 65. How many subjects were there in the examination? A. 12
B. 11
C. 13
D. 14
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the number of subjects = x Then, Total marks he scored for all subjects = 63x
If he had obtained 20 more marks for his Geography and 2 more marks for his history, his average would have been 65 => Total marks he would have scored for all subjects = 65x Now we can form the equation as 65x - 63x = the additional marks of the student = 20 + 2 = 22 => 2x = 22
=> x =
22 = 11 2
Banker's Discount A. Important Concepts Banker's Discount Assume that a merchant A purchases goods worth, say Rs.1000 from another merchant B at a credit of say 4 months. Then B prepares a bill called bill of exchange (also called Hundi). On receipts of goods, A gives an agreement by signing on the bill allowing B to withdraw the money from A’s bank exactly after 4 months of the date of the bill. The date exactly after 4 months is known as nominally due date. Three more days (called grace days) are added to this date to get a date known as legally due date. The amount given on the bill is called the Face Value (F)which is Rs.1000 in this case. Assume that B needs this money before the legally due date. He can approach a banker or broker who pays him the money against the bill, but somewhat less than the face value. The banker deducts the simple interest on the face value for the unexpired time. This deduction is known as Bankers Discount (BD). In another words, Bank Discount (BD) is the simple interest on the face value for the period from the date on which the bill was discounted and the legally due date. The present value is the amount which, if placed at a particular rate for a specified period will amount to that sum of money at the end of the specified period. The interest on the present value is called the True Discount (TD). If the banker deducts the true discount on the face value for the unexpired time, he will not gain anything. Banker’s Gain (BG) is the difference between banker’s discount and the true discount for the unexpired time. Note: When the date of bill is not given, grace days are not to be added.
B. Important Formulas Banker's Discount
Let F = Face Value of the Bill, TD = True Discount, BD = Bankers Discount, BG = Banker’s Gain, R = Rate of Interest, PW = True Present Worth and T = Time in Years
BD = Simple Interest on the face value of the bill for unexpired time = F
PW =
1+T(
R ) 100
TD = Simple Interest on the present value for unexpired time =
TD =
BD × 100 100 + TR
PW = F - TD F =
BD × TD (BD – TD)
(TD) 2 BG = BD – TD = Simple Interest on TD = PW −−−−−−−−− TD = √ PW × BG TD =
FTR 100
BG × 100 TR
PW × TR 100
=
FTR 100 + (TR)
C. A Simple Example to understand the Basic Concepts of Banker's Discount
What is the present value, true discount, banker's discount and banker's gain on a bill of Rs.104500 due in 9 months at 6% per annum?
F = Rs. 104500
T = 9 months =
9 3 years = years 12 4
R = 6%
3 104500 × × 6 FTR 3 4 = = 1045 × × 6 = Rs. 4702.50 Banker's Discount, BD = 100 100 4 Present value, PW =
True Discount, TD =
F
=
R ) 1+T( 100 PW × TR 100
=
104500 3 6 ) 1+ ( ) ( 4 100
= Rs. 100000
3 ×6 3 4 = 1000 × × 6 = Rs. 4500 100 4
100000 ×
Banker's Gain, BG = BD – TD = 4702.50 4500 = Rs.202.50
1. The banker's discount on a bill due 4 months hence at 15% is Rs. 420. What is the true discount? A. Rs. 410
B. Rs. 400
C. Rs. 390
D. Rs. 380
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
TD =
=
BD × 100 = 100 + TR
420 × 100 4 100 + ( × 15) 12
=
420 × 100 1 100 + ( × 15) 3
420 × 100 420 × 100 84 × 100 = = = 4 × 100 = 400 100 + 5 105 21
2. The banker's discount on a certain amount due 2 years hence is 11∕10 of the true discount. What is the rate percent? A. 1%
B. 5%
C. 10%
D. 12%
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let TD = Rs. 1
Then BD =
11 11 × 1= Rs. 10 10
T = 2
R = ?
F =
BD × TD = (BD – TD)
BD = ⇒
(
11 × 1) 10
(
11 − 1) 10
FTR 100
11 11 × 2 × R = 10 100
⇒ 110 = 22R ⇒R=
110 = 5% 22
11 10 = = Rs. 11 1 10
3. The present worth of a sum due sometimes hence is Rs.5760 and the baker's gain is Rs.10. What is the true discount? A. Rs. 480
B. Rs. 420
C. Rs. 120
D. Rs. 240
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
−−−−−−−−− −−−−−−−− −−−−− TD = √ PW × BG = √ 5760 × 10 = √ 57600 = Rs. 240
4. What is the banker's discount if the true discount on a bill of Rs.540 is Rs.90 ? A. Rs. 108
B. Rs. 120
C. Rs. 102
D. Rs. 106
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Present Worth, PW = F TD = 540 90 = Rs. 450 Simple Interest on the Present Worth = True Discount Hence Simple Interest on 450 = 90 (Equation 1) Simple Interest on the face value = Bankers Discount => Simple Interest on 540 = Bankers Discount
From Equation 1, Simple Interest on 450 = 90
Hence, Simple Interest on 540 =
90 540 × 540 = = Rs. 108 450 5
=> Bankers Discount = Rs. 108
5. A bill for Rs. 3000 is drawn on 14th July at 5 months. It is discounted on 5th October at 10%. What is the Banker's Discount? A. Rs. 60
B. Rs. 82
C. Rs. 90
D. Rs. 120
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option A
Explanation : F = Rs. 3000 R = 10% Date on which the bill is drawn = 14th July at 5 months Nominally Due Date = 14th December Legally Due Date = 14th December + 3 days = 17th December Date on which the bill is discounted = 5th October Unexpired Time = [6th to 31st of October] + [30 Days in November] + [1st to 17th of December] = 26 + 30 + 17 = 73 Days
=
73 1 year = year 365 5
BD = Simple Interest on the face value of the bill for unexpired time =
=
FTR 100
1 × 10 1 5 = 30 × × 10 = Rs. 60 100 5
3000 ×
6. The bankers discount and the true discount of a sum at 10% per annum simple interest for the same time are Rs.100 and Rs.80 respectively. What is the sum and the time? A. Sum = Rs.400 and Time = 5 years
B. Sum = Rs.200 and Time = 2.5 years
C. Sum = Rs.400 and Time = 2.5 years
D. Sum = Rs.200 and Time = 5 years
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : BD = Rs.100 TD = Rs.80 R = 10%
F =
BD × TD 100 × 80 100 × 80 = = = Rs. 400 20 (BD – TD) (100– 80)
BD = Simple Interest on the face value of the bill for unexpired time = ⇒ 100 =
FTR 100
400 × T × 10 100
⇒ 100 = 4 × T × 10 ⇒ 10 = 4 × T ⇒T =
10 = 2.5 years 4
7. The banker's gain on a sum due 6 years hence at 12% per annum is Rs. 540. What is the banker's discount? A. 1240
B. 1120
C. 1190
D. 1290
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
TD =
BG × 100 540 × 100 90 × 100 15 × 100 = = = = Rs. 750 TR 6 × 12 12 2
BG = BD – TD
=> 540 = BD 750
=> BD = 540 + 750 = 1290
8. The present worth of a certain bill due sometime hence is Rs. 1296 and the true discount is Rs. 72. What is the banker's discount? A. Rs. 76
B. Rs. 72
C. Rs. 74
D. Rs. 4
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
(TD) 2 72 2 72 × 72 12 × 12 12 BG = = = = = = Rs. 4 PW 1296 1296 36 3 BG = BD – TD
=> 4 = BD 72
=> BD = 72 + 4 = Rs. 76
9. The banker's discount of a certain sum of money is Rs. 36 and the true discount on the same sum for the same time is Rs. 30. What is the sum due? A. Rs. 180
B. Rs. 120
C. Rs. 220
D. Rs. 200
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
F =
BD × TD 36 × 30 36 × 30 = = = 36 × 5 = Rs. 180 6 (BD – TD) (36– 30)
10. The banker's gain on a bill due 1 year hence at 10% per annum is Rs. 20. What is the true discount? A. Rs. 200
B. Rs. 100
C. Rs. 150
D. Rs. 250
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
TD =
BG × 100 20 × 100 = = Rs. 200 TR 1 × 10
11. The banker's gain of a certain sum due 3 years hence at 10% per annum is Rs. 36. What is the present worth ? A. Rs. 400
B. Rs. 300
C. Rs. 500
D. Rs. 350
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : T = 3 years R = 10%
TD =
BG × 100 36 × 100 = = 12 × 10 = Rs. 120 TR 3 × 10
TD =
PW × TR 100
⇒ 120 =
PW × 3 × 10 100
⇒ 1200 = PW × 3 PW =
1200 = Rs. 400 3
12. The present worth of a certain sum due sometime hence is Rs. 3400 and the true discount is Rs. 340. The banker's gain is: A. Rs. 21
B. Rs. 17
C. Rs. 18
D. Rs. 34
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
(TD) 2 (340) 2 340 × 340 340 BG = = = = = Rs. 34 PW 3400 3400 10
13. The banker's discount on Rs. 1600 at 15% per annum is the same as true discount on Rs. 1680 for the same time and at the same rate. What is the time? A. 3 months
B. 4 months
C. 5 months
D. 6 months
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Bankers Discount, BD = Simple Interest on the face value of the bill for unexpired time. True Discount, TD = Simple Interest on the present value for unexpired time. Simple Interest on Rs. 1600 = True Discount on Rs.1680
=> Rs. 1600 is the Present Worth (PW) of Rs. 1680
=> Rs. 80 is the Simple Interest of Rs.1600 at 15%
⇒ 80 =
1600 × T × 15 100
⇒ 80 = 16 × T × 15 ⇒ 5 = T × 15 ⇒1=T ×3 => T =
1 12 year = months = 4 months 3 3
14. The banker's gain on a certain sum due 2 rate percent?
1 9 years hence is of the banker's discount. What is the 2 25
1 % 3 1 C. 24 % 3
1 % 2 1 D. 22 % 2
A. 18
B. 18
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
T = 2
1 5 years = years 2 2
Let the banker's discount, BD = Rs. 1
Then, banker's gain, BG =
9 9 × 1 = Rs. 25 25
BG = BD – TD ⇒
9 = 1 - TD 25
⇒ TD = 1 −
9 16 = 25 25
16 BD × TD 25 F = = (BD – TD) 16 (1– ) 25 1×
16 16 25 = = Rs. 9 9 25
BD = Simple Interest on the face value of the bill for unexpired time =
FTR 100
16 5 × ×R 9 2 ⇒1= 100 ⇒ 100 =
16 5 × × R 9 2
⇒ 100 =
16 × 5 × R 9×2
⇒ 100 =
8×5×R 9
⇒ R = =
100 × 9 100 × 9 5×9 = = 8×5 40 2
45 1 = 22 % 2 2
15. The banker's gain on a sum due 3 years hence at 12% per annum is Rs. 360. The banker's discount is: A. Rs. 1360
B. Rs. 1000
C. Rs. 360
D. Rs. 640
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : BG = Rs. 360 T = 3 years R = 12%
TD =
BG × 100 360 × 100 = = Rs. 1000 TR 3 × 12
BG = BD – TD
=> BD = BG + TD = 360 + 1000 = Rs. 1360
16. The true discount on a certain sum due 6 months hence at 15% is Rs. 240. What is the banker's discount on the same sum for the same time at the same rate? A. None of these
B. Rs. 278
C. Rs. 228
D. Rs. 258
Hide Answer | Notebook | Notebook | Discuss
Here is the answer and explanation Answer : Option D Explanation : TD = Rs. 240
T = 6 months =
1 year 2
R = 15%
TD =
BG × 100 TR
⇒ 240 =
BG =
BG × 100 1 ( × 15) 2
240 × 15 120 × 15 = = Rs. 18 100 × 2 100
BG = BD – TD
=> 18 = BD 240
=> BD = 18 + 240 = Rs. 258
17. A bill is discounted at 10% per annum. If banker's discount is allowed, at what rate percent should the proceeds be invested so that nothing will be lost? A. 10
1 % 9
C. 11
1 % 9 2 % D. 10 9 B. 11
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the amount = Rs.100 Then BD = Rs.10 (∵ banker's discount, BD is the simple Interest on the face value of the bill for unexpired time and bill is discounted at 10% per annum) Proceeds = Rs.100 – Rs.10 = Rs.90 Hence we should get Rs.10 as the interest of Rs.90 for 1 year so that nothing will be lost
⇒ 10 = ⇒R=
90 × 1 × R 100 10 × 100 100 1 = = 11 % 90 9 9
18. A banker paid Rs.5767.20 for a bill of Rs.5840, drawn of Apr 4 at 6 months. If the rate of interest was 7%, what was the day on which the bill was discounted? A. 3rd March
B. 3rd September
C. 3rd October
D. 3rd August
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : F = Rs.5840 R = 7% BD = 5840 5767.20 = Rs.72.8
BD =
FTR 100
⇒ 72.8 = ⇒ T = =
5840 × T × 7 100
72.8 × 100 10.4 × 100 1040 104 13 = = = = years 7 × 5840 5840 5840 584 73
13 × 365 Days = 65 Days 73
⇒ Unexpired Time = 65 Days Given that Date of Draw of the bill = 4th April at 6 months
=> Nominally Due Date = 4th October
=> Legally Due Date = (4th October + 3 days) = 7th October
Hence, The date on which the bill was discounted
= (7th October 65 days)
= (7th October 7 days in October 30 days in September 28 days in August)
= 3rd August
19. The banker's discount on a sum of money for 3 years is Rs. 1116. The true discount on the same sum for 4 years is Rs. 1200. What is the rate per cent? A. 8%
B. 12%
C. 10%
D. 6%
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : BD for 3 years = Rs. 1116
BD for 4 years =
1116 × 4 = Rs. 1488 3
TD for 4 years = Rs. 1200
F =
BD × TD 1488 × 1200 1488 × 1200 124 × 1200 124 × 100 = = = = = 62 × 100 = Rs. 6200 288 24 2 (BD – TD) (1488– 1200)
=> Rs.1488 is the simple interest on Rs. 6200 for 4 years
⇒ 1488 = ⇒R= =
6200 × 4 × R 100
1488 × 100 372 × 100 = 6200 × 4 6200
372 = 6% 62
20. The true discount on a bill of Rs. 2160 is Rs. 360. What is the banker's discount? A. Rs. 432
B. Rs. 422
C. Rs. 412
D. Rs. 442
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : F = Rs. 2160 TD = Rs. 360 PW = F TD = 2160 360 = Rs. 1800 True Discount is the Simple Interest on the present value for unexpired time
=>Simple Interest on Rs. 1800 for unexpired time = Rs. 360 Banker's Discount is the Simple Interest on the face value of the bill for unexpired time = Simple Interest on Rs. 2160 for unexpired time
=
360 1 × 2160 = × 2160 = Rs. 432 1800 5
21. The banker's gain of a certain sum due 2 years hence at 10% per annum is Rs. 24. What is the present worth? A. Rs. 600
B. Rs. 500
C. Rs. 400
D. Rs. 300
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : T = 2 years R = 10%
TD =
BG × 100 24 × 100 = = 12 × 10 = Rs. 120 TR 2 × 10
TD =
PW × TR 100
⇒ 120 =
PW × 2 × 10 100
⇒ 1200 = PW × 2 PW =
1200 = Rs. 600 2
22. The true discount on a bill for Rs. 2520 due 6 months hence at 10% per annum is A. Rs. 180
B. Rs. 140
C. Rs. 80
D. Rs. 120
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : F= Rs. 2520
T = 6 months =
1 year 2
R = 10%
TD =
=
1 × 10 2 = 1 100 + ( × 10) 2 2520 ×
FTR 100 + (TR)
1 × 10 2 = 1 100 + ( × 10) 2 2520 ×
1260 × 10 12600 2520 = = = Rs. 120 100 + 5 105 21
23. What is the present worth of a bill of Rs.1764 due 2 years hence at 5% compound interest is A. Rs. 1600
B. Rs. 1200
C. Rs. 1800
D. Rs. 1400
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Since the compound interest is taken here, 2
5 ) PW (1 + 100 1 ) PW (1 + 20 PW (
21 ) 20
PW ×
2
2
= 1764
= 1764
= 1764
441 = 1764 400
⇒ PW =
1764 × 400 = 4 × 400 = Rs. 1600 441
24. If the discount on Rs. 498 at 5% simple interest is Rs.18, when is the sum due? A. 8 months
B. 11 months
C. 10 months
D. 9 months
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : F = Rs. 498 TD = Rs. 18
PW = F TD = 498 18 = Rs. 480 R = 5%
TD =
PW × TR 100
⇒ 18 =
480 × T × 5 100
⇒ 18 = 24 × T ⇒ T =
18 3 12 × 3 = years = months = 9 months 24 4 4
25. What is the difference between the banker's discount and the true discount on Rs.8100 for 3 months at 5% A. Rs. 2
B. Rs. 1.25
C. Rs. 2.25
D. Rs. 0.5
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : F = Rs. 8100 R = 5%
T = 3 months =
BD =
TD =
=
FTR = 100
1 years 4 1 ×5 2025 405 4 = = = Rs. 101.25 100 20 4
8100 ×
FTR 100 + (TR)
=
8100 × 100 + (
1 ×5 4
1 × 5) 4
=
2025 × 5 5 100 + ( ) 4
2025 × 5 × 4 2025 × 5 × 4 405 × 5 × 4 45 × 5 × 4 = = = 400 + 5 405 81 9
= 5 × 5 × 4= Rs. 100 BD TD = Rs. 101.25 Rs. 100 = Rs. 1.25
26. The B.G. on a certain sum 4 years hence at 5% is Rs. 200. What is the present worth? A. Rs. 4500
B. Rs. 6000
C. Rs. 5000
D. Rs. 4000
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : T = 4 years R = 5% Banker's Gain, BG = Rs.200
BG × 100 200 × 100 = = Rs. 1000 TR 4×5 −−−−−−−−− TD = √ PW × BG
TD =
−−−−−−−−− 1000 = √ PW × 200 1000000 = PW × 200 PW =
1000000 = Rs. 5000 200
27. The B.D. and T.D. on a certain sum is Rs.200 and Rs.100 respectively. Find out the sum. A. Rs. 400
B. Rs. 300
C. Rs. 100
D. Rs. 200
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
F =
BD × TD 200 × 100 200 × 100 = = = Rs. 200 200 − 100 100 (BD – TD)
28. The banker's discount on a bill due 6 months hence at 6% is Rs. 18.54. What is the true discount? A. Rs. 24
B. Rs. 12
C. Rs. 36
D. Rs. 18
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
T = 6 months = R = 6%
1 year 2
TD =
=
BD × 100 = 100 + TR
18.54 × 100 1 100 + ( × 6) 2
=
18.54 × 100 103
1854 = Rs. 18 103
29. The banker's discount on a sum of money for 1 sum for 2 years is Rs.150. What is the rate per cent?
1 % 3 2 C. 3 % 3
1 years is Rs. 120. The true discount on the same 2 1 % 3 2 D. 4 % 3
A. 3
B. 4
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
BD for 1
1 years = Rs. 120 2
BD for 2 years = 120 ×
2 × 2 = Rs.160 3
TD for 2 years = Rs. 150
F =
BD × TD 160 × 150 160 × 150 = = = Rs. 2400 160– 150 10 (BD – TD)
=> Rs.160 is the simple interest on Rs. 2400 for 2 years
⇒ 160 = ⇒R= =
2400 × 2 × R 100
160 × 100 160 = 2400 × 2 48
20 10 1 = =3 % 6 3 3
30. The present worth of a certain bill due sometime hence is Rs. 400 and the true discount is Rs. 20. What is the banker's discount? A. Rs. 19
B. Rs. 22
C. Rs. 20
D. Rs. 21
Hide Answer | Notebook | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
(TD) 2 20 2 BG = = = Rs. 1 PW 400 BG = BD – TD
=> 1 = BD 20
=> BD = 1 + 20 = Rs. 21
Important Formulas Boats and Streams 1. Dow nstream In water, the direction along the stream is called downstream. 2. U pstream In water, the direction against the stream is called upstream. Let the speed of a boat in still water be u km/hr and the speed of the stream be v km/hr, then
3.
Speed downstream = (u + v) km/hr Speed upstream = (u - v) km/hr.
Let the speed downstream be a km/hr and the speed upstream be b km/hr, then 4.
Som e m ore short-cut m ethods Assume that a man can row at the speed of x km/hr in still water and he rows the same distance up and down in a stream which flows at a rate of y km/hr. Then his average speed throughout the journey 5.
Let the speed of a man in still water be x km/hr and the speed of a stream be y km/hr. If he takes t hours more in upstream than to go downstream for the same distance, the distance
6.
A man rows a certain distance downstream in t1 hours and returns the same distance upstream in t2 hours. If the speed of the stream is y km/hr, then the speed of the man in still water 7.
A man can row a boat in still water at x km/hr. In a stream flowing at y km/hr, if it takes him t hours to row a place and come back, then the distance between the two places
8.
A man takes n times as long to row upstream as to row downstream the river. If the speed of the man is x km/hr and the speed of the stream is y km/hr, then 9.
1. A man's speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man's speed against the current is: A. 8.5 km/hr
B. 10 km/hr.
C. 12.5 km/hr
D. 9 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Man's speed with the current = 15 km/hr =>speed of the man + speed of the current = 15 km/hr speed of the current is 2.5 km/hr Hence, speed of the man = 15 - 2.5 = 12.5 km/hr man's speed against the current = speed of the man - speed of the current = 12.5 - 2.5 = 10 km/hr
2. A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is: A. 10
B. 6
C. 5
D. 4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Speed of the motor boat = 15 km/hr Let speed of the stream = v Speed downstream = (15+v) km/hr Speed upstream = (15-v) km/hr
3. In one hour, a boat goes 14 km/hr along the stream and 8 km/hr against the stream. The speed of the boat in still water (in km/hr) is: A. 12 km/hr
B. 11 km/hr
C. 10 km/hr
D. 8 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
------------------------------------------------------------------Solution 1 : U sing Form ula -------------------------------------------------------------------
Let the speed downstream be a km/hr and the speed upstream be b km/hr, then
[Read more ...]
------------------------------------------------------------------Solution 2 : U sing P rinciples ------------------------------------------------------------------Let speed of the boat in still water = a and speed of the stream = b Then a + b = 14 a - b = 8 Adding these two equations, we get 2a = 22 => a = 11 ie, speed of boat in still water = 11 km/hr
4. A man rows to a place 48 km distant and come back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. The rate of the stream is: A. 1 km/hr.
B. 2 km/hr.
C. 1.5 km/hr.
D. 2.5 km/hr.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Assume that he moves 4 km downstream in x hours
Given that he can row 4 km with the stream in the same time as 3 km against the stream
He rows to a place 48 km distant and come back in 14 hours
Now we can use the below formula to find the rate of the stream
Let the speed downstream be a km/hr and the speed upstream be b km/hr, then
[Read more ...]
5. A boatman goes 2 km against the current of the stream in 2 hour and goes 1 km along the current in 20 minutes. How long will it take to go 5 km in stationary water? A. 2 hr 30 min
B. 2 hr
C. 4 hr
D. 1 hr 15 min
Hide Answer | Notebook | Discuss
Here is the answer and explanation Answer : Option A Explanation :
6. Speed of a boat in standing water is 14 kmph and the speed of the stream is 1.2 kmph. A man rows to a place at a distance of 4864 km and comes back to the starting point. The total time taken by him is: A. 700 hours
B. 350 hours
C. 1400 hours
D. 1010 hours
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Speed downstream = (14 + 1.2) = 15.2 kmph Speed upstream = (14 - 1.2) = 12.8 kmph
7. The speed of a boat in still water in 22 km/hr and the rate of current is 4 km/hr. The distance travelled downstream in 24 minutes is: A. 9.4 km
B. 10.2 km
C. 10.4 km
D. 9.2 km
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Speed downstream = (22 + 4) = 26 kmph
8. A boat covers a certain distance downstream in 1 hour, while it comes back in 11∕2 hours. If the speed of the stream be 3 kmph, what is the speed of the boat in still water? A. 14 kmph
B. 15 kmph
C. 13 kmph
D. 12 kmph
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the speed of the water in still water = x Given that speed of the stream = 3 kmph Speed downstream = (x+3) kmph Speed upstream = (x-3) kmph He travels a certain distance downstream in 1 hour and come back in 11∕2 hour. ie, distance travelled downstream in 1 hour = distance travelled upstream in 11∕2 hour since distance = speed × time, we have
=> 2(x + 3) = 3(x-3) => 2x + 6 = 3x - 9 => x = 6+9 = 15 kmph
9. A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively? A. 5 : 6
B. 6 : 5
C. 8 : 3
D. 3 : 8
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the rate upstream of the boat = x kmph and the rate downstream of the boat = y kmph Distance travelled upstream in 8 hrs 48 min = Distance travelled downstream in 4 hrs. Since distance = speed × time, we have
Now consider the formula given below
Let the speed downstream be a km/hr and the speed upstream be b km/hr, then
[Read more ...]
10. A boat can travel with a speed of 22 km/hr in still water. If the speed of the stream is 5 km/hr, find the time taken by the boat to go 54 km downstream A. 5 hours
B. 4 hours
C. 3 hours
D. 2 hours
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Speed of the boat in still water = 22 km/hr speed of the stream = 5 km/hr Speed downstream = (22+5) = 27 km/hr Distance travelled downstream = 54 km
11. A boat running downstream covers a distance of 22 km in 4 hours while for covering the same distance upstream, it takes 5 hours. What is the speed of the boat in still water? A. 5 kmph
B. 4.95 kmph
C. 4.75 kmph
D. 4.65
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
12. A man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. The ratio of the speed of the boat (in still water) and the stream is: A. 3 : 1
B. 1 : 3
C. 1 : 2
D. 2 : 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let speed upstream = x Then, speed downstream = 2x
13. A man can row at 5 kmph in still water. If the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back, how far is the place? A. 3.2 km
B. 3 km
C. 2.4 km
D. 3.6 km
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Speed in still water = 5 kmph Speed of the current = 1 kmph Speed downstream = (5+1) = 6 kmph Speed upstream = (5-1) = 4 kmph Let the requited distance be x km Total time taken = 1 hour
=> 2x + 3x = 12 => 5x = 12 => x = 2.4 km
14. A man can row three-quarters of a kilometre against the stream in 111∕4 minutes and down the
stream in 71∕2minutes. The speed (in km/hr) of the man in still water is: A. 4 kmph
B. 5 kmph
C. 6 kmph
D. 8 kmph
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
15. A boat takes 90 minutes less to travel 36 miles downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 mph, the speed of the stream is: A. 4 mph
B. 2.5 mph
C. 3 mph
D. 2 mph
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Speed of the boat in still water = 10 mph Let speed of the stream be x mph Then, speed downstream = (10+x) mph speed upstream = (10-x) mph
Since x can not be negative, x = 2 mph
16. Tap 'A' can fill the tank completely in 6 hrs while tap 'B' can empty it by 12 hrs. By mistake, the person forgot to close the tap 'B', As a result, both the taps, remained open. After 4 hrs, the person realized the mistake and immediately closed the tap 'B'. In how much time now onwards, would the tank be full? A. 2 hours
B. 4 hours
C. 5 hours
D. 1 hour
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Tap A can fill the tank completely in 6 hours => I n 1 hour, Tap A can fill 1 ∕6 of the tank Tap B can empty the tank completely in 12 hours => I n 1 hour, Tap B can em pty 1 ∕1 2 of the tank i.e., In one hour, Tank A and B together can effectively fill 1∕6- 1∕12 = 1∕12 of the tank => In 4 hours, Tank A and B can effectively fill 1∕12 × 4 = 1∕3of the tank.
17. A Cistern is filled by pipe A in 8 hrs and the full Cistern can be leaked out by an exhaust pipe B in 12 hrs. If both the pipes are opened in what time the Cistern is full? A. 12 hrs
B. 24 hrs
C. 16 hrs
D. 32 hrs
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Pipe A can fill 1∕8 of the cistern in 1 hour. Pipe B can empty 1∕12 of the cistern in 1 hour Both Pipe A and B together can effectively fill 1∕8-1∕12= 1∕24 of the cistern in 1 hour i.e, the cistern will be full in 24 hrs.
18. In a river flowing at 2 km/hr, a boat travels 32 km upstream and then returns downstream to the starting point. If its speed in still water be 6 km/hr, find the total journey time. A. 10 hours
B. 12 hours
C. 14 hours
D. 16 hours
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : -----------------------------------------------------------Solution 1 -----------------------------------------------------------speed of the boat = 6 km/hr Speed downstream = (6+2) = 8 km/hr Speed upstream = (6-2) = 4 km/hr Distance travelled downstream = Distance travelled upstream = 32 km Total time taken = Time taken downstream + Time taken upstream
-----------------------------------------------------------Solution 2 ------------------------------------------------------------
A man can row a boat in still water at x km/hr. In a stream flowing at y km/hr, if it takes him t hours to row a place and come back, then the distance between the two places
[Read more ...]
x = 6 km/hr y = 2 km/hr distance = 32 km As per the formula, we have
19. Two pipes A and B can fill a tank in 10 hrs and 40 hrs respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank? A. 8 hours
B. 6 hours
C. 4 hours
D. 2 hours
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Pipe A can fill 1∕10 of the tank in 1 hr Pipe B can fill 1∕40 of the tank in 1 hr Pipe A and B together can fill 1∕10 + 1∕40 = 1∕8 of the tank in 1
hr i.e., Pipe A and B together can fill the tank in 8 hours
20. A boat covers a certain distance downstream in 4 hours but takes 6 hours to return upstream to the starting point. If the speed of the stream be 3 km/hr, find the speed of the boat in still water A. 15 km/hr
B. 12 km/hr
C. 13 km/hr
D. 14 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : -----------------------------------------------------------------------Solution 1 -----------------------------------------------------------------------Let the speed of the water in still water = x Given that speed of the stream = 3 kmph Speed downstream = (x+3) kmph Speed upstream = (x-3) kmph He travels a certain distance downstream in 4 hour and come back in 6 hour. ie, distance travelled downstream in 4 hour = distance travelled upstream in 6 hour since distance = speed × time, we have
=> (x + 3)2 = (x - 3)3 => 2x + 6 = 3x - 9 => x = 6+9 = 15 kmph -----------------------------------------------------------------------Solution 2 ------------------------------------------------------------------------
A man rows a certain distance downstream in t1 hours and returns the same distance upstream in t2 hours. If the speed of the stream is y km/hr, then the speed of the man in still water
[Read more ...]
t1 = 4 hour t2 = 6 hour y = 3 km/hr By using the the abov formula, Speed of the boat in still water
21. If a man rows at the rate of 5 kmph in still water and his rate against the current is 3 kmph, then the man's rate along the current is: A. 5 kmph
B. 7 kmph
C. 12 kmph
D. 8 kmph
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the rate along with the current is x km/hr
=> x + 3 = 10 => x = 7 kmph
22. A man can row 8 km/hr in still water. If the river is running at 3 km/hr, it takes 3 hours more in upstream than to go downstream for the same distance. How far is the place? A. 32.5 km
B. 25 km
C. 27.5 km
D. 22.5 km
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : -------------------------------------------------------Solution 1 -------------------------------------------------------Let the speed downstream = x and speed updstream = y Then
=> x + y = 16 ---(Equation 1)
=> x - y = 6 ---(Equation 2) (Equation 1 + Equation 2) = > 2x = 22 => x = 11 km/hr (Equation 1 - Equation 2) = > 2y = 10 =>y = 5 km/hr Time taken to travel upstream = Time taken to travel downstream + 3 Let distance be x km Then
=> 11x = 5x + 165 => 6x = 165 => 2x = 55 => x = 27.5 -------------------------------------------------------Solution 2
---------------------------------------------------------
Let the speed of a man in still water be x km/hr and the speed of a stream be y km/hr. If he takes t hours more in upstream than to go downstream for the same distance, the distance
[Read more ...]
x = 8 km/hr y = 3 km/hr t = 3 hours As per the formula, we have
23. A man can row 4 kmph is still water. If the river is running at 2 kmph it takes 90 min to row to a place and back. How far is the place? A. 2 km
B. 4 km
C. 5 km
D. 2.25 km
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Speed in still water = 4 kmph Speed of the stream = 2 kmph Speed upstream = (4-2)= 2 kmph Speed downstream = (4+2)= 6 kmph Total time = 90 minutes = 90∕60 hour = 3∕2 hour Let L be the distance. Then
=> L + 3L = 9 => 4L = 9 => L = 9∕4= 2.25 km
24. At his usual rowing rate, Rahul can travel 12 miles downstream in a certain river in 6 hours less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for his 24-mile round trip, the downstream 12 miles would then take only one hour less than the upstream 12 miles. What is the speed of the current in miles per hour? A.
mph
B.
mph
C.
mph
D.
mph
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the speed of Rahul in still water be x mph and the speed of the current be y mph Then, Speed upstream = (x - y) mph Speed downstream = (x + y) mph Distance = 12 miles Time taken to travel upstream - Time taken to travel downstream = 6 hours
Now he doubles his speed. i.e., his new speed = 2x Now, Speed upstream = (2x - y) mph Speed downstream = (2x + y) mph
In this case, Time taken to travel upstream - Time taken to travel downstream = 1 hour
25. A man can row 40 kmph in still water and the river is running at 10 kmph. If the man takes 1 hr to row to a place and back, how far is the place? A. 16.5 kmph
B. 12.15 kmph
C. 2.25 kmph
D. 18.75 kmph
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the distance be x Speed upstream = (40 - 10) = 30 kmph Speed downstream = (40 + 10) = 50 kmph
Total time taken = 1 hr
26. A boatman can row 96 km downstream in 8 hr. If the speed of the current is 4 km/hr, then find in what time will be able to cover 8 km upstream? A. 6 hr
B. 2 hr
C. 4 hr
D. 1 hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Speed downstream = 96∕8 = 12 kmph Speed of current = 4 km/hr Speed of the boatman in still water = 12-4 = 8 kmph Speed upstream = 8-4 = 4 kmph Time taken to cover 8 km upstream = 8∕4 = 2 hours
27. The speed of a boat in still water is 10 km/hr. If it can travel 78 km downstream and 42 km upstream in the same time, the speed of the stream is A. 3 km/hr
B. 12 km/hr
C. 1.5 km/hr
D. 4.4 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the speed of the stream be x km/hr. Then Speed upstream = (10-x) km/hr Speed downstream = (10+x) km/hr
Time taken to travel 78 km downstream = Time taken to travel 42 km upstream
28. A man can row at a speed of 12 km/hr in still water to a certain upstream point and back to the starting point in a river which flows at 3 km/hr. Find his average speed for total journey. A.
km/hr
B.
km/hr
C.
km/hr
D.
km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Assume that a man can row at the speed of x km/hr in still water and he rows the same distance up and down in a stream which flows at a rate of y km/hr. Then his average speed throughout the journey
[Read more ...]
Speed of the man in still water = 12 km/hr Speed of the stream = 3 km/hr Speed downstream = (12+3) = 15 km/hr Speed upstream = (12-3) = 9 km/hr
29. A boatman can row 3 km against the stream in 20 minutes and return in 18 minutes. Find the rate of current A. 1∕2 km/hr
B. 1 km/hr
C. 1∕3 km/hr
D. 2∕3 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
30. A boat takes 38 hours for travelling downstream from point A to point B and coming back to point C midway between A and B. If the velocity of the stream is 4 kmph and the speed of the boat in still water is 14 kmph, what is the distance between A and B? A. 240 km
B. 120 km
C. 360 km
D. 180 km
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : velocity of the stream = 4 kmph Speed of the boat in still water is 14 kmph Speed downstream = (14+4) = 18 kmph Speed upstream = (14-4) = 10 kmph Let the distance between A and B be x km Time taken to travel downstream from A to B + Time taken to travel upstream from B to C(mid of A and B) = 38 hours
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31. If a man's rate with the current is 15 km/hr and the rate of the current is 11∕2 km/hr, then his rate against the current is A. 12 km/hr
B. 10 km/hr
C. 10.5 km/hr
D. 12.5 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Speed downstream = 15 km/hr Rate of the current= 11∕2 km/hr Speed in still water = 15 - 11∕2 = 131∕2 km/hr Rate against the current = 131∕2 km/hr - 11∕2 = 12 km/hr
32. The speed of the boat in still water in 12 kmph. It can travel downstream through 45 kms in 3 hrs. In what time would it cover the same distance upstream? A. 8 hours
B. 6 hours
C. 4 hours
D. 5 hours
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Speed of the boat in still water = 12 km/hr Speed downstream = 45∕3 = 15 km/hr Speed of the stream = 15-12 = 3 km/hr Speed upstream = 12-3 = 9 km/hr Time taken to cover 45 km upstream = 45∕9 = 5 hours
33. A boat goes 8 km upstream in 24 minutes. The speed of stream is 4 km/hr. The speed of boat in still water is: A. 25 km/hr
B. 26 km/hr
C. 22 km/hr
D. 24 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Speed upstream =
8 24 ( ) 60
= 20 km/hr
Speed of the stream = 4 km/hr speed of boat in still water = (20+4) = 24 km/hr
34. The Speed of a boat in still water is 25 kmph. If it can travel 10 km upstream in 1 hr, What time it would take to travel the same distance downstream? A. 22 minutes
B. 30 minutes
C. 40 minutes
D. 15 minutes
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Speed of boat in still water = 25 km/hr Speed upstream = 10∕1 = 10 km/hr Speed of the stream = (25-10) = 15 km/hr Speed downstream = (25+15) = 40 km/hr
Time taken to travel 10 km downstream =
10 10 × 60 hours = = 15 minutes 40 40
35. A boat can travel with a speed of 12 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 68 km downstream. A. 4 hr
B. 4.25 hr
C. 4.5 hr
D. 6 hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
speed of boat in still water = 12 km/hr speed of the stream = 4 km/hr Speed downstream = (12+4) = 16 km/hr Time taken to travel 68 km downstream = 68∕16 = 17∕4 = 4.25 hours
36. A man can row 7.5 kmph in still water and he finds that it takes him twice as long to row up as to row down the river. Find the rate of stream. A. 10 km/hr
B. 2.5 km/hr
C. 5 km/hr
D. 7.5 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Given that, time taken to travel upstream = 2 × time taken to travel downstream When distance is constant, speed is inversely proportional to the time Hence, 2 × speed upstream = speed downstream Let speed upstream = x Then speed downstream = 2x
we have, ⇒
1 (x + 2x) = speed in still water 2
1 (3x) = 7.5 2
3x = 15 x = 5 i.e., speed upstream = 5 km/hr
Rate of stream =
1 x 5 (2x − x) = = = 2.5 km/hr 2 2 2
37. A boat moves downstream at the rate of one km in 5 minutes and upstream at the rate of 4 km an hour. What is the velocity of the current?
A. 4 km/hr
B. 2 km/hr
C. 3 km/hr
D. 1 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Speed downstream =
1 5 ( ) 60
= 12 km/hr
4 = 4 km/hr 1 1 velocity of the current = (12 − 4) = 4 km/hr 2 Speed upstream =
38. The speed of a boat in still water is 15 km/hr and the rate of current is 3 km/hr. The distance travelled downstream in 24 minutes is A. 3.6 km
B. 2.4 km
C. 3.2 km
D. 7.2 km
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : speed of a boat in still water = 15 km/hr Speed of the current = 3 km/hr Speed downstream = (15+3) = 18 km/hr
Distance travelled downstream in 24 minutes =
24 2 × 18 × 18 = = 7.2 km 60 5
39. The current of a stream runs at the rate of 2 km per hr. A motor boat goes 10 km upstream and back again to the starting point in 55 min. Find the speed of the motor boat in still water? A. 22 km/hr
B. 12 km/hr
C. 20 km/hr
D. 16 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Let the speed of the boat in still water = x km/hr Speed of the current = 2 km/hr Then, speed downstream = (x+2) km/hr speed upstream = (x-2) km/hr Total time taken to travel 10 km upstream and back = 55 minutes = 55∕60 hour = 11∕12 hour
⇒
10 10 11 + = x−2 x+2 12
120(x + 2) + 120(x − 2) = 11(x 2 − 4) 240x = 11x 2 − 44 11x 2 − 240x − 44 = 0 11x 2 − 242x + 2x − 44 = 0 11x(x − 22) + 2(x − 22) = 0 (x − 22)(11x + 2) = 0 x = 22 or
−2 11
Since x can not be negative, x = 22 i.e., speed of the boat in still water = 22 km/hr
40. The speed of a boat in still water is 8 kmph. If it can travel 1 km upstream in 1 hr, What time it would take to travel the same distance downstream? A. 1 minute
B. 2 minutes
C. 3 minutes
D. 4 minutes
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Speed of the boat in still water = 8 km/hr Speed upstream = 1∕1 = 1 km/hr
Speed of the stream = 8-1 = 7 km/hr Speed downstream = (8+7) = 15 km/hr
Time taken to travel 1 km downstream =
1 1 × 60 hr = = 4 minutes 15 15
1. What day of the week does May 28 2006 fall on A. Saturday
B. Monday
C. Sunday
D. Thursday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : 28th May 2006 = (2005 years + period from 1-Jan-2006 to 28May-2006)
We know that number of odd days in 400 years = 0 Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)
Number of odd days in the period 2001-2005 = 4 normal years + 1 leap year = 4 x 1 + 1 x 2 = 6
Days from 1-Jan-2006 to 28-May-2006 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 28(may) = 148 148 days = 21 weeks + 1 day = 1 odd day
Total number of odd days = (0 + 6 + 1) = 7 odd days = 0 odd day 0 odd day = Sunday
Hence May 28 2006 is Sunday.
2. What will be the day of the week 15th August, 2010? A. Thursday
B. Sunday
C. Monday
D. Saturday
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option B Explanation : 15th Aug 2010 = (2009 years + period from 1-Jan-2010 to 15Aug-2010)
We know that number of odd days in 400 years = 0 Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)
Number of odd days in the period 2001-2009 = 7 normal years + 2 leap year = 7 x 1 + 2 x 2 = 11 = (11 - 7x1) odd day = 4 odd day
Days from 1-Jan-2010 to 15-Aug-2010 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 31(may) + 30(Jun) + 31(Jul) + 15(Aug) = 227 227 days = 32 weeks + 3 day = 3 odd day
Total number of odd days = (0 + 4 + 3) = 7 odd days = 0 odd day 0 odd day = Sunday
Hence 15th August, 2010 is Sunday.
3. Today is Monday. After 61 days, it will be A. Thursday
B. Sunday
C. Monday
D. Saturday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : 61 days = 8 weeks 5 days = 5 odd days
Hence if today is Monday, After 61 days, it will be = (Monday + 5 odd days) = Saturday
4. On what dates of April, 2001 did Wednesday fall? A. 2nd, 9th, 16th, 23rd
B. 4th, 11th, 18th, 25th
C. 3rd, 10th, 17th, 24th
D. 1st, 8th, 15th, 22nd, 29th
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : We need to find out the day of 01-Apr-2001
01-Apr-2001 = (2000 years + period from 1-Jan-2001 to 01Apr-2001)
We know that number of odd days in 400 years = 0 Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)
Days from 1-Jan-2001 to 01-Apr-2001 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 1(Apr) = 91 91 days = 13 weeks = 0 odd day
Total number of odd days = (0 + 0) = 0 odd days
0 odd day = Sunday. Hence 01-Apr-2001 is Sunday.
Hence first Wednesday of Apr 2011 comes in 04th and successive Wednesdays come in 11th, 18th and 25th
5. How many days are there in x weeks x days A. 14x
B. 8x
C. 7x2
D. 7
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
x weeks x days = (7 × x) + x = 7x + x = 8x days
6. The calendar for the year 2007 will be the same for the year A. 2017
B. 2018
C. 2014
D. 2016
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : For a year to have the same calendar with 2007 ,the total odd days from 2007 should be 0.
Y ear
: 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017
Odd : 1 Days
2
1
1
1
2
1
Take the year 2014 given in the choice. Total odd days in the period 2007-2013 = 5 normal years + 2 leap year = 5 x 1 + 2 x 2 = 9 odd days = 2 odd day (As we can reduce multiples of 7 from odd days which will not change anything)
Take the year 2016 given in the choice. Number of odd days in the period 2007-2015 = 7 normal years + 2 leap year
1
1
2
1
= 7 x 1 + 2 x 2 = 11 odd days = 4 odd days (Even if the odd days were 0, calendar of 2007 will not be same as the calendar of 2016 because 2007 is not a leap year whereas 2016 is a leap year. In fact, you can straight away ignore this choice due to this fact without even bothering to check the odd days) Take the year 2017 given in the choice. Number of odd days in the period 2007-2016 = 7 normal years + 3 leap year = 7 x 1 + 3 x 2 = 13 odd days = 6 odd days
Take the year 2018 given in the choice. Number of odd days in the period 2007-2017 = 8 normal years + 3 leap year = 8 x 1 + 3 x 2 = 14 odd days = 0 odd day (As we can reduce multiples of 7 from odd days which will not change anything) Also, both 2007 and 2018 are not leap years. Since total odd days in the period 2007-2017 = 0 and both 2007 and 2018 are of same type, 2018 will have the same calendar as that of 2007
7. Which of the following is not a leap year? A. 1200
B. 800
C. 700
D. 2000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Remember the leap year rule (Given in the formulas)
1. Every year divisible by 4 is a leap year, if it is not a
century. 2. Every 4th century is a leap year, but no other century is a leap year.
800,1200 and 2000 comes in the category of 4th century (such as 400,800,1200,1600,2000 etc). Hence 800,1200 and 2000 are leap years
700 is not a 4th century, but it is a century. Hence it is not a leap year
8. 01-Jan-2007 was Monday. What day of the week lies on 01-Jan-2008? A. Wednesday
B. Sunday
C. Friday
D. Tuesday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Given that January 1, 2007 was Monday.
Odd days in 2007 = 1 (we have taken the complete year 2007 because we need to find out the odd days from 01-Jan-2007 to 31-Dec-2007, that is the whole year 2007)
Hence January 1, 2008 = (Monday + 1 Odd day) = Tuesday
9. 8th Dec 2007 was Saturday, what day of the week was it on 8th Dec, 2006? A. Sunday
B. Tuesday
C. Friday
D. Tuesday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Given that 8th Dec 2007 was Saturday
Number of days from 8th Dec, 2006 to 7th Dec 2007 = 365 days 365 days = 1 odd day
Hence 8th Dec 2006 was = (Saturday - 1 odd day) = Friday
10. On 8th Feb, 2005 it was Tuesday. What was the day of the week on 8th Feb, 2004? A. Sunday
B. Friday
C. Saturday
D. Monday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Given that 8th Feb, 2005 was Tuesday
Number of days from 8th Feb, 2004 to 7th Feb, 2005 = 366 (Since Feb 2004 has 29 days as it is a leap year)
366 days = 2 odd days
Hence 8th Feb, 2004 = (Tuesday - 2 odd days) = Sunday
11. The last day of a century cannot be A. Monday
B. Wednesday
C. Tuesday
D. Friday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : We know that number of odd days in 100 years = 5 Hence last day of the first century is Friday
Number of odd days in 200 years = 5 x 2 = 10 = 3 (As we can reduce multiples of 7 from odd days which will not change anything) Hence last day of the 2nd century is Wednesday
Number of odd days in 300 years = 5 x 3 = 15 = 1 (As we can reduce multiples of 7 from odd days which will not change anything) Hence last day of the 3rd century is Monday
We know that umber of odd days in 400 years = 0. (? 5 x 4 + 1 = 21 = 0) Hence last day of the 4th century is Sunday
Now this cycle will be repeated.
Hence last day of a century will not be Tuesday or Thursday or Saturday.
its better to learn this by heart which will be helpful to save time in objective type exams
12. January 1, 2008 is Tuesday. What day of the week lies on Jan 1, 2009? A. Saturday
B. Wednesday
C. Thursday
D. Saturday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Number of odd days in 2008 = 2 (since it is a leap year) (we have taken the complete year 2008 because we need to find out the odd days from
01-Jan-2008 to 31-Dec-2008, that is the whole year 2008)
Given that January 1, 2008 is Tuesday
Hence January 1, 2009 = (Tuesday + 2 odd days) = Thursday
13. If Jan 1, 2006 was a Sunday, What was the day of the week Jan 1, 2010? A. Friday
B. Thursday
C. Tuesday
D. Saturday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Given that Jan 1 2006 was a Sunday
Number of odd days in the period 2006-2009 = 3 normal years + 1 leap year = 3 x 1 + 1 x 2 = 5 (note that we have taken the complete year 2006 because the period in 2006 is from 01-Jan-2006 to 31-Dec-2006, which is the whole year 2006. Then the complete years 2007, 2008 and 2009 are also involved)
Hence Jan 1 2010 = (Sunday + 5 odd days) = Friday
14. What was the day of the week on 17th June 1998? A. Monday
B. Sunday
C. Wednesday
D. Friday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
17 Jun 1998 = (1997 years + period from 1-Jan-1998 to 17Jun-1998)
We know that number of odd days in 400 years = 0 Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)
Number of odd days in the period 1601-1900 = Number of odd days in 300 years = 5 x 3 = 15 = 1 (As we can reduce perfect multiples of 7 from odd days without affecting anything)
Number of odd days in the period 1901-1997 = 73 normal years + 24 leap year = 73 x 1 + 24 x 2 = 73 + 48 = 121 = (121 - 7 x 17) = 2 odd days
Number of days from 1-Jan-1998 to 17-Jun-1998 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 31(may) + 17(Jun) = 168 168 days = 0 odd day
Total number of odd days = (0 + 1 + 2 + 0) = 3 3 odd days = Wednesday
Hence 17th June 1998 is Wednesday.
15. 6th March, 2005 is Monday, what was the day of the week on 6th March, 2004? A. Friday
B. Saturday
C. Wednesday
D. Sunday
Hide Answer | Notebook | Discuss
Here is the answer and explanation Answer : Option D Explanation : Number of days from 6th March, 2004 to 5th March 2005 = 365 days (Though Feb 2004 has 29 days as it is a leap year, it will not come in the required period)
365 days = 1 odd day
Given that 6th March, 2005 is Monday
Hence 6th March, 2004 = (Monday - 1 odd day) = Sunday
16. What day of the week was 1 January 1901 A. Monday
B. Tuesday
C. Saturday
D. Friday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 1 Jan 1901 = (1900 years + 1-Jan-1901)
We know that number of odd days in 400 years = 0 Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)
Number of odd days in the period 1601-1900 = Number of odd days in 300 years = 5 x 3 = 15 = 1 (As we can reduce perfect multiples of 7 from odd days without affecting anything)
1-Jan-1901 = 1 odd day
Total number of odd days = (0 + 1 + 1) = 2 2 odd days = Tuesday
Hence 1 January 1901 is Tuesday.
17. What day of the week will 22 Apr 2222 be? A. Monday
B. Tuesday
C. Sunday
D. Thursday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 22 Apr 2222 = (2221 years + period from 1-Jan-2222 to 22Apr-2222)
We know that number of odd days in 400 years = 0 Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)
Number of odd days in the period 2001-2200 = Number of odd days in 200 years = 5 x 2 = 10 = 3 (As we can reduce perfect multiples of 7 from odd days without affecting anything)
Number of odd days in the period 2201-2221 = 16 normal years + 5 leap years = 16 x 1 + 5 x 2 = 16 + 10 = 26 = 5 odd days
Number of days from 1-Jan-2222 to 22 Apr 2222 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 22(Apr) = 112
112 days = 0 odd day
Total number of odd days = (0 + 3 + 5 + 0) = 8 = 1 odd day 1 odd days = Monday
Hence 22 Apr 2222 is Monday.
18. Today is Thursday. The day after 59 days will be? A. Monday
B. Tuesday
C. Saturday
D. Sunday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : 59 days = 8 weeks 3 days = 3 odd days
Hence if today is Thursday, After 59 days, it will be = (Thursday + 3 odd days) = Sunday
19. What is the year next to 1990 which will have the same calendar as that of the year 1990? A. 1992
B. 2001
C. 1995
D. 1996
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : For a year to have the same calendar with 1990 ,total odd days from 1990 should be 0. Take the year 1992 from the given choices. Total odd days in the period 1990-1991= 2 normal years ≡ 2 x 1 = 2 odd days Take the year 1995 from the given choices. Number of odd days in the period 1990-1994 = 4 normal years + 1 leap year
≡ 4 x 1 + 1 x 2 = 6 odd days Take the year 1996 from the given choices. Number of odd days in the period 1990-1995= 5 normal years + 1 leap year ≡ 5 x 1 + 1 x 2 = 7 odd days ≡ 0 odd days (As we can reduce multiples of 7 from odd days which will not change anything) Though number of odd days in the period 1990-1995 is 0, there is a catch here. 1990 is not a leap year whereas 1996 is a leap year. Hence calendar for 1990 and 1996 will never be the same. Take the year 2001 from the given choices. Number of odd days in the period 1990-2000= 8 normal years + 3 leap years ≡ 8 x 1 + 3 x 2 = 14 odd days ≡ 0 odd days Also, both 1990 and 2001 are normal years. Hence 1990 will have the same calendar as that of 2001
20. January 1, 2004 was a Thursday, what day of the week lies on January 1 2005. A. Saturday
B. Monday
C. Saturday
D. Tuesday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Given that January 1, 2004 was Thursday.
Odd days in 2004 = 2 (because 2004 is a leap year) (Also note that we have taken the complete year 2004 because we need to find out the odd days from 01-Jan-2004 to 31-Dec-2004, that is the whole year 2004)
Hence January 1, 2005 = (Thursday + 2 Odd days) = Saturday
21. If the first day of a year (other than leap year) was Friday, then which was the last day of that year? A. Saturday
B. Friday
C. Tuesday
D. Monday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Given that first day of a normal year was Friday
Odd days of the mentioned year = 1 (Since it is an ordinary year)
Hence First day of the next year = (Friday + 1 Odd day) = Saturday ? Last day of the mentioned year = Friday
22. If 1st October is Sunday, then 1st November will be A. Saturday
B. Thursday
C. Wednesday
D. Tuesday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Given that 1st October is Sunday
Number of days in October = 31 31 days = 3 odd days (As we can reduce multiples of 7 from odd days which will not change anything)
Hence 1st November = (Sunday + 3 odd days) = Wednesday
23. Arun went for a movie nine days ago. He goes to watch movies only on Thursdays. What day of the week is today? A. Wednesday
B. Saturday
C. Friday
D. Sunday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Clearly it can be understood from the question that 9 days ago was a Thursday
Number of odd days in 9 days = 2 (As 9-7 = 2, reduced perfect multiple of 7 from total days)
Hence today = (Thursday + 2 odd days) = Saturday
24. 1.12.91 is the first Sunday. Which is the fourth Tuesday of December 91? A. 20.12.91
B. 22.12.91
C. 24.12.91
D. 25.12.91
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Given that 1.12.91 is the first Sunday Hence we can assume that 3.12.91 is the first Tuesday
If we add 7 days to 3.12.91, we will get second Tuesday If we add 14 days to 3.12.91, we will get third Tuesday If we add 21 days to 3.12.91, we will get fourth Tuesday
=> fourth Tuesday = (3.12.91 + 21 days) = 24.12.91
25. If the day before yesterday was Thursday, when will Sunday be? A. Day after tomorrow
B. Tomorow
C. Two days after today
D. Today
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B
Explanation : Day before yesterday was Thursday =>Yesterday was a Friday => Today is a Saturday => Tomorrow is a Sunday
26. The second day of a month is Friday, What will be the last day of the next month which has 31 days? A. Friday
B. Saturday
C. Wednesday
D. Data inadequate
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : We cannot find out the answer because the number of days of the current month is not given.
27. How many days will there be from 26th January,1996 to 15th May,1996(both days included)? A. 102
B. 103
C. 111
D. 120
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Number of days from 26-Jan-1996 to 15-May-1996 (both days included) = 6(Jan) + 29(Feb) + 31 (Mar) + 30(Apr)+ 15(May) = 111
28. If 25th of August in a year is Thursday, the number of Mondays in that month is A. 4
B. 5
C. 2
D. 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B
Explanation : Given that 25th August = Thursday
Hence 29th August = Monday So 22nd,15th and 8th and 1st of August will be Mondays Number of Mondays in August = 5
29. If the seventh day of a month is three days earlier than Friday, What day will it be on the nineteenth day of the month? A. Saturday
B. Monday
C. Sunday
D. Wednesday
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Given that seventh day of a month is three days earlier than Friday => Seventh day is Tuesday => 14th is Tuesday => 19th is Sunday
30. Every second Saturday and all Sundays are holidays. How many working days will be there in a month of 30 days beginning on a Saturday? A. 24
B. 23
C. 18
D. 21
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Mentioned month has begins on a Saturday and has 30 days
Sundays = 2nd, 9th, 16th, 23rd, 30th
=> Total Sundays = 5
Second Saturdays = 8th and 22nd Total Second Saturdays = 2
Total Holidays = Total Sundays + Total Second Saturdays = 5 + 2 = 7 Total days in the month = 30 Total working days = 30 - 7 = 23
I mportant Formulas - Calendar 1.
Odd Days Number of days more than the complete weeks are called odd days in a given period
2.
Leap Y ear A leap year has 366 days. In a leap year, the month of February has 29 days a.
Every year divisible by 4 is a leap year, if it is not a century. Examples: 1952, 2008, 1680 etc. are leap years. 1991, 2003 etc. are not leap years
b.
Every 4th century is a leap year and no other century is a leap year. Examples: 400, 800, 1200 etc. are leap years. 100, 200, 1900 etc. are not leap years
3.
Ordinary Y ear The year which is not a leap year is an ordinary year. An ordinary year has 365 days
4.
Counting odd days and Calculating the day of any particular date I.
1 ordinary year ≡ 365 days ≡ (52 weeks + 1 day) Hence number of odd days in 1 ordinary year= 1.
II.
1 leap year ≡ 366 days ≡ (52 weeks + 2 days) Hence number of odd days in 1 leap year= 2.
III.
100 years ≡ (76 ordinary years + 24 leap years ) ≡ (76 x 1 + 24 x 2) odd days ≡ 124 odd days. ≡ (17 weeks + 5 days)
≡ 5 odd days. Hence number of odd days in 100 years = 5.
IV.
Number of odd days in 200 years = (5 x 2) = 10 ≡ 3 odd days
V.
Number of odd days in 300 years = (5 x 3) = 15 ≡ 1 odd days
VI.
Number of odd days in 400 years = (5 x 4 + 1) = 21 ≡ 0 odd days Similarly, the number of odd days in all 4th centuries (400, 800, 1200 etc.) = 0
VII.
Mapping of the number of odd day to the day of the week
Number of Odd : 0 Days Day of the w eek
1
2
3
4
5
6
: Sunday Monday Tuesday Wednesday Thursday Friday Saturday
5.
Last day of a century cannot be Tuesday or Thursday or Saturday.
6.
For the calendars of two different years to be the same, the following conditions must be satisfied. i.
ii.
Both years must be of the same type. i.e., both years must be ordinary years or both years must be leap years. 1st January of both the years must be the same day of the week.
1. If the cost of x metres of wire is d rupees, then what is the cost of y metres of wire at the same rate?
xd ) y yd ) C. Rs. ( x A. Rs. (
B. Rs. xd D. Rs. yd
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : cost of x metres of wire = Rs. d
d cost of 1 metre of wire = Rs. ( ) x cost of y metre of wire = Rs. (y ×
d yd ) = Rs. ( ) x x
2. In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat one bag of husk? A. 1
B. 40
C. 20
D. 26
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Assume that in x days, one cow will eat one bag of husk. More cows, less days (Indirect proportion) More bags, more days (direct proportion) Hence we can write as
Cows Bags
40 : 1 . 1 : 40
.
:: x : 40
⇒ 40 × 1 × 40 = 1 × 40 × x ⇒ x = 40
3. If 7 spiders make 7 webs in 7 days, then how many days are needed for 1 spider to make 1 web? A. 1
B. 7
C. 3
D. 14
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let, 1 spider make 1 web in x days. More spiders, Less days (Indirect proportion) More webs, more days (Direct proportion) Hence we can write as
Spiders
7:1 :: x : 7
Webs
1:7
⇒7×1×7=1×7×x ⇒x=7
4. 4 mat-weavers can weave 4 mats in 4 days. At the same rate, how many mats would be woven by 8 mat-weavers in 8 days? A. 4
B. 16
C. 8
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the required number of mats be x More mat-weavers, more mats (direct proportion) More days, more mats (direct proportion) Hence we can write as
Mat-weavers
4:8 :: 4 : x
Days
4:8
⇒4×4×x=8×8×4 ⇒ x = 2 × 2 × 4 = 16
5. If a quarter kg of potato costs 60 paise, how many paise does 200 gm cost? A. 65 paise
B. 70 paise
C. 52 paise
D. 48 paise
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : -------------------------------------------------------------------Solution 1 (Chain Rule) -------------------------------------------------------------------Let 200 gm potato costs x paise Cost of ¼ Kg potato = 60 Paise => Cost of 250 gm potato = 60 Paise (∵ 1 Kg = 1000 gm => ¼ Kg = 1000/4 gm = 250 gm) More quantity, More Paise (direct proportion) Hence we can write as
Quantity
⇒ ⇒ ⇒ ⇒
200 : 250 } :: x : 60
200 × 60 = 250 × x 4 × 60 = 5 × x 4 × 12 = x x = 48
-------------------------------------------------------------------Solution 2 -------------------------------------------------------------------Cost of ¼ Kg potato = 60 Paise => Cost of 250 gm potato = 60 Paise (∵ 1 Kg = 1000 gm => ¼ Kg = 1000/4 gm = 250 gm)
=> Cost of 200 gm potato =
60 × 200 60 × 4 = = 48paise 250 5
6. In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with remaining meal? A. 50
B. 30
C. 40
D. 10
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Meal for 200 children = Meal for 120 men
120 men 200 120 × 150 Meal for 150 children = Meal for men = Meal for 90 men 200 Meal for 1 child = Meal for
Total mean available = Meal for 120 men Renaming meal = Meal for 120 men - Meal for 90 men = Meal for 30 men
7. 36 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work? A. 26
B. 22
C. 12
D. 24
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : -------------------------------------------------------------------Solution 1 (Chain Rule) -------------------------------------------------------------------Let the required number of days be x More men, less days (indirect proportion) Hence we can write as
Men
36 : 27 } :: x : 18
⇒ ⇒ ⇒ ⇒
36 × 18 = 27 × x 12 × 18 = 9 × x 12 × 2 = x x = 24
-------------------------------------------------------------------Solution 2 (Using Time and Work) -------------------------------------------------------------------Amount of work 36 men can do in 1 day = 1/18
Amount of work 1 man can do in 1 day =
1 18 × 36
Amount of work 27 men can do in 1 day =
27 3 1 = = 18 × 36 18 × 4 24
⇒ 27 men can complete the work in 24 days
-------------------------------------------------------------------Solution 3 (Using Time and Work) --------------------------------------------------------------------
If M 1 men can do W 1 work in D 1 days working H 1 hours per day and M 2 men can do W 2 work in D 2 days working H 2 hours per day (where all men work at the same rate), then M1D 1H 1 M2D 2H 2 = W1 W2
In this case, M1 = 36, M2 = 27 D1 = 18, D2 = x W1 = W2 H1 = H2 (∵ We can assume like this as these vales are not explicitly given)
Hence, M 1 D 1 = M 2 D 2 ⇒ 36 × 18 = 27x ⇒x=
36 × 18 4 × 18 = = 4 × 6 = 24 27 3
8. A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. If the smaller wheel has made 21 revolutions, what will be the number of revolutions made by the larger wheel? A. 15
B. 12
C. 21
D. 9
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the number of revolutions made by the larger wheel be x More cogs, less revolutions (Indirect proportion) Hence we can write as
Cogs
⇒ ⇒ ⇒ ⇒
6 : 14 } :: x : 21
6 × 21 = 14 × x 6×3=2×x 3×3=x x=9
9. 3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day should 4 pumps work in order to empty the tank in 1 day? A. 10
B. 12
C. 8
D. 15
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the required hours needed be x More pumps, less hours (Indirect proportion)
More Days, less hours (Indirect proportion) Hence we can write as
Pumps
3:4 :: x : 8
Days
2:1
⇒3×2×8=4×1×x ⇒3×2×2=x ⇒ x = 12
10. 39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons, working 6 hours a day, complete the work? A. 9
B. 12
C. 10
D. 13
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : -------------------------------------------------------------------Solution 1 (Chain Rule) -------------------------------------------------------------------Let the required number of days be x More persons, less days (indirect proportion) More hours, less days (indirect proportion) Hence we can write as
Persons Hours
⇒ ⇒ ⇒ ⇒
39 : 30 } :: x : 12 5:6
39 × 5 × 12 = 30 × 6 × x 39 × 5 × 2 = 30 × x 39 = 3 × x x = 13
--------------------------------------------------------------------
Solution 2 (Using Time and Work) -------------------------------------------------------------------Amount of work 39 persons can do in 1 day, working 5 hours a day = 1/12
Amount of work 1 person can do in 1 day, working 5 hours a day =
1 12 × 39
Amount of work 1 person can do in 1 day, working 1 hours a day =
1 12 × 39 × 5
Amount of work 30 person can do in 1 day, working 1 hours a day =
30 12 × 39 × 5
Amount of work 30 person can do in 1 day, working 6 hours a day =
30 × 6 12 × 39 × 5
=
30 3 1 = = 2 × 39 × 5 39 13
⇒ 30 persons can complete the work ,working 6 hours a day in 13 days
-------------------------------------------------------------------Solution 3 (Using Time and Work) --------------------------------------------------------------------
If M 1 men can do W 1 work in D 1 days working H 1 hours per day and M 2 men can do W 2 work in D 2 days working H 2 hours per day (where all men work at the same rate), then M1D 1H 1 M2D 2H 2 = W1 W2
In this case, M1 = 39, M2 = 30 D1 = 12, D2 = x W1 = W2 H1 = 5, H2 = 6
Hence, M 1 D 1 H 1 = M 2 D 2 H 2 ⇒ 39 × 12 × 5 = 30 × x × 6 ⇒ 39 × 2 × 5 = 30 × x ⇒ 39 = 3 × x ⇒x=
39 = 13 3
11. A certain industrial loom weaves 0.128 meters of cloth every second. Approximately how many seconds will it take for the loom to weave 25 meters of cloth? A. 205
B. 200
C. 180
D. 195
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the required number of seconds be x More cloth, More time, (direct proportion) Hence we can write as
Cloth
0.128 : 25 } :: 1 : x
⇒ 0.128x = 25 25 25000 3125 ⇒x= = = ≈ 195 .128 128 16
12. A contract is to be completed in 56 days if 104 persons work, each working at 8 hours a day. After 30 days, 2/5 of the work is completed. How many additional persons should be deployed so that the work will be completed in the scheduled time,each persons now working 9 hours a day. A. 160
B. 150
C. 24
D. 56
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D
Explanation : -------------------------------------------------------------------Solution 1 (Chain Rule) -------------------------------------------------------------------Persons worked = 104 Number of hours each person worked per day = 8 Number of days they worked = 30 Work completed = 2/5 Remaining days = 56 - 30 = 26 Remaining Work to be completed = 1 - 2/5 = 3/5 Let the total number of persons who do the remaining work = x Number of hours each person needs to be work per day = 9 More days, less persons(indirect proportion) More hours, less persons(indirect proportion) More work, more persons(direct proportion) Hence we can write as
Days
30 : 26
Hours
8:9
Work
3 2 : 5 5
:: x : 104
3 2 × 104 = 26 × 9 × × x 5 5 3 30 × 8 × × 104 30 × 8 × 3 × 104 5 ⇒x= = 2 26 × 9 × 2 26 × 9 × 5 ⇒ 30 × 8 ×
=
=
30 × 8 × 104 26 × 3 × 2
30 × 8 × 4 = 5 × 8 × 4 = 160 3×2
Number of additional persons required = 160 - 104 = 56
-------------------------------------------------------------------Solution 2 (Using Time and Work) --------------------------------------------------------------------
Persons worked = 104 Number of hours each person worked per day = 8 Number of days they worked = 30 Work completed = 2/5
Remaining days = 56 - 30 = 26 Remaining Work to be completed = 1 - 2/5 = 3/5 Let the total number of persons who do the remaining work = x Number of hours each person needs to be work per day = 9
2 ( ) 5 Amount of work 1 person did in 1 day, working 1 hours a day = 104 × 30 × 8 3 ( ) 5 Now, the amount of work x person should do in 1 day, working 1 hours a day = 26 × 9 3 ( ) 5 26 × 9 ⇒x= 2 ( ) 5 104 × 30 × 8
=
=
3 × 104 5 2 26 × 9 × 5
30 × 8 ×
=
30 × 8 × 3 × 104 26 × 9 × 2
30 × 8 × 4 = 5 × 8 × 4 = 160 3×2
Number of additional persons required = 160 - 104 = 56
-------------------------------------------------------------------Solution 3 (Using Time and Work) --------------------------------------------------------------------
=
30 × 8 × 104 26 × 3 × 2
If M 1 men can do W 1 work in D 1 days working H 1 hours per day and M 2 men can do W 2 work in D 2 days working H 2 hours per day (where all men work at the same rate), then M1D 1H 1 M2D 2H 2 = W1 W2
Persons worked (M1) = 104 Number of hours each person worked per day (H1) = 8 Number of days they worked (D1) = 30 Work completed (W1)= 2/5
Remaining days (D2)= 56 - 30 = 26 Remaining Work to be completed (W2)= 1 - 2/5 = 3/5 Let the total number of persons who do the remaining work (M2) = x Number of hours each person needs to be work per day (H2) = 9
M1D 1H 1 M2D 2H 2 = W1 W2 ⇒
⇒
104 × 30 × 8 x × 26 × 9 = 2 3 ( ) ( ) 5 5 104 × 30 × 8 x × 26 × 9 = 2 3
⇒ 52 × 30 × 8 = x × 26 × 3 ⇒ 2 × 30 × 8 = 3x ⇒ x = 2 × 10 × 8 = 160 Number of additional persons required = 160 - 104 = 56
13. x men working x hours per day can do x units of a work in x days. How much work can be completed by y men working y hours per day in y days?
x2 y2 x3 C. y2
A.
y3 B. x2 y2 D. x2
units units
units units
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : -------------------------------------------------------------------Solution 1 (Chain Rule) -------------------------------------------------------------------Let amount of work completed by y men working y hours per in y days = w units More men, more work(direct proportion) More hours, more work(direct proportion) More days, more work(direct proportion) Hence we can write as
Men
x:y
Hours
x:y
Days
x:y
:: x : w
⇒ x 3w = y 3x y 3x y3 ⇒w= 3 = 2 x x
-------------------------------------------------------------------Solution 2 (Using Time and Work) --------------------------------------------------------------------
Amount of work completed by 1 man in 1 day, working 1 hours a day =
x 1 = 2 3 x x
1 y3 Amount of work y men in y days, working y hours a day = y × 2 = 2 x x 3
-------------------------------------------------------------------Solution 3 (Using Time and Work) --------------------------------------------------------------------
If M 1 men can do W 1 work in D 1 days working H 1 hours per day and M 2 men can do W 2 work in D 2 days working H 2 hours per day (where all men work at the same rate), then M1D 1H 1 M2D 2H 2 = W1 W2
M1 = x H1 = x D1 = x W1 = x
M2 = y D2= y H2 = y Let W2= w
M1D 1H 1 M2D 2H 2 = W1 W2 ⇒
x3 y3 = x w
y3 ⇒x = w 2
⇒w=
y3 x2
14. 21 goats eat as much as 15 cows. How many goats each as much as 35 cows?
A. 49
B. 32
C. 36
D. 41
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
15 cows ≡ 21 goats 21 1 cow ≡ goats 15 21 × 35 21 × 7 35 cows ≡ goats ≡ goats ≡ 7 × 7 goats ≡ 49 goats 15 3
15. A flagstaff 17.5 m high casts a shadow of length 40.25 m. What will be the height of a building, which casts a shadow of length 28.75 m under similar conditions? A. 12.5 m
B. 10.5 m
C. 14
D. 12
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : -------------------------------------------------------------------Solution 1 (Chain Rule) -------------------------------------------------------------------Let the required height of the building be x meter More shadow length, More height(direct proportion) Hence we can write as
shadow length
40.25 : 28.75 } :: 17.5 : x
⇒ 40.25 × x = 28.75 × 17.5 ⇒x=
28.75 × 17.5 2875 × 175 2875 × 7 2875 575 = = = = = 12.5 40.25 40250 1610 230 46
-------------------------------------------------------------------Solution 2 --------------------------------------------------------------------
shadow of length 40.25 m ≡ 17.5 m high 17.5 shadow of length 1 m ≡ m high 40.25 28.75 × 17.5 shadow of length 28.75 m ≡ m high = 12.5 m high 40.25
16. Running at the same constant rate, 6 identical machines can produce a total of 270 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes? A. 1800
B. 900
C. 2500
D. 2700
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let required number of bottles be x More machines, more bottles(direct proportion) More minutes, more bottles(direct proportion) Hence we can write as
machines
6 : 10 :: 270 : x
minutes
1:4
⇒ 6 × 1 × x = 10 × 4 × 270 ⇒x=
10 × 4 × 270 10 × 4 × 90 = = 10 × 4 × 45 = 1800 6 2
17. A person works on a project and completes 5/8 of the job in 10 days. At this rate, how many more days will he it take to finish the job? A. 7
B. 6
C. 5
D. 4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B
Explanation : ------------------------------------------------------------------Solution 1 (Chain Rule) -------------------------------------------------------------------Number of days he worked = 10 Work completed = 5/8 Let the required number of days = x Remaining Work to be completed = 1 - 5/8 = 3/8 More work, more days(direct proportion) Hence we can write as
Work
⇒
5 3 : } :: 10 : x 8 8
5 3 × x = × 10 8 8
⇒ 5 × x = 3 × 10 ⇒x=3×2=6
-------------------------------------------------------------------Solution 2 (Using Time and Work) -------------------------------------------------------------------Number of days he worked = 10 Work completed = 5/8
Let the required number of days = x Remaining Work to be completed = 1 - 5/8 = 3/8
5 ( ) 8 Amount of work 1 person did in 1 day = 10 Now, the amount of work 1 person should do in x days =
⇒x=
3 ( ) 8 5 ( ) 8 10
=
3 5 ( ) 10
=
3 8
3 × 10 =6 5
-------------------------------------------------------------------Solution 3 (Using Time and Work) --------------------------------------------------------------------
If M 1 men can do W 1 work in D 1 days working H 1 hours per day and M 2 men can do W 2 work in D 2 days working H 2 hours per day (where all men work at the same rate), then M1D 1H 1 M2D 2H 2 = W1 W2
Here, M1 = M2 H1 = H2
D1 = 10 W1 = 5/8
Let D2 = x W2= 1 - 5/8 = 3/8
Hence, the equation can be written as D1 D2 = W1 W2 ⇒
⇒
10 5 ( ) 8
=
x 3 ( ) 8
10 x = 5 3
⇒2=
x 3
⇒x=2×3=6
18. A fort had provision of food for 150 men for 45 days. After 10 days, 25 men left the fort. Find out the number of days for which the remaining food will last. A. 44
B. 42
C. 40
D. 38
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Given that fort had provision of food for 150 men for 45 days Hence, after 10 days, the remaining food is sufficient for 150 men for 35 days Remaining men after 10 days = 150 - 25 = 125 Assume that after 10 days,the remaining food is sufficient for 125 men for x days More men, Less days (Indirect Proportion)
⇒ Men
150 : 125 } :: x : 35
⇒ 150 × 35 = 125x ⇒ 6 × 35 = 5x ⇒ x = 6 × 7 = 42 ⇒ The remaining food is sufficient for 125 men for 42 days
19. If the price of 357 apples is Rs.1517.25, what will be the approximate price of 49 dozens of such apples? A. Rs. 2500
B. Rs. 2300
C. Rs. 2200
D. Rs. 1400
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : -------------------------------------------------------------------Solution 1 (Chain Rule) -------------------------------------------------------------------Let the required price be x More apples, More price(direct proportion) Hence we can write as
apples
357 : (49 × 12) } :: 1517.25 : x
⇒ 357x = (49 × 12) × 1517.25 ⇒x= =
49 × 12 × 1517.25 7 × 12 × 1517.25 = 357 51
7 × 4 × 1517.25 = 7 × 4 × 89.25 ≈ 2500 17
-------------------------------------------------------------------Solution 2 --------------------------------------------------------------------
price of 357 apples = Rs.1517.25 price of 1 apple = Rs.
1517.25 357
price of 49 dozens apples = Rs. (
49 × 12 × 1517.25 ) ≈ Rs. 2500 357
20. On a scale of a map 0.6 cm represents 6.6km. If the distance between two points on the map is 80.5 cm , what is the the actual distance between these points? A. 885.5 km
B. 860 km
C. 892.5 km
D. 825 km
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : -------------------------------------------------------------------Solution 1 (Chain Rule) -------------------------------------------------------------------Let the required actual distance be x km More scale distance, More actual distance(direct proportion) Hence we can write as
scale distance
.6 : 80.5 } :: 6.6 : x
⇒ .6x = 80.5 × 6.6 ⇒ .1x = 80.5 × 1.1 ⇒ x = 80.5 × 11 = 885.5
-------------------------------------------------------------------Solution 2 --------------------------------------------------------------------
.6 cm in map ≡ actual distance of 6.6 km 1 cm in map ≡
6.6 km .6
80.5 cm in map ≡
80.5 × 6.6 km = 885.5 km .6
21. A rope can make 70 rounds of the circumference of a cylinder whose radius of the base is 14cm. how many times can it go round a cylinder having radius 20 cm? A. 49 rounds
B. 42 rounds
C. 54 rounds
D. 52 rounds
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the required number of rounds be x More radius, less rounds(Indirect proportion) Hence we can write as
radius
14 : 20 } :: x : 70
⇒ 14 × 70 = 20x ⇒ 14 × 7 = 2x ⇒ x = 7 × 7 = 49
22. 8 persons can build a wall 140m long in 42 days. In how many days can 30 persons complete a similar wall 100 m long? A. 12
B. 10
C. 8
D. 6
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
-------------------------------------------------------------------Solution 1 (Chain Rule) -------------------------------------------------------------------More persons, less days(indirect proportion) More length of the wall, more days(direct proportion) Hence we can write as
persons
8 : 30 :: x : 42
Length of the wall
100 : 140
⇒ 8 × 100 × 42 = 30 × 140 × x ⇒x=
8 × 100 × 42 8 × 100 × 14 8 × 100 = = =8 30 × 140 10 × 140 10 × 10
-------------------------------------------------------------------Solution 2 (Using Time and Work) --------------------------------------------------------------------
Work done by 8 persons working 42 days = 140 Work done by 1 person working 42 days = Work done by 1 person working 1 day =
140 8
140 8 × 42
Work done by 30 persons working 1 day =
30 × 140 100 = 8 × 42 8
Assume that 30 persons working x days complete a similar wall 100 m ⇒ Work done by 30 persons working x days =100
Hence x =
100 100 ( ) 8
=8
-------------------------------------------------------------------Solution 3 (Using Time and Work)
--------------------------------------------------------------------
If M 1 men can do W 1 work in D 1 days working H 1 hours per day and M 2 men can do W 2 work in D 2 days working H 2 hours per day (where all men work at the same rate), then M1D 1H 1 M2D 2H 2 = W1 W2
M1 = 8 D1 = 42 W1= 140
M2 = 30 Let D2 = x W2= 100
Here H1 = H2
M1D 1 M2D 2 = W1 W2 ⇒
8 × 42 30 × x = 140 100
⇒
8 × 42 30 × x = 14 10
⇒8×3=3×x ⇒x=8
23. A certain number of persons can finish a piece of work in 100 days. If there were 10 persons less, it would take 10 more days finish the work. How many persons were there originally? A. 90
B. 100
C. 110
D. 120
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : -------------------------------------------------------------------Solution 1 (Chain Rule) -------------------------------------------------------------------Assume that x persons can finish a piece of work in 100 days Also it is given that (x-10) persons can finish a piece of work in 110 days (∵ 100 + 10 = 110) More persons, less days(indirect proportion) Hence we can write as
persons
x : (x − 10) } :: 110 : 100
⇒ 100x = 110(x − 10) ⇒ 100x = 110x − 1100 ⇒ 10x = 1100 ⇒x=
1100 = 110 10
-------------------------------------------------------------------Solution 2 (Using Time and Work) --------------------------------------------------------------------
Assume that x persons can finish the work in 100 days 1 ⇒ Work done by 1 person in 1 day = .......(Equation 1) 100x
Also it is given that (x-10) persons can finish the work in 110 days(∵ 100 + 10 = 110) 1 ⇒ Work done by 1 person in 1 day = .......(Equation 2) 110(x − 10)
But (Equation 1) = (Equation 2) ⇒
1 1 = 100x 110(x − 10)
⇒ 110(x − 10) = 100x ⇒ 110x − 1100 = 100x ⇒ 10x = 1100 x=
1100 = 110 10
-------------------------------------------------------------------Solution 3 (Using Time and Work) --------------------------------------------------------------------
If M 1 men can do W 1 work in D 1 days working H 1 hours per day and M 2 men can do W 2 work in D 2 days working H 2 hours per day (where all men work at the same rate), then M1D 1H 1 M2D 2H 2 = W1 W2
Assume that x persons can finish a piece of work in 100 days Also it is given that (x-10) persons can finish a piece of work in 110 days (∵ 100 + 10 = 110)
M1 = x D1 = 100
M2 = (x-10) D2 = 110
Here H1 = H2 and W1 = W2
Hence M 1 D 1 = M 2 D 2 ⇒ 100x = 110(x − 10) ⇒ 100x = 110x − 1100 ⇒ 10x = 1100 x=
1100 = 110 10
24. 9 examiners can examine a certain number of answer books in 12 days by working 5 hours a day. How many hours in a day should 4 examiners work to examine twice the number of answer books in 30 days? A. 9
B. 10
C. 11
D. 12
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : -------------------------------------------------------------------Solution 1 (Chain Rule) -------------------------------------------------------------------Let required number of hours be x More examiners, less hours(indirect proportion) More days, less hours(indirect proportion) More answer books, more hours(direct proportion)
Hence we can write as
examiners
9:4
days
12 : 30
answer books
2:1
:: x : 5
⇒ 9 × 12 × 2 × 5 = 4 × 30 × 1 × x ⇒x=
9 × 12 × 2 × 5 4 × 30 × 1
=
9×3×2×5 30
=
9×2×5 =9 10
-------------------------------------------------------------------Solution 2 (Using Time and Work) --------------------------------------------------------------------
Given that Work done by 9 examiners in 12 day in working 5 hours a day = 1 ⇒ Work done by 1 examiner in 1 day in 1 hour =
1 ......(Equation 1) 9 × 12 × 5
⇒ Work needs to be done by 4 examiners in 30 days working x hours a day = 2 (∵ twice work to be completed) ⇒ Work needs to be done by 1 examiner in 1 day working x hours a day =
2 ...(Equation 2) 4 × 30
From (Equation 1) and (Equation 2), ( ⇒x= (
2 ) 4 × 30
1 ) 9 × 12 × 5
=
2 × 9 × 12 × 5 4 × 30
=
9 × 12 × 5 9×6×5 = =9 2 × 30 30
--------------------------------------------------------------------
Solution 3 (Using Time and Work) --------------------------------------------------------------------
If M 1 men can do W 1 work in D 1 days working H 1 hours per day and M 2 men can do W 2 work in D 2 days working H 2 hours per day (where all men work at the same rate), then M1D 1H 1 M2D 2H 2 = W1 W2
M1 = 9 D1 = 12 H1 = 5 W1 = 1
M2 = 4 D2 = 30 H2 = x W2 = 2
M1D 1H 1 M2D 2H 2 = W1 W2 ⇒
9 × 12 × 5 4 × 30 × x = 1 2
⇒ 9 × 12 × 5 = 2 × 30 × x ⇒ 3 × 12 × 5 = 2 × 10 × x ⇒ 3 × 6 × 5 = 10 × x ⇒3×3×5=5×x ⇒x=3×3=9
25. 9 engines consume 24 metric tonnes of coal, when each is working 8 hours day. How much coal is required for 8 engines, each running 13 hours a day, if 3 engines of former type consume as much as 4 engines of latter type? A. 20 metric tonnes
B. 22 metric tonnes
C. 24 metric tonnes
D. 26 metric tonnes
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : -------------------------------------------------------------------Solution 1 (Chain Rule) -------------------------------------------------------------------Let required amount of coal be x metric tonnes More engines, more amount of coal (direct proportion) If 3 engines of first type consume 1 unit, then 1 engine will consume 1/3 unit which is its the rate of consumption. If 4 engines of second type consume 1 unit, then 1 engine will consume 1/4 unit which is its the rate of consumption More rate of consumption, more amount of coal (direct proportion) More hours, more amount of coal(direct proportion) Hence we can write as
engines
9:8
rate of consumption
1 1 : 3 4
hours
8 : 13
⇒9×
:: 24 : x
1 1 × 8 × x = 8 × × 13 × 24 3 4
⇒ 3 × 8 × x = 8 × 6 × 13 ⇒ 3 × x = 6 × 13 ⇒ x = 2 × 13 = 26
26. A garrison had provisions for a certain number of days. After 10 days, 1/5 of the men desert and it is found that the provisions will now last just as long as before. How long was that? A. 50 days
B. 30 days
C. 40 days
D. 60 days
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : ------------------------------------------------------------Solution 1 -------------------------------------------------------------
Assume that initially garrison had provisions for x men for y days So, after 10 days, garrison had provisions for x men for (y-10) days
Also, garrison had provisions for
4x x 4x ) men for y days (∵ x − = 5 5 5
More men, Less days (Indirect Proportion)
⇒ Men
x:
⇒ x(y − 10) = ⇒ (y − 10) =
4x } :: y : (y − 10) 5
4xy 5 4y 5
⇒ 5(y − 10) = 4y ⇒ 5y − 50 = 4y ⇒ y = 50
------------------------------------------------------------Solution 2 ------------------------------------------------------------Assume that amount of food eaten by 1 man in 1 day = x, total men = y and the garrison had provisions for z days
Then total quantity of food = xyz amount of food eaten by y men in 10 days = 10xy Remaining food = xyz - 10xy = xy(z - 10) -------(Equation 1)
After 10 days, total men = 4y/5 (∵ y - y/5 = 4y/5) food taken by 4y/5 men in 1 day = 4xy/5 -------(Equation 2)
From Equations 1 and 2, Time taken for 4y/5 men to complete xy(z - 10) food
=
xy(z − 10) (
4xy ) 5
=
5(z − 10) 4
Given that number of days remain the same
⇒
5(z − 10) =z 4
⇒ 5z − 50 = 4z ⇒ z = 50
27. A garrison of 500 persons had provisions for 27 days. After 3 days a reinforcement of 300 persons arrived. For how many more days will the remaining food last now? A. 12 days
B. 16 days
C. 14 days
D. 15 days
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : ------------------------------------------------------------Solution 1 ------------------------------------------------------------Given that fort had provision for 500 persons for 27 days Hence, after 3 days, the remaining food is sufficient for 500 persons for 24 days Remaining persons after 3 days = 500 + 300 = 800
Assume that after 10 days,the remaining food is sufficient for 800 persons for x days More men, Less days (Indirect Proportion)
⇒ Men
500 : 800 } :: x : 24
⇒ 500 × 24 = 800x ⇒ 5 × 24 = 8x ⇒ x = 5 × 3 = 15
------------------------------------------------------------Solution 2 ------------------------------------------------------------Assume that amount of food taken by 1 man in 1 day = x
Given that the garrison had provisions for 500 persons for 27 days
⇒ Total quantity of food = 500 × 27 × x = 13500x Amount of food eaten by 500 persons in 3 days = 500 × 3 × x = 1500x Remaining food = 13500x − 1500x = 12000x........(Equation 1)
After 10 days, total persons = 500 + 300 = 800 Food eaten by 800 persons 1 day = 800x ...................(Equation 2)
From Equations 1 and 2, Time taken for 800 persons to complete 12000x food
=
12000x 120 = = 15 800x 8
28. A hostel had provisions for 250 men for 40 days. If 50 men left the hostel, how long will the food last at the same rate?
A. 48 days
B. 50 days
C. 45 days
D. 60 days
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : ------------------------------------------------------------Solution 1 ------------------------------------------------------------A hostel had provisions for 250 men for 40 days If 50 men leaves the hostel, remaining men = 250 - 50 = 200 We need to find out how long the food will last for these 200 men. Let the required numer of days = x days More men, Less days (Indirect Proportion)
⇒ Men
250 : 200 } :: x : 40
⇒ 250 × 40 = 200x ⇒ 5 × 40 = 4x ⇒ x = 5 × 10 = 50
------------------------------------------------------------Solution 2 ------------------------------------------------------------Assume that amount of food taken by 1 man in 1 day = x
Given that the hostel had provisions for 250 men for 40 days
⇒ Total quantity of food = 250 × 40 × x = 10000x...(Equation 1)
If 50 men leaves the hostel, remaining men = 250 - 50 = 200
Food eaten by 200 men 1 day = 200x ...................(Equation 2)
From Equations 1 and 2, Time taken for 200 men to complete 10000x food
=
10000x 100 = = 50 200x 2
29. in a camp, food was was sufficient for 2000 people for 54 days. After 15 days, more people came and the food last only for 20 more days. How many people came? A. 1900
B. 1800
C. 1940
D. 2000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : ------------------------------------------------------------Solution 1 ------------------------------------------------------------Given that food was sufficient for 2000 people for 54 days Hence, after 15 days, the remaining food was sufficient for 2000 people for 39 days (∵ 54 - 15 = 39) Let x number of people came after 15 days. Then, total number of people after 15 days = (2000 + x) Then, the remaining food was sufficient for (2000 + x) people for 20 days More men, Less days (Indirect Proportion)
⇒ Men
2000 : (2000 + x) } :: 20 : 39
⇒ 2000 × 39 = (2000 + x)20 ⇒ 100 × 39 = (2000 + x) ⇒ 3900 = 2000 + x x = 3900 − 2000 = 1900
------------------------------------------------------------Solution 2 ------------------------------------------------------------Assume that amount of food eaten by 1 person in 1 day = k
Given that food was sufficient for 2000 people for 54 days Hence,Total quantity of food = 2000 × 54 × k = 108000k
Amount of food eaten by 2000 persons in 15 days = 2000 × 15 × k = 30000k
Remaining food = 108000k - 30000k = 78000k ............................(Equation 1)
Let x number of people came after 15 days. Then, total number of people after 15 days = (2000 + x) Food eaten by (2000 + x) persons 1 day = (2000 + x)k ...................(Equation 2)
From Equations 1 and 2, Time taken for (2000 + x) persons to complete 78000k food
=
78000k 78000 = (2000 + x)k (2000 + x)
Given that food lasted only for 20 more days
⇒
78000 = 20 (2000 + x)
⇒ 78000 = 20(2000 + x) ⇒ 3900 = 2000 + x ⇒ x = 3900 − 2000 = 1900
30. If 40 men can make 30 boxes in 20 days, How many more men are needed to make 60 boxes in 25 days? A. 28
B. 24
C. 22
D. 26
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : -------------------------------------------------------------------Solution 1 (Chain Rule) -------------------------------------------------------------------Given that 40 men can make 30 boxes in 20 days Let x more men are needed to make 60 boxes in 25 days Then (40 + x) men can make 60 boxes in 25 days More boxes, more men(direct proportion) More days, less men(indirect proportion) Hence we can write as
boxes
30 : 60 :: 40 : (40 + x)
days
25 : 20
⇒ 30 × 25 × (40 + x) = 60 × 20 × 40 ⇒ 25 × (40 + x) = 2 × 20 × 40 ⇒ 5 × (40 + x) = 2 × 4 × 40 ⇒ (40 + x) = 2 × 4 × 8 = 64 ⇒ x = 64 − 40 = 24
-------------------------------------------------------------------Solution 2 (Using Time and Work) -------------------------------------------------------------------Let x more men are needed to make 60 boxes in 25 days Then (40 + x) men can make 60 boxes in 25 days
Work done by 40 men in 20 days = 30 Work done by 1 man in 1 day =
30 40 × 20
Work done by (40 + x) men in 25 days =
30(40 + x) × 25 ...(Equation 1) 40 × 20
If (40 + x) men can make 60 boxes in 25 days, ⇒ Work done by (40 + x) men in 25 days =60.....................(Equation 2)
Hence, from equation 1 and equation 2,
30(40 + x) × 25 40 × 20
= 60
⇒ 30(40 + x) × 25 = 60 × 40 × 20 ⇒ (40 + x) × 25 = 2 × 40 × 20 ⇒ (40 + x) × 5 = 2 × 8 × 20 ⇒ (40 + x) = 2 × 8 × 4 = 64 ⇒ x = 64 − 40 = 24
-------------------------------------------------------------------Solution 3 (Using Time and Work) --------------------------------------------------------------------
If M 1 men can do W 1 work in D 1 days working H 1 hours per day and M 2 men can do W 2 work in D 2 days working H 2 hours per day (where all men work at the same rate), then M1D 1H 1 M2D 2H 2 = W1 W2
M1 = 40 D1 = 20 W1= 30
Let M2 = x Let D2 = 25 W2= 60
Here H1 = H2
Hence, ⇒
M1D 1 M2D 2 = W1 W2
40 × 20 x × 25 = 30 60
⇒ 2(40 × 20) = x × 25 ⇒ 2(8 × 20) = x × 5 ⇒ x = 2(8 × 4) = 64 Hence, additional men required = 64 - 40 = 24
I mportant Formulas - Chain Rule 1.
Direct P roportion Two quantities are said to be directly proportional, if on the increase or decrease of the one, the other increases or decreases the same extent. Examples i. Cost of the goods is directly proportional to the number of goods. (More goods, More cost) ii. Amount of work done is directly proportional to the number of persons who did the work. (More persons, More Work)
2.
I ndirect P roportion (inverse proportion) Two quantities are said to be indirectly proportional (inversely proportional) if on the increase of the one, the other decreases to the same extent and vice-versa. Examples i. Number of days needed to complete a work is indirectly proportional (inversely proportional) with the number of persons who does the work (More Persons, Less Days needed) ii. The time taken to travel a distance is indirectly proportional (inversely proportional) with the speed in which one is travelling (More Speed, Less Time)
1. An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon? A. 154°
B. 180°
C. 170°
D. 160°
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : We know that Angle traced by hour hand in 12 hrs = 360° From 8 to 2, there are 6 hours
The angle traced by the hour hand in 6 hours = 6 ×
360 = 180° 12
2. A clock is started at noon. By 10 minutes past 5, the hour hand has turned through A. 155°
B. 145°
C. 152°
D. 140°
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : We know that Angle traced by hour hand in 12 hrs = 360°
Time duration from noon to 10 minutes past 5 = 5 hours 10 minutes 10 31 =5 hour = hours 60 6 Hence the angle traced by hour hand from noon to 10 minutes past 5 = 31 360 31 × = × 30 = 31 × 5 = 155° 6 12 6
3. At what time between 7 and 8 o'clock will the hands of a clock be in the same straight line but, not together? A. 5 minutes past 7 C. 5
1 minutes past 7 11
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D
3 miinutes past 7 11 5 D. 5 minutes past 7 11 B. 5
Explanation :
The two hands of a clock will be in the same straight line but not together between H and (H + 1) o' clock at . (5H − 30) 12 minutes past H, when H > 6 11 . (5H + 30) 12 minutes past H, when H Time taken by the minute hand to gain 60 min spaces in a normal clock = 60 12 720 5 × 60 = × 60 = = 65 min 55 11 11 11 In the given watch, hands coincide every 64 minutes. In another words, minute hand
gains 60 min spaces in every 64 minutes for the given watch.
5 5 16 − 64 = 1 = minute 11 11 11 16 1 1 1 360 8 Loss in 24 hours = × × 24 × 60 = × × 24 × 60 = = 32 minute 11 64 11 4 11 11
Loss in 64 min = 65
13. At what time between 9 and 10 o' clock will the hands of a clock be together?
2 min past 9 9 1 C. 48 min past 9 12
A. 45
1 min past 9 11 2 D. 47 min past 9 15 B. 49
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
The two hands of a clock will be together between H and (H+1) o' clock at (
60H ) 11
minutes past H o' clock. Here H = 9. Hands will be together at =
60 × 9 minutes past 9 11
540 1 minutes past 9 = 49 minutes past 9 11 11
--------------------------------------------------------------------------------------Solution 2 --------------------------------------------------------------------------------------Its better to use formula as it can save lots of time in exams. However its better to understand the basics. Please find the method given below to solve the same problem in the traditional way.
At 9 o' clock, the hands are 45 minute spaces apart Hence minute hand needs to gain 45 more minute spaces so that the hands will coincide
each other?
We know that 55 min spaces are gained by minute hand (with respect to hour hand) in 60 min
Hence time taken for gaining 45 minute spaces by minute hand
60 12 540 1 × 45 minute = × 45 minute = minute = 49 minute 55 11 11 11 1 Hence hands will coincide at 49 minute past 9 11
=
14. At what time between 4 and 5 o'clock will the hands of a watch point in opposite directions?
6 minutes past 4 11 6 C. 54 minutes past 4 11
A. 53
7 minutes past 4 11 7 D. 54 minutes past 4 11 B. 53
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
The two hands of a clock will be in the same straight line but not together between H and (H + 1) o' clock at 12 (5H − 30) minutes past H, when H > 6 11
(5H + 30)
12 minutes past H, when H The given watch gains
34 mins in 170 hours 5
Given that the said watch was 2 minutes low at Monday noon. Hence when it gained 2 minutes, the time was correct.
34 mins in 170 hours 5 5 => The given watch gains 2 mins in 170 × × 2 hours 34 = 5 × 5 × 2 hours = 50 hours
The given watch gains
Hence the time was correct after 50 hours from Monday noon = after 2 days 2 hours from Monday noon = 2 pm on Wednesday
18. What is the reflex angle between the hands of a clock at 10.25? A. 195° C. 180°
1 ° 2 1 D. 193 ° 2 B. 197
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Angle between Hands of a clock
When the minute hand is behind the hour hand, the angle between the two hands at M minutes past H'o clock M M )+ = 30 (H − degree 5 2
When the minute hand is ahead of the hour hand, the angle between the two hands at M minutes past H'o clock M M = 30 ( − H) − degree 5 2 Here H = 10 , M = 25 and the minute hand is behind the hour hand. Hence the angle
M M 25 25 )+ )+ = 30 (10 − = 30 (10 − 5) + 12.5 5 2 5 2 = 30 × 5 + 12.5 = 150 + 12.5 = 162.5°
= 30 (H −
But the question is to find out the reflex angle.
Reflex angle = 360 - 162.5 = 197.5 °
--------------------------------------------------------------------------------------Solution 2 --------------------------------------------------------------------------------------Its better to use formula as it can save lots of time in exams. However we should understand the basics for sure. Please find the method given below to solve the same problem in the traditional way.
10 hour 25 minutes = 10
25 5 125 hour = 10 hour = hour 60 12 12
Angle traced by hour hand in 12 hrs = 360°
Angle traced by hour hand in = 10 ×
125 360 125 125 hour = × = 30 × 12 12 12 12
125 = 10 × 31.25 = 312.5° 4
Angle traced by minute hand in 60 min. = 360°.
Angle traced by minute hand in 25 min. =
360 × 25 = 6 × 25 = 150°. 60
Angle between the hands of the clock when the time is 10.25 = 312.5 ° -150 ° =162.5 °.
Reflex angle = 360 - 162.5 = 197.5 °
19. The angle between the minute hand and the hour hand of a clock when the time is 4.20 is A. 10°
B. 5°
C. 0°
D. 1°
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Angle between Hands of a clock
When the minute hand is behind the hour hand, the angle between the two hands at M minutes past H'o clock M M )+ = 30 (H − degree 5 2
When the minute hand is ahead of the hour hand, the angle between the two hands at M minutes past H'o clock M M = 30 ( − H) − degree 5 2 Here H = 4, M = 20 the minute hand is slightly behind the hour hand. Hence the angle
M M 20 20 )+ )+ = 30 (4 − = 30 (4 − 4) + 10 5 2 5 2 = 30 × 0 + 10 = 10°
= 30 (H −
-------------------------------------------------------------------------------------Solution 2 --------------------------------------------------------------------------------------Its better to use formula as it can save lots of time in exams. However we should understand the basics for sure. Please find the method given below to solve the same problem in the traditional way.
4 hour 20 minutes = 4
1 13 hour = hour 3 3
Angle traced by hour hand in 12 hrs = 360°
Angle traced by hour hand in = 10 × 13 = 130°
13 360 13 13 hour = × = 30 × 3 12 3 3
Angle traced by minute hand in 60 min. = 360°.
Angle traced by minute hand in 20 min. =
360 × 20 = 6 × 20 = 120°. 60
Angle between the hands of the clock when the time is 4.20 = 130° - 120° =10°.
20. A clock is set at 5 am. If the clock loses 16 minutes in 24 hours, what will be the true time when the clock indicates 10 pm on 4th day? A. 9.30 pm
B. 10 pm
C. 10.30 pm
D. 11 pm
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Time from 5 am to 10 pm on the 4th day = 3 days 17 hours = 3 × 24 + 17 = 89 hours Given that clock loses 16 minutes in 24 hours ⇒> 23 hour 44 minutes of the given clock = 24 hours in a normal clock ⇒ 23
44 hours of the given clock = 24 hours in a normal clock 60
⇒ 23
11 hours of the given clock = 24 hours in a normal clock 15
⇒
356 hours of the given clock = 24 hours in a normal clock 15
⇒ 89 hours of the given clock = 24 × = 24 ×
15 × 89 hours in a normal clock 356
15 = 6 × 15 = 90 hours 4
So the correct time is 90 hours after 5 am = 3 days 18 hours after 5 am = 11 pm on the 4th day
21. What is the angle between the hour and the minute hand of a clock when the time is 3.25? A. 47 C. 46 Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
1 2 1 D. 47 2 B. 46
Angle between Hands of a clock
When the minute hand is behind the hour hand, the angle between the two hands at M minutes past H'o clock M M )+ = 30 (H − degree 5 2
When the minute hand is ahead of the hour hand, the angle between the two hands at M minutes past H'o clock M M = 30 ( − H) − degree 5 2 Here H = 3, M = 25 and the minute hand is ahead of the hour hand. Hence the angle
M M 25 25 − H) − = 30 ( − 3) − = 30 (5 − 3) − 12.5 5 2 5 2 = 30 × 2 − 12.5 = 60 − 12.5 = 47.5°
= 30 (
--------------------------------------------------------------------------------------Solution 2 --------------------------------------------------------------------------------------Its better to use formula as it can save lots of time in exams. However we should understand the basics for sure. Please find the method given below to solve the same problem in the traditional way.
3 hour 25 minutes = 3
25 5 41 hour = 3 hour = hour 60 12 12
Angle traced by hour hand in 12 hrs = 360°
Angle traced by hour hand in = 10 ×
41 360 41 41 hour = × = 30 × 12 12 12 12
41 = 10 × 10.25 = 102.5° 4
Angle traced by minute hand in 60 min. = 360°.
Angle traced by minute hand in 25 min. =
360 × 25 = 6 × 25 = 150°. 60
Angle between the hands of the clock when the time is 10.25 = 150° - 102.5° = 47.5 °.
22. At what time between 8 and 9 o'clock will the hands of a clock are in the same straight line but not together?
8 minutes past 8 11 10 C. 11 minutes past 8 11
A. 11
8 minutes past 8 11 10 D. 10 minutes past 8 11
B. 10
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
The two hands of a clock will be in the same straight line but not together between H and (H + 1) o' clock at 12 (5H − 30) minutes past H, when H > 6 11
(5H + 30)
12 minutes past H, when H 1/4 Hence,
1 2 4 5 6 7 , , , , , 4 5 7 6 7 8
Similarly,
is not in descending order
1 2 4 1 3 7 , , , , , 4 5 7 6 7 8
is also not in descending order
7/8 = 0.8 (Taken only one digit after the decimal point) 4/7 = 0.5 (Taken only one digit after the decimal point) 3/7 = 0.42 (Taken only two digits after the decimal point) 2/5 = 0.4
1/4 = 0.2 (Taken only one digit after the decimal point) 1/6 = 0.1 (Taken only one digit after the decimal point)
Hence,
32.
7 4 3 2 1 1 , , , , , is in descending order 8 7 7 5 4 6
4 × 0.30 − 3 × 0.500 0.003
A. 10
B. 10
C. 100
D. 100
= ?
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
4 × 0.30 − 3 × 0.500 0.003
=
1.2 − 1.5 −0.3 −300 = = = −100 0.003 0.003 3
Recurring Decimals (or Repeating Decimals) 1.
Recurring Decimal
If a figure or a set of figures is repeated continuously in a decimal fraction, it is called a recurring decimal (also known as repeating decimal and circulating decimal) A repeating decimal is denoted by placing a horizontal line above the repeated numerals. Another way to denote a recurring decimal is placing dots over the first and last digits Examples
2.
(a)
1 = 0.333... = 0.¯ 3 = 0.˙3 3
(b)
1 ¯¯¯¯¯¯¯¯¯ = 0.˙1 4285˙7 = 0.142857142857142857... = 0.¯142857 7
(c)
7 = 0.58333... = 0.58¯ 3 = 0.58˙3 12
Type of Recurring Decimals Recurring Decimals can be classified into two categories Pure Recurring Decimals and Mixed Recurring Decimals i.
Pure Recurring Decimals
Pure Recurring Decimal is a decimal fraction in which all the figures after the decimal point are repeated. Examples
ii.
(a)
1 ¯ = 0.3 ˙ = 0.333... = 0.3 3
(b)
4 ¯¯¯ = 0.˙3 ˙6 = 0.36363636... = 0.¯36 11
Mixed Recurring Decimals
Pure Recurring Decimal is a decimal fraction in which some figures are not repeated and some figures are repeated. Examples (a)
7 = 0.58333... = 0.58¯ 3 = 0.58˙3 12
(b)
87 ¯¯¯ = 1.3˙1 ˙8 = 1.3181818... = 1.3¯18 66
1. Two numbers are in the ratio 2 : 3. If their L.C.M. is 48. what is sum of the numbers? A. 28
B. 40
C. 64
D. 42
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the numbers be 2x and 3x LCM of 2x and 3x = 6x
(∵ LCM of 2 and 3 is 6. Hence LCM
of 2x and 3x is 6x) Given that LCM of 2x and 3x is 48 => 6x = 48 => x = 48∕6 = 8 Sum of the numbers = 2x + 3x = 5x = 5 × 8 = 40
2. What is the greatest number of four digits which is divisible by 15, 25, 40 and 75 ? A. 9800
B. 9600
C. 9400
D. 9200
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Greatest number of four digits = 9999 LCM of 15, 25, 40 and 75 = 600 9999 ÷ 600 = 16, remainder = 399 Hence, greatest number of four digits which is divisible by 15, 25, 40 and 75 = 9999 - 399 = 9600
3. Three numbers are in the ratio of 2 : 3 : 4 and their L.C.M. is 240. Their H.C.F. is: A. 40
B. 30
C. 20
D. 10
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the numbers be 2x, 3x and 4x LCM of 2x, 3x and 4x = 12x => 12x = 240 => x = 240∕12 = 20 H.C.F of 2x, 3x and 4x = x = 20
4. What is the lowest common multiple of 12, 36 and 20? A. 160
B. 220
C. 120
D. 180
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : 2 12, 36, 20 2 6, 18, 10 3 3, 9, 5 1, 3, 5
LCM = 2 × 2 × 3 × 1 × 3 × 5 = 180
5. What is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder? A. 1108
B. 1683
C. 2007
D. 3363
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : -----------------------------------------------------------------------
--------Solution 1 ------------------------------------------------------------------------------LCM of 5, 6, 7 and 8 = 840 Hence the number can be written in the form (840k + 3) which is divisible by 9 If k = 1, number = (840 × 1) + 3 = 843 which is not divisible by 9 If k = 2, number = (840 × 2) + 3 = 1683 which is divisible by 9 Hence 1683 is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder ------------------------------------------------------------------------------Solution 2 - Hit and Trial Method ------------------------------------------------------------------------------Just see which of the given choices satisfy the given condtions Take 3363. This is not even divisible by 9. Hence this is not the answer Take 1108. This is not even divisible by 9. Hence this is not the answer Take 2007. This is divisible by 9. 2007 ÷ 5 = 401, remainder = 2 . Hence this is not the answer Take 1683. This is divisible by 9. 1683 ÷ 5 = 336, remainder = 3 1683 ÷ 6 = 280, remainder = 3 1683 ÷ 7 = 240, remainder = 3 1683 ÷ 8 = 210, remainder = 3 Hence 1683 is the answer
6. The H.C.F. of two numbers is 5 and their L.C.M. is 150. If one of the numbers is 25, then the other is:
A. 30
B. 28
C. 24
D. 20
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Product of two numbers = Product of their HCF and LCM.
Let one number = x ⇒ 25 × x = 5 × 150 ⇒x=
5 × 150 = 30 25
7. 504 can be expressed as a product of primes as A. 2 × 2 × 3 × 3 × 7 × 7
B. 2 × 3 × 3 × 3 × 7 × 7
C. 2 × 3 × 3 × 3 × 3 × 7
D. 2 × 2 × 2 × 3 × 3 × 7
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : It is clear that 504 = 2 × 2 × 2 × 3 × 3 × 7
8. Which of the following integers has the most number of divisors? A. 101
B. 99
C. 182
D. 176
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : 99 = 1 × 3 × 3 × 11 => Divisors of 99 are 1, 3, 11, 9, 33 and 99
101 = 1 × 101 => Divisors of 101 are 1 and 101 182 = 1 × 2 × 7 × 13 => Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182 176 = 1 × 2 × 2 × 2 × 2 × 11 => Divisors of 176 are 1, 2, 11, 4, 22, 8, 44, 16, 88, 176 Hence 176 has most number of divisors
9. The least number which should be added to 28523 so that the sum is exactly divisible by 3, 5, 7 and 8 is A. 41
B. 42
C. 32
D. 37
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : LCM of 3, 5, 7 and 8 = 840 28523 ÷ 840 = 33 remainder = 803 Hence the least number which should be added = 840 - 803 = 37
10. What is the least number which when doubled will be exactly divisible by 12, 14, 18 and 22 ? A. 1286
B. 1436
C. 1216
D. 1386
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : LCM of 12, 14, 18 and 22 = 2772 Hence the least number which will be exactly divisible by 12, 14, 18 and 22 = 2772 2772 ÷ 2 = 1386
=> 1386 is the number which when doubled, we get 2772 Hence, 1386 is the least number which when doubled will be exactly divisible by 12, 14, 18 and 22 ?
11. What is the greatest possible length which can be used to measure exactly the lengths 8 m, 4 m 20 cm and 12 m 20 cm? A. 10 cm
B. 30 cm
C. 25 cm
D. 20 cm
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Required length = HCF of 800 cm, 420 cm, 1220 cm = 20 cm
12. Which of the following fraction is the largest ?
11 12 21 C. 40 A.
41 50 5 D. 6 B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : ----------------------------------------------------------------------Solution 1 -----------------------------------------------------------------------
LCM of 6, 40, 50 and 12 = 600 5 500 = 6 600 21 315 = 40 600 41 492 = 50 600 11 550 = 12 600 Clearly
550 11 = is the largest among these fractions 600 12
----------------------------------------------------------------------Solution 2 -----------------------------------------------------------------------
LCM of 6, 40, 50 and 12 = 600 5 = .83 6 21 = .52 40 41 = .82 50 11 = .92 12 Clearly .92 =
11 is the largest among these 12
13. The product of two 2 digit numbers is 2028 and their HCF is 13. What are the numbers ? A. 26, 78
B. 39, 52
C. 13, 156
D. 36, 68
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the numbers be 13x and 13y
(∵ HCF of the numbers =
13) 13x × 13y = 2028 => xy = 12 co-primes with product 12 are (1, 12) and (3, 4) need to take only
(∵ we
co-primes with product 12. If we take two numbers with product 12, but not co-prime, the HCF will not remain as 13) (Reference : Co-prime Numbers) Hence the numbers with HCF 13 and product 2028 = (13 × 1, 13 × 12) and (13 × 3, 13 × 4) = (13, 156) and (39, 52) Given that the numbers are 2 digit numbers Hence numbers are 39 and 52
14. The product of two numbers is 2028 and their HCF is 13. What are the number of such pairs? A. 4
B. 3
C. 2
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the numbers be 13x and 13y 13)
(∵ HCF of the numbers =
13x × 13y = 2028 => xy = 12 co-primes with product 12 are (1, 12) and (3, 4) need to take only
(∵ we
co-primes with product 12. If we take two numbers with product 12, but not co-prime, the HCF will not remain as 13)
(Reference : Co-prime Numbers) Hence the numbers with HCF 13 and product 2028 = (13 × 1, 13 × 12) and (13 × 3, 13 × 4) = (13, 156) and (39, 52) So, there are 2 pairs of numbers with HCF 13 and product 2028
15. N is the greatest number which divides 1305, 4665 and 6905 and gives the same remainder in each case. What is the sum of the digits in N? A. 4
B. 3
C. 6
D. 5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
If the remainder is same in each case and remainder is not given, HCF of the differences of the numbers is the required greatest number 6905 - 1305 = 5600
6905 - 4665 = 2240
4665 - 1305 = 3360
Hence, the greatest number which divides 1305, 4665 and 6905 and gives the same remainder, N
= HCF of 5600, 2240, 3360
= 1120
Sum of digits in N
= Sum of digits in 1120
= 1 + 1 + 2 + 0
= 4
16. A boy divided the numbers 7654, 8506 and 9997 by a certain largest number and he gets same remainder in each case. What is the common remainder? A. 156
B. 199
C. 211
D. 231
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
If the remainder is same in each case and remainder is not given, HCF of the differences of the numbers is the required largest number 9997 - 7654 = 2343
9997 - 8506 = 1491
8506 - 7654 = 852
Hence, the greatest number which divides 7654, 8506 and 9997 and leaves same remainder
= HCF of 2343, 1491, 852
= 213
Now we need to find out the common remainder.
Take any of the given numbers from 7654, 8506 and 9997, say 7654
7654 ÷ 213 = 35, remainder = 199
17. Find the greatest common divisor of 24 and 16 A. 6
B. 2
C. 4
D. 8
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : 16) 24 (1 16 8) 16 (2 16 0 Hence, greatest common divisor of 24 and 16 = 8
18. A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ? A. 36 minutes 22 seconds
B. 46 minutes 22 seconds
C. 36 minutes 12 seconds
D. 46 minutes 12 seconds
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : LCM of 252, 308 and 198 = 2772 Hence they all will be again at the starting point after 2772 seconds or 46 minutes 12 seconds
19. The ratio of two numbers is 4 : 5. If the HCF of these numbers is 6, what is their LCM? A. 30
B. 60
C. 90
D. 120
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option D Explanation : Let the numbers be 4k and 5k HCF of 4 and 5 = 1 Hence HCF of 4k and 5k = k Given that HCF of 4k and 5k = 6 => k = 6 Hence the numbers are (4 × 6) and (5 × 6) = 24 and 30 LCM of 24 and 30 = 120
20. What is the HCF of 2.04, 0.24 and 0.8 ? A. 1
B. 2
C. 0.02
D. 0.04
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Reference : How to calculate LCM and HCF for Decimals Step 1 : Make the same number of decimal places in all the given numbers by suffixing zero(s) in required numbers as needed. => 2.04, 0.24 and 0.80 Step 2 : Now find the HCF of these numbers without decimal. =>HCF of 204, 24 and 80 = 4 Step 3 : Put the decimal point in the result obtained in step 2 leaving as many digits on its right as there are in each of the numbers. i.e., here, we need to put decimal point in the result obtained in step 2 leaving two digits on its right.
=> HCF of 2.04, 0.24 and 0.8 = 0.04
21. If HCF of two numbers is 11 and the product of these numbers is 363, what is the the greater number? A. 9
B. 22
C. 33
D. 11
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the numbers be 11a and 11b 11a × 11b = 363 => ab = 3 co-primes with product 3 are (1, 3) (Reference : Co-prime Numbers) Hence the numbers with HCF 11 and product 363 = (11 × 1, 11 × 3) = (11, 33) Hence numbers are 11 and 33 The greater number = 33
22. What is the greatest number which on dividing 1223 and 2351 leaves remainders 90 and 85 respectively? A. 1133
B. 127
C. 42
D. 1100
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Required number = HCF of (1223 - 90) and (2351 - 85)
= HCF of 1133 and 2266 = 1133
23. What is the least multiple of 7 which leaves a remainder of 4 when divided by 6, 9, 15 and 18 ? A. 364
B. 350
C. 343
D. 371
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : LCM of 6, 9, 15 and 18 = 90 Required Number = (90k + 4) which is a multiple of 7 Put k = 1. We get number as (90 × 1) + 4 = 94. But this is not a multiple of 7 Put k = 2. We get number as (90 × 2) + 4 = 184. But this is not a multiple of 7 Put k = 3. We get number as (90 × 3) + 4 = 274. But this is not a multiple of 7 Put k = 4. We get number as (90 × 4) + 4 = 364. This is a multiple of 7 Hence 364 is the answer.
24. Three numbers which are co-prime to each other are such that the product of the first two is 119 and that of the last two is 391. What is the sum of the three numbers? A. 47
B. 43
C. 53
D. 51
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Since the numbers are co-prime, their HCF = 1 Product of first two numbers = 119
Product of last two numbers = 391 The middle number is common in both of these products. Hence if we take HCF of 119 and 391, we get the common middle number HCF of 119 and 391 = 17 => Middle Number = 17 First Number = 119∕17 = 7 Last Number = 391∕17 = 23 Sum of the three numbers = 7 + 17 + 23 = 47
25. Reduce
7 13 13 C. 14 A.
4329 to its lowest terms 4662 13 B. 17 7 D. 12
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : We need to find out HCF of 4329 and 4662 4329) 4662 (1 4329 333) 4329 (13 4329 0 Hence, HCF of 4329 and 4662 = 333
4329 ÷ 333 = 13
4662 ÷ 333 = 14
Hence
4329 13 = 4662 14
26. What is the greatest number which divides 24, 28 and 34 and leaves the same remainder in each case? A. 1
B. 2
C. 3
D. 4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
If the remainder is same in each case and remainder is not given, HCF of the differences of the numbers is the required greatest number 34 - 24 = 10
34 - 28 = 6
28 - 24 = 4
Hence, the greatest number which divides 24, 28 and 34 and gives the same remainder
= HCF of 10, 6, 4
= 2
27. Six bells start ringing together and ring at intervals of 4, 8, 10, 12, 15 and 20 seconds respectively. how many times will they ring together in 60 minutes ? A. 31
B. 15
C. 16
D. 30
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : LCM of 4, 8, 10, 12, 15 and 20 = 120 120 seconds = 2 minutes
Hence all the six bells will ring together in every 2 minutes Hence, number of times they will ring together in 60 minutes
=1+
60 = 31 2
28. What is the least number which when divided by 8, 12, 15 and 20 leaves in each case a remainder of 5 ? A. 125
B. 117
C. 132
D. 112
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : LCM of 8, 12, 15 and 20 = 120 Required Number = 120 + 5 = 125
29. The HCF of two numbers is 23 and the other two factors of their LCM are 13 and 14. What is the largest number? A. 312
B. 282
C. 299
D. 322
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
The HCF of a group of numbers will be always a factor of their LCM HCF is the product of all common prime factors using the least power of each common prime factor. LCM is the product of highest powers of all prime factors
HCF of the two numbers = 23 => Highest Common Factor in the numbers = 23
Since HCF will be always a factor of LCM, 23 is a factor of the LCM. Other two factors in the LCM are 13 and 14.
Hence factors of the LCM are 23, 13, 14
So, numbers can be taken as (23 × 13) and (23 × 14)
= 299 and 322
Hence, largest number = 322
[A more detailed explanation ...
we can take the numbers as (23 × 13) and (23 × 14) because of the following reasons
HCF is given as 23.
The HCF of a group of numbers will be always a factor of their LCM.
Hence, 23 is a factor of the LCM
Given that other two factors of the LCM are 13 and 14.
Hence factors of the LCM are 23, 13, 14
Now assume that we take the numbers are (23 × 13) and (23 × 14).
If we write the numbers as the product of prime factors,
first number = (23 × 13)
second numbers = (23 × 14) = (23 × 2 × 7)
HCF = product of all common prime factors using the least power of each common prime factor
= 23
LCM is the product of highest powers of all prime factors
= 23 × 13 × 2 × 7 = 23 × 13 × 14
Clearly we get HCF as 23 and the factors in the LCM as 13, 14 and 23.
Hence every conditions are satisfied.
30. What is the smallest number which when diminished by 12, is divisible 8, 12, 22 and 24? A. 276
B. 264
C. 272
D. 268
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Required Number = (LCM of 8, 12, 22 and 24) + 12 = 264 + 12 = 276
1 2 1 , and ? 3 3 4 1 B. 3 1 D. 12
31. What is the HCF of
2 3 1 C. 4 A.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
HCF for fractions =
HCF of Numerators LCM of Denominators
Read More ...
HCF of
HCF (1, 2, 1) 1 2 1 1 , and = = 3 3 4 12 LCM (3, 3, 4)
2 5 4 , and ? 3 6 9 3 B. 20 20 D. 3
32. What is the LCM of
3 10 10 C. 3 A.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
LCM for fractions =
LCM of Numerators HCF of Denominators
Read More ...
LCM of
LCM (2, 5, 4) 2 5 4 20 , and = = 3 6 9 3 HCF (3, 6, 9)
I mportant Concepts and Formulas - Numbers (Directly take me to HCF and LCM section) I.
N umb er Sets and P rop erties of N umbers a.
Counting Num bers (Natural num bers) : 1, 2, 3 ...
b.
W hole Num bers : 0, 1, 2, 3 ...
c.
I ntegers : -3, -2, -1, 0, 1, 2, 3 ...
d.
Rational Num bers Rational numbers can be expressed as Examples:
11 4 −8 , , 0 , 2 2 11
a where a and b are integers and b ≠ 0 b
etc.
All integers, fractions and terminating or recurring decimals are rational numbers.
e.
I rrational Num bers Any number which is not a rational number is an irrational number. In other words, an irrational number is a number which cannot be expressed as
a where a and b are integers. b
For instance, numbers whose decimals do not terminate and do not repeat cannot be written as a fraction and hence they are irrational numbers.
Example : π , √
2 , (3 + √ 5 ) , 4√ 3 (meaning 4 × √ 3 ), √3 6 etc
Please note that the value of π = 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59230 78164 06286 20899 86280 34825 34211 70679... We cannot π as a simple fraction (The fraction 22/7 = 3.14.... is just an approximate value of π ) f.
Real Num bers Real numbers include counting numbers, whole numbers, integers, rational numbers and irrational numbers.
g.
Surds Let a be any rational number and n be any positive integer such that √n √n a is a surd. Example : √
a is irrational. Then
−− 3 , √6 10 , 4√ 3 etc
Please note that numbers like √ numbers
−− 9 , √3 27 etc are not surds because they are not irrational
Every surd is an irrational number. But every irrational number is not a surd. (eg : π , e etc are not surds though they are irrational numbers.)
II.
A dd ition, Sub traction and Multiplication Rules for Even and Odd N umb ers
Addition Rules for Even and Odd Num bers 1. The sum of any number of even numbers is always even 2. The sum of even number of odd numbers is always even 3. The sum of odd number of odd numbers is always odd
Subtraction Rules for Even and Odd Num bers 1. The difference of two even numbers is always even 2. The difference of two odd numbers is always even
Multiplication Rules for Even and Odd Num bers 1. The product of even numbers is always even 2. The product of odd numbers is always odd 3. If there is at least one even number multiplied by any number of odd numbers, the product is always even III.
Divisib ility a.
Divisible By One whole number is divisible by another if the remainder we get after the division is zero. Examples i.
36 is divisible by 4 because 36 ÷ 4 = 9 with a remainder of 0.
ii.
36 is divisible by 6 because 36 ÷ 6 = 6 with a remainder of 0.
iii.
b.
36 is not divisible by 5 because 36 ÷ 5 = 7 with a remainder of 1.
Divisibility Rules
By using divisibility rules we can easily find out whether a given number is divisible by another number without actually performing the division. This helps to save time especially when working with numbers.
Divisibility Rule
Description
Ex am ples
Example1: Check if 64 is divisible by 2.
Divisibility by 2
A number is divisible by 2 if the last digit is even. i.e., if the last digit is 0 or 2 or 4 or 6 or 8
The last digit of 64 is 4 (even). Hence 64 is divisible by 2 Example2: Check if 69 is divisible by 2. The last digit of 69 is 9 (not even). Hence 69 is not divisible by 2
A number is divisible by 3 if the sum of the digits is divisible by 3 Divisibility
Example1: Check if 387 is divisible by 3. 3 + 8 + 7 = 18. 18 is divisible by 3. Hence 387 is divisible by 3
by 3
(Please note that we can apply this rule to
Example2: Check if 421 is divisible by 3.
the answer again and
4 + 2 + 1 = 7. 7 is not divisible by 3. Hence 421 is not divisible by 3
again if we need)
Example1: Check if 416 is divisible by 4.
Divisibility by 4
A number is divisible by 4 if the number formed by the last two digits is divisible by 4.
Number formed by the last two digits = 16. 16 is divisible by 4. Hence 416 is divisible by 4 Example2: Check if 481 is divisible by 4. Number formed by the last two digits = 81. 81 is not divisible by 4. Hence 481 is not divisible by 4 Example1: Check if 305 is divisible by 5. Last digit is 5. Hence 305 is divisible by 5.
Divisibility by 5
A number is divisible by 5 if the last digit is either 0 or 5.
Example2: Check if 420 is divisible by 5. Last digit is 0. Hence 420 is divisible by 5. Example3: Check if 312 is divisible by 5. Last digit is 2. Hence 312 is not divisible by 5. Example1: Check if 546 is divisible by 6. 546 is divisible by 2. 546 is also divisible by 3. (Check the divisibility rule of 2 and 3 to find out this) Hence 546 is divisible by 6 Example2: Check if 633 is divisible by 6. 633 is not divisible by 2 though 633 is divisible by 3. (Check the divisibility rule of 2 and 3 to find out this)
Divisibility by 6
A number is divisible by 6 if it is divisible by both 2 and 3.
Hence 633 is not divisible by 6 Example3: Check if 635 is divisible by 6. 635 is not divisible by 2. 635 is also not divisible by 3. (Check thedivisibility rule of 2 and 3 to find out this) Hence 635 is not divisible by 6 Example4: Check if 428 is divisible by 6. 428 is divisible by 2 but 428 is not divisible by 3.(Check the divisibility rule of 2 and 3 to find out this) Hence 428 is not divisible by 6 Example1: Check if 349 is divisible by 7.
Divisibility by 7
Given number = 349 34 - (9 × 2) = 34 - 18 = 16 16 is not divisible by 7. Hence 349 is not divisible by 7
To find out if a number is divisible by 7, double the last digit and subtact it from the number formed by the remaining digits.
Example2: Check if 364 is divisible by 7.
Repeat this process until we get at a smaller number whose divisibility we know.
Given number = 364 36 - (4 × 2) = 36 - 8 = 28 28 is divisible by 7. Hence 364 is also divisible by 7
If this smaller number is 0 or divisible by 7, the original number is also divisible by 7.
Example3: Check if 3374 is divisible by 7. Given number = 3374 337 - (4 × 2) = 337 - 8 = 329 32 - (9 × 2) = 32 - 18 = 14 14 is divisible by 7. Hence 329 is also divisible by 7. Hence 3374 is also divisible by 7. Example1: Check if 7624 is divisible by 8. The number formed by the last three digits of 7624 = 624. 624 is divisible by 8. Hence 7624 is
also divisible by 8.
Divisibility by 8
A number is divisible by 8 if the number formed by the last three digits is divisible by 8.
Example2: Check if 129437464 is divisible by 8. The number formed by the last three digits of 129437464 = 464. 464 is divisible by 8. Hence 129437464 is also divisible by 8. Example3: Check if 737460 is divisible by 8. The number formed by the last three digits of 737460 = 460. 460 is not divisible by 8. Hence 737460 is also not divisible by 8. Example1: Check if 367821 is divisible by 9. 3 + 6 + 7 + 8 + 2 + 1 = 27 27 is divisible by 9. Hence 367821 is also divisible by 9.
Divisibility by 9
A number is divisible by 9 if the sum of its digits is divisible by 9. (Please note that we can apply this rule to the answer again and again if we need)
Example2: Check if 47128 is divisible by 9. 4 + 7 + 1 + 2 + 8 = 22 22 is not divisible by 9. Hence 47128 is not divisible by 9. Example3: Check if 4975291989 is divisible by 9. 4 + 9+ 7 + 5 + 2 + 9 + 1 + 9 + 8 + 9= 63 Since 63 is big, we can use the same method to see if it is divisible by 9. 6 + 3 = 9 9 is divisible by 9. Hence 63 is also divisible by 9. Hence 4975291989 is also divisible by 9. Example1: Check if 2570 is divisible by 10.
Divisibility by 10
A number is divisible by 10 if the last digit is 0.
Last digit is 0. Hence 2570 is divisible by 10. Example2: Check if 5462 is divisible by 10. Last digit is not 0. Hence 5462 is not
divisible by 10 Example1: Check if 85136 is divisible by 11. 8 + 1 + 6 = 15 5 + 3 = 8 15 - 8 = 7 7 is not divisible by 11. Hence 85136 is not divisible by 11. To find out if a number is divisible by 11, find the sum of the odd numbered digits and the sum of the even numbered digits. Divisibility by 11
Now substract the lower number obtained from the bigger number obtained. If the number we get is 0 or divisible by 11, the original number is also divisible by 11.
Example2: Check if 2737152 is divisible by 11. 2 + 3 + 1 + 2 = 8 7 + 7 + 5 = 19 19 - 8 = 11 11 is divisible by 11. Hence 2737152 is also divisible by 11. Example3: Check if 957 is divisible by 11. 9 + 7 = 16 5 = 5 16 - 5 = 11 11 is divisible by 11. Hence 957 is also divisible by 11. Example4: Check if 9548 is divisible by 11. 9 + 4 = 13 5 + 8 = 13 13 - 13 = 0 We got the difference as 0. Hence 9548 is divisible by 11. Example1: Check if 720 is divisible by 12. 720 is divisible by 3 and 720 is also divisible by 4. (Check the divisibility rules of 3 and 4 to find out this) Hence 720 is also divisible by 12 Example2: Check if 916 is divisible by 12. 916 is not divisible by 3 , though 916 is divisible by 4.(Check thedivisibility rules of 3 and 4 to find out this)
A number is divisible by 12
Hence 916 is not divisible by 12
Divisibility by 12
if the number is divisible by both 3 and 4
Example3: Check if 921 is divisible by 12. 921 is divisible by 3. But 921 is not divisible by 4.(Check the divisibility rules of 3 and 4 to find out this) Hence 921 is not divisible by 12 Example4: Check if 827 is divisible by 12. 827 is not divisible by 3. 827 is also not divisible by 4.(Check thedivisibility rules of 3 and 4 to find out this) Hence 827 is not divisible by 12 Example1: Check if 349 is divisible by 13. Given number = 349 34 + (9 × 4) = 34 + 36 = 70 70 is not divisible by 13. Hence 349 is not divisible by 349
To find out if a number is divisible by 13, multiply the last digit by 4 and add it to the number formed by the remaining digits. Divisibility by 13
Repeat this process until we get at a smaller number whose divisibility we know. If this smaller number is divisible by 13, the original number is also divisible by 13.
Example2: Check if 572 is divisible by 13. Given number = 572 57 + (2 × 4) = 57 + 8 = 65 65 is divisible by 13. Hence 572 is also divisible by 13 Example3: Check if 68172 is divisible by 13. Given number = 68172 6817 + (2 × 4) = 6817 + 8 = 6825 682 + (5 × 4) = 682 + 20 = 702 70 + (2 × 4) = 70 + 8 = 78 78 is divisible by 13. Hence 68172 is also divisible by 13. Example4: Check if 651 is divisible by 13. Given number = 651 65 + (1 × 4) = 65 + 4 = 69 69 is not divisible by 13. Hence 651 is not divisible by 13
Example1: Check if 238 is divisible by 14 238 is divisible by 2 . 238 is also divisible by 7. (Please check thedivisibility rule of 2 and 7 to find out this) Hence 238 is also divisible by 14 Example2: Check if 336 is divisible by 14 336 is divisible by 2 . 336 is also divisible by 7. (Please check thedivisibility rule of 2 and 7 to find out this) Hence 336 is also divisible by 14
Divisibility by 14
A number is divisible by 14 if it is divisible by both 2 and 7.
Example3: Check if 342 is divisible by 14. 342 is divisible by 2 , but 342 is not divisible by 7.(Please check thedivisibility rule of 2 and 7 to find out this) Hence 342 is not divisible by 12 Example4: Check if 175 is divisible by 14. 175 is not divisible by 2 , though it is divisible by 7.(Please check thedivisibility rule of 2 and 7 to find out this) Hence 175 is not divisible by 14 Example5: Check if 337 is divisible by 14. 337 is not divisible by 2 and also by 7 (Please check the divisibility rule of 2and 7 to find out this) Hence 337 is not divisible by 14 Example1: Check if 435 is divisible by 15 435 is divisible by 3 . 435 is also divisible by 5. (Please check thedivisibility rule of 3 and 5 to find out this) Hence 435 is also divisible by 15 Example2: Check if 555 is divisible
by 15 555 is divisible by 3 . 555 is also divisible by 5. (Please check thedivisibility rule of 3 and 5 to find out this) Hence 555 is also divisible by 15
Divisibility by 15
A number is divisible by 15 If it is divisible by both 3 and 5.
Example3: Check if 483 is divisible by 15. 483 is divisible by 3 , but 483 is not divisible by 5. (Please check thedivisibility rule of 3 and 5 to find out this) Hence 483 is not divisible by 15 Example4: Check if 485 is divisible by 15. 485 is not divisible by 3 , though it is divisible by 5. (Please check thedivisibility rule of 3 and 5 to find out this) Hence 485 is not divisible by 15 Example5: Check if 487 is divisible by 15. 487 is not divisible by 3 . It is also not divisible by 5 (Please check thedivisibility rule of 3 and 5 to find out this) Hence 487 is not divisible by 15 Example1: Check if 5696512 is divisible by 16. The number formed by the last four digits of 5696512 = 6512 6512 is divisible by 16. Hence 5696512 is also divisible by 16.
Divisibility by 16
A number is divisible by 16 if the number formed by the last four digits is divisible by 16.
Example2: Check if 3326976 is divisible by 16. The number formed by the last four digits of 3326976 = 6976 6976 is divisible by 16. Hence 3326976 is also divisible by 16. Example3: Check if 732374360 is divisible by 16.
The number formed by the last three digits of 732374360 = 4360 4360 is not divisible by 16. Hence 732374360 is also not divisible by 16. To find out if a number is divisible by 17, multiply the last digit by 5 and subtract it from the number formed by the remaining digits. Divisibility by 17
Repeat this process until you arrive at a smaller number whose divisibility you know. If this smaller number is divisible by 17, the original number is also divisible by 17.
Example1: Check if 500327 is divisible by 17. Given Number = 500327 50032 - (7 × 5 )= 50032 - 35 = 49997 4999 - (7 × 5 ) = 4999 - 35 = 4964 496 - (4 × 5 ) = 496 - 20 = 476 47 - (6 × 5 ) = 47 - 30 = 17 17 is divisible by 17. Hence 500327 is also divisible by 17 Example2: Check if 521461 is divisible by 17. Given Number = 521461 52146 - (1 × 5 )= 52146 -5 = 52141 5214 - (1 × 5 ) = 5214 - 5 = 5209 520 - (9 × 5 ) = 520 - 45 = 475 47 - (5 × 5 ) = 47 - 25 = 22 22 is not divisible by 17. Hence 521461 is not divisible by 17 Example1: Check if 31104 is divisible by 18. 31104 is divisible by 2. 31104 is also divisible by 9. (Please check thedivisibility rule of 2 and 9 to find out this) Hence 31104 is divisible by 18 Example2: Check if 1170 is divisible by 18.
Divisibility by 18
A number is divisible by 18 if it is divisible by both 2 and 9.
1170 is divisible by 2. 1170 is also divisible by 9. (Please check thedivisibility rule of 2 and 9 to find out this) Hence 1170 is divisible by 18 Example3: Check if 1182 is divisible by 18. 1182 is divisible by 2 , but 1182 is not divisible by 9. (Please check thedivisibility rule of 2 and 9 to find out this) Hence 1182 is not divisible by 18
Example4: Check if 1287 is divisible by 18. 1287 is not divisible by 2 though it is divisible by 9. (Please check thedivisibility rule of 2 and 9 to find out this) Hence 1287 is not divisible by 18 To find out if a number is divisible by 19, multiply the last digit by 2 and add it to the number formed by the remaining digits. Divisibility by 19
Repeat this process until you arrive at a smaller number whose divisibility you know. If this smaller number is divisible by 19, the original number is also divisible by 19.
Example1: Check if 74689 is divisible by 19. Given Number = 74689 7468 + (9 × 2 )= 7468 + 18 = 7486 748 + (6 × 2 ) = 748 + 12 = 760 76 + (0 × 2 ) = 76 + 0 = 76 76 is divisible by 19. Hence 74689 is also divisible by 19 Example2: Check if 71234 is divisible by 19. Given Number = 71234 7123 + (4 × 2 )= 7123 + 8 = 7131 713 + (1 × 2 )= 713 + 2 = 715 71 + (5 × 2 )= 71 + 10 = 81 81 is not divisible by 19. Hence 71234 is not divisible by 19 Example1: Check if 720 is divisible by 20 720 is divisible by 10. (Please check the divisibility rule of 10 to find out this). The tens digit = 2 = even digit. Hence 720 is also divisible by 20 Example2: Check if 1340 is divisible by 20
A number is divisible by 20 if it is divisible by 10 and the tens digit is even. Divisibility by 20
(There is one more rule to see if a number is divisible by 20 which is given below. A number is divisible by 20 if the number is divisible by
1340 is divisible by 10. (Please check the divisibility rule of 10 to find out this). The tens digit = 2 = even digit. Hence 1340 is divisible by 20 Example3: Check if 1350 is divisible by 20 1350 is divisible by 10. (Please check the divisibility rule of 10 to find out
both 4 and 5)
this). But the tens digit = 5 = not an even digit. Hence 1350 is not divisible by 20 Example4: Check if 1325 is divisible by 20 1325 is not divisible by 10 (Please check the divisibility rule of 10 to find out this) though the tens digit = 2 = even digit. Hence 1325 is not divisible by 20
IV.
W hat are Factors of a N umb er and how to find it out? a.
Factors of a num ber If one number is divisible by a second number, the second number is a factor of the first number The lowest factor of any positive number = 1 The highest factor of any positive number = the number itself Ex am ple The factors of 36 are 1, 2, 3, 4, 6, 9 12, 18, 36 because each of these numbers divides 36 with a remainder of 0
b.
How to find out factors of a num ber i. ii.
Write down 1 and the number itself (lowest and highest factors). Check if the given number is divisible by 2 (Reference: Divisibility by 2 rule) If the number is divisible by 2, write down 2 as the second lowest factor and divide the given number by 2 to get the second highest factor
iii.
Check for divisibility by 3, 4,5, and so on. till the beginning of the list reaches the end
Ex am ple1: Find out the factors of 72 i.
Write down 1 and the number itself (72) as lowest and highest factors. 1 . . . 72
ii.
72 is divisible by 2 (Reference: Divisibility by 2 Rule). 72 ÷ 2 = 36. Hence 2nd lowest factor = 2 and 2ndhighest factor = 36. So we can write as 1, 2 . . . 36, 72
iii.
72 is divisible by 3 (Reference: Divisibility by 3 Rule). 72 ÷ 3 = 24 . Hence 3rd lowest factor = 3 and 3rdhighest factor = 24. So we can write as 1, 2, 3, . . . 24, 36, 72
iv.
72 is divisible by 4 (Reference: Divisibility by 4 Rule). 72 ÷ 4 = 18. Hence 4th lowest factor = 4 and 4thhighest factor = 18. So we can write as 1, 2, 3, 4, . . . 18, 24, 36, 72
v.
vi.
72 is not divisible by 5 (Reference: Divisibility by 5 Rule)
72 is divisible by 6 (Reference: Divisibility by 6 Rule). 72 ÷ 6 = 12. Hence 5th lowest factor = 6 and 5thhighest factor = 12. So we can write as 1, 2, 3, 4, 6, . . . 12, 18, 24, 36, 72
vii.
viii.
72 is not divisible by 7 (Reference: Divisibility by 7 Rule)
72 is divisible by 8 (Reference: Divisibility by 8 Rule). 72 ÷ 8 = 9. Hence 6th lowest factor = 8 and 6thhighest factor = 9. Now our list is complete and the factors of 72 are 1, 2, 3, 4, 6, 8, 9 12, 18, 24, 36, 72
Ex am ple2: Find out the factors of 22 i.
Write down 1 and the number itself (22) as lowest and highest factors 1 . . . 22
ii.
22 is divisible by 2 (Reference: Divisibility by 2 Rule). 22 ÷ 2 = 11. Hence 2nd lowest factor = 2 and 2ndhighest factor = 11. So we can write as 1, 2 . . . 11, 22
iii.
22 is not divisible by 3 (Reference: Divisibility by 3 Rule).
iv.
22 is not divisible by 4 (Reference: Divisibility by 4 Rule).
v.
22 is not divisible by 5 (Reference: Divisibility by 5 Rule).
vi.
22 is not divisible by 6 (Reference: Divisibility by 6 Rule).
vii.
22 is not divisible by 7 (Reference: Divisibility by 7 Rule).
viii.
22 is not divisible by 8 (Reference: Divisibility by 8 Rule).
ix.
22 is not divisible by 9 (Reference: Divisibility by 9 Rule).
x.
22 is not divisible by 10 (Reference: Divisibility by 10 Rule). Now our list is complete and the factors of 22 are 1, 2, 11, 22
c.
I m portant P roperties of Factors If a number is divisible by another number, then it is also divisible by all the factors of that number. Example : 108 is divisible by 36 because 106 ÷ 38 = 3 with remainder of 0. The factors of 36 are 1, 2, 3, 4, 6, 9 12, 18, 36 because each of these numbers divides 36 with a remainder of 0. Hence, 108 is also divisible by each of the numbers 1, 2, 3, 4, 6, 9, 12, 18, 36.
V.
W hat are P rime N umb ers and Composite N umbers? a.
P rim e Num bers A prime number is a positve integer that is divisible byitself and 1 only. Prime numbers will have exactly two integer factors. Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, etc. Please note the following facts Zero is not a prime number because zero is divisible by more than two factors. Zero can be divided by 1, 2, 3 etc. (0 ÷ 1 = 0, 0÷ 2 = 0 ...) One is not a prime number because it does not have two factors. It is divisible by only 1
b.
Com posite Num bers Composite numbers are numbers that have more than two factors. A composite number is divisible by at least one number other than 1 and itself. Examples: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, etc. Please note that zero and 1 are neither prime numbers nor composite numbers. Every whole number is either prime or composite, with two exceptions 0 and 1 which are neither prime nor composite
VI.
W hat are P rime Factorization and P rime factors ? a.
P rim e factor
The factors which are prime numbers are called prime factors b.
P rim e factorization Prime factorization of a number is the expression of the number as the product of its prime factors Ex am ple 1: Prime factorization of 280 can be written as 280 = 2 × 2 × 2 × 5 × 7 = 23 × 5 × 7 and the prime factors of 280 are 2, 5 and 7 Ex am ple 2: Prime factorization of 72 can be written as 72 = 2 × 2 × 2 × 3 × 3 = 23 × 32 and the prime factors of 72 are 2 and 3
c.
How to find out prim e factorization and prim e factors of a num ber Repeated Division Method : In order to find out the prime factorization of a number, repeatedly divide the number by the smallest prime number possible(2,3,5,7,11, ...) until the quotient is 1. Ex am ple 1: Find out Prime factorization of 280 2 280 2 140 2 70 5 35 7 7 1 Hence, prime factorization of 280 can be written as 280 = 2 × 2 × 2 × 5 × 7 = 23 × 5 × 7 and the prime factors of 280 are 2, 5 and 7
Ex am ple 2: Find out Prime factorization of 72 2 72 2 36 2 18 3 9 3 3 1 Hence, prime factorization of 72 can be written as 72 = 2 × 2 × 2 × 3 × 3 = 23 × 32 and the prime factors of 72 are 2 and 3 d.
I m portant P roperties Every whole number greater than 1 can be uniquely expressed as the product of its prime
factors. For example, 700 = 22 × 52 × 7
VII.
Multip les Multiples of a whole number are the products of that number with 1, 2, 3, 4, and so on Example : Multiples of 3 are 3, 6, 9, 12, 15, ... If a number x divides another number y exactly with a remainder of 0, we can say that x is a factor of y and y is a multiple of x For instance, 4 divides 36 exactly with a remainder of 0. Hence 4 is a factor of 36 and 36 is a multiple of 4
VIII.
W hat is L east Common Multip le (LCM) and how to find LCM a.
Least Com m on Multiple (LCM) Least Common Multiple (LCM) of two or more numbers is the smallest number that is a multiple of all the numbers Example: LCM of 3 and 4 = 12 because 12 is the smallest multiple which is common to 3 and 4 (In other words, 12 is the smallest number which is divisible by both 3 and 4) We can find out LCM using prime factorization method or division method
b.
How to find out LCM using prim e factorization m ethod Step1 : Express each number as a product of prime factors. Step2 : LCM = The product of highest powers of all prime factors Ex am ple 1 : Find out LCM of 8 and 14 Step1 : Express each number as a product of prime factors. (Reference: Prime Factorization and how to find out Prime Factorization) 8 = 23 14 = 2 × 7 Step2 : LCM = The product of highest powers of all prime factors Here the prime factors are 2 and 7 The highest power of 2 here = 23 The highest power of 7 here = 7 Hence LCM = 23 × 7 = 56 Ex am ple 2 : Find out LCM of 18, 24, 9, 36 and 90 Step1 : Express each number as a product of prime factors (Reference: Prime Factorization and how to find out Prime Factorization).
18 = 2 × 32 24 = 23 × 3 9 = 32 36 = 23 × 32 90 = 2 × 5 × 32 Step2 : LCM = The product of highest powers of all prime factors Here the prime factors are 2, 3 and 5 The highest power of 2 here = 23 The highest power of 3 here = 32 The highest power of 5 here = 5 Hence LCM = 23 × 32 × 5 = 360 c.
How to find out LCM using Division Method (shortcut) Step 1 : Write the given numbers in a horizontal line separated by commas. Step 2 : Divide the given numbers by the smallest prime number which can exactly divide at least two of the given numbers. Step 3 : Write the quotients and undivided numbers in a line below the first. Step 4 : Repeat the process until we reach a stage where no prime factor is common to any two numbers in the row. Step 5 : LCM = The product of all the divisors and the numbers in the last line. Ex am ple 1 : Find out LCM of 8 and 14 2 8, 14 4, 7 Hence Least common multiple (L.C.M) of 8 and 14 = 2 × 4 × 7 = 56 Ex am ple 2 : Find out LCM of 18, 24, 9, 36 and 90 2 18, 24, 9, 36, 90 2 9, 12, 9, 18, 45 3 9, 6, 9, 9, 45 3 3, 2, 3, 3, 15 1, 2, 1, 1, 5 Hence Least common multiple (L.C.M) of 18, 24, 9, 36 and 90 = 2 × 2 × 3 × 3 × 2 × 5 = 360
IX.
W hat is H ig hest Common Factor (H CF) or Greatest Common Measure (GCM) or Greatest Common Divisor (GCD) and H ow to find it out ? a.
Highest Com m on Factor(H.C.F) or Greatest Com m on Measure(G.C.M) or Greatest Com m on Divisor (G.C.D) Highest Common Factor(H.C.F) or Greatest Common Measure(G.C.M) or Greatest Common Divisor (G.C.D) of two or more numbers is the greatest number which divides each of them exactly. Example : HCF or GCM or GCD of 60 and 75 = 15 because 15 is the highest number which divides both 60 and 75 exactly. We can find out HCF using prime factorization method or division method
b.
How to find out HCF using prim e factorization m ethod Step1 : Express each number as a product of prime factors. (Reference: Prime Factorization and how to find out Prime Factorization) Step2 : HCF is the product of all common prime factors using the least power of each common prime factor. Ex am ple 1 : Find out HCF of 60 and 75 (Reference:Prime Factorization and how to find out Prime Factorization) Step1 : Express each number as a product of prime factors. 60 = 22 × 3 × 5 75 = 3 × 52 Step2 : HCF is the product of all common prime factors using the least power of each common prime factor. Here, common prime factors are 3 and 5 The least power of 3 here = 3 The least power of 5 here = 5 Hence, HCF = 3 × 5 = 15 Ex am ple 2 : Find out HCF of 36, 24 and 12 Step1 : Express each number as a product of prime factors. (Reference: Prime Factorization and how to find out Prime Factorization) 36 = 22 × 32 24 = 23 × 3 12 = 22 × 3 Step2 : HCF is the product of all common prime factors using the least power of each common
prime factor. Here 2 and 3 are common prime factors. The least power of 2 here = 22 The least power of 3 here = 3 Hence, HCF = 22 × 3 = 12 Ex am ple 3 : Find out HCF of 36, 27 and 80 Step1 : Express each number as a product of prime factors. (Reference: Prime Factorization and how to find out Prime Factorization) 36 = 22 × 32 27 = 33 80 = 24 × 5 Step2 : HCF = HCF is the product of all common prime factors using the least power of each common prime factor. Here you can see that there are no common prime factors. Hence, HCF = 1 c.
How to find out HCF using prim e factorization m ethod - By dividing the num bers (shortcut) Step 1 : Write the given numbers in a horizontal line separated by commas. Step 2 : Divide the given numbers by the smallest prime number which can exactly divide all of the given numbers. Step 3 : Write the quotients in a line below the first. Step 4 : Repeat the process until we reach a stage where no common prime factor exists for all of the numbers. Step 5 :We can see that the factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. Their product is the HCF Ex am ple 1 : Find out HCF of 60 and 75 3 60, 75 5 20, 25 4, 5 We can see that the prime factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. no common prime factor is exists for the numbers came at the bottom. Hence HCF = 3 × 5 =15. Ex am ple 2 : Find out HCF of 36, 24 and 12
2 36, 24, 12 2 18, 12, 6 3 9, 6, 3 3, 2, 1 We can see that the prime factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. no common prime factor is exists for the numbers came at the bottom. Hence HCF = 2 × 2 × 3 = 12. Ex am ple 3 : Find out HCF of 36, 24 and 48 2 36, 24, 48 2 18, 12, 24 3 9, 6, 12 3, 2, 4 We can see that the prime factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. no common prime factor is exists for the numbers came at the bottom. Hence HCF = 2 × 2 × 3 = 12. d.
How to find out HCF using division m ethod(shortcut)
i.
To find out HCF of tw o given num bers using division m ethod Step 1: Divide the larger number by the smaller number Step 2: Divisor of step 1 is divided by its remainder Step 3: Divisor of step 2 is divided by its remainder. Continue this process till we get 0 as remainder. Step 4: Divisior of the last step is the HCF.
ii.
To find out HCF of three given num bers using division m ethod Step 1: Find out HCF of any two numbers. Step 2: Find out the HCF of the third number and the HCF obtained in step 1 Step 3: HCF obtained in step 2 will be the HCF of the three numbers
iii.
To find out HCF of m ore than three num bers using division m ethod In a similar way as explained for three numbers, we can find out HCF of more than three
numbers also
Ex am ple 1 : Find out HCF of 60 and 75 60) 75 (1 60 15) 60 (4 60 0 Hence HCF of 60 and 75 = 15 Ex am ple 2 : Find out HCF of 12 and 48 12) 48 (4 48 0 Hence HCF of 12 and 48 = 12 Ex am ple 3 : Find out HCF of 3556 and 3224 3224) 3556 (1 3224 332) 3224 (9 2988 236) 332 (1 236 96) 236 (2 192 44) 96 (2 88 8) 44 (5 40 4) 8 (2 8 0 Hence HCF of 3556 and 3224 = 4 Ex am ple 3 : Find out HCF of 9, 27, and 48 Taken any two numbers and find out their HCF first. Say, let's find out HCF of 9 and 27 initially.
9) 27 (3 27 0 Hence HCF of 9 and 27 = 9 HCF of 9 ,27, 48 = HCF of [(HCF of 9, 27) and 48] = HCF of [9 and 48] 9) 48 (5 45 3) 9 (3 9 0 Hence, HCF of 9 ,27, 48 = 3 Ex am ple 4 : Find out HCF of 5 and 7 5) 7 (1 5 2) 5 (2 4 1) 2 (2 2 0 Hence HCF of 5 and 7 = 1 X.
H ow to calculate L CM and H CF for fractions Least Com m on Multiple (L.C.M.) for fractions
LCM for fractions =
LCM of Numerators HCF of Denominators
Exam ple 1: Find out LCM of
LCM =
LCM (1, 3, 3) 3 = 2 HCF (2, 8, 4)
Exam ple 2: Find out LCM of
LCM =
1 3 3 , , 2 8 4
LCM (2, 3) 6 = 5 HCF (5, 10)
2 3 , 5 10
XI.
H ig hest Common Multip le (H .C.F) for fractions
HCF for fractions =
HCF of Numerators LCM of Denominators
Exam ple 1: Find out HCF of
HCF =
HCF (3, 6, 9) 3 = 220 LCM (5, 11, 20)
Exam ple 2: Find out HCF of
HCF =
XII.
3 6 9 , , 5 11 20
4 2 , 5 3
HCF (4, 2) 2 = 15 LCM (5, 3)
H ow to calculate L CM and H CF for Decimals Step 1 : Make the same number of decimal places in all the given numbers by suffixing zero(s) in required numbers as needed. Step 2 : Now find the LCM/HCF of these numbers without decimal. Step 3 : Put the decimal point in the result obtained in step 2 leaving as many digits on its right as there are in each of the numbers. Exam ple1 : Find the LCM and HCF of .63, 1.05, 2.1 Step 1 : Make the same number of decimal places in all the given numbers by suffixing zero(s) in required numbers as needed. i.e., the numbers can be writtten as .63, 1.05, 2.10 Step 2 : Now find the LCM/HCF of these numbers without decimal. Without decimal, the numbers can be written as 63, 105 and 210 . LCM (63, 105 and 210) = 630 HCF (63, 105 and 210) = 21 Step 3 : Put the decimal point in the result obtained in step 2 leaving as many digits on its right as there are in each of the numbers. i.e., here, we need to put decimal point in the result obtained in step 2 leaving two digits on its right. i.e., the LCM (.63, 1.05, 2.1) = 6.30 HCF (.63, 1.05, 2.1) = .21
XIII.
H ow to comp are fractions? a.
Type 1 : Fractions w ith sam e denom inators. Compare
3 1 and 5 5
These fractions have same denominator. So just compare the numerators. Bigger the numerator, bigger the number. 3 > 1. Hence
3 1 > 5 5
Ex am ple 2: Compare
2 3 8 and and 7 7 7
These fractions have same denominator. So just compare the numerators. Bigger the numerator, bigger the number. 8 > 3 > 2. Hence
b.
8 3 2 > > 7 7 7
Type 2 : Fractions w ith sam e num erators. Ex am ple 1: Compare
3 3 and 5 8
These fractions have same numerator. So just compare the denominators. Bigger the denominator, smaller the number. 8 > 5. Hence
3 3 < 8 5
Ex am ple 2: Compare
7 7 7 and and 8 2 5
These fractions have same numerator. So just compare the denominators. Bigger the denominator, smaller the number. 8 > 5 > 2. Hence
c.
7 7 7 < < 8 5 2
Type 3 : Fractions w ith different num erators and denom inators. Ex am ple 1: Compare
3 4 and 5 7
To compare such fractions, find out LCM of the denominators. Here, LCM(5, 7) = 35 Now , convert each of the given fractions into an equivalent fraction with 35 (LCM) as the denominator. The denominator of
3 3×7 21 = = 5 5×7 35
3 is 5. 5 needs to be multiplied with 7 to get 35. Hence, 5
The denominator of
4 is 7. 7 needs to be multiplied with 5 to get 35. Hence, 7
4 4×5 20 = = 7 7×5 35 21 20 > 35 35 Hence,
3 4 > 5 7
Or Convert the fractions to decimals
3 = .6 5 4 = .5. . . 7
(Need not find out the complete decimal value; just find out up to what is required for
comparison. In this case the first digit itself is sufficient to do the comparison) .6 > .5... Hence,
XIV.
3 4 > 5 7
Co-p rime N umb ers or Relatively P rime N umbers Two numbers are said to be co-prime (also spelled coprime) or relatively prime if they do not have a common factor other than 1. i.e., if their HCF is 1. Exam ple1: 3, 5 are co-prime numbers (Because HCF of 3 and 5 = 1) Exam ple2: 14, 15 are co-prime numbers (Because HCF of 14 and 15 = 1) A set of numbers is said to be pairwise co-prime (or pairwise relatively prime) if every two distinct numbers in the set are co-prime Exam ple1 : The numbers 10, 7, 33, 13 are pairwise co-prime, because HCF of any pair of the numbers in this is 1. HCF (10, 7) = HCF (10, 33) = HCF (10, 13) = HCF (7, 33) = HCF (7, 13) = HCF (33, 13) = 1. Exam ple2 : The numbers 10, 7, 33, 14 are not pairwise co-prime because HCF(10, 14) = 2 ≠ 1 and HCF(7, 14) = 7 ≠ 1. If a number is divisible by two co-prime numbers, then the number is divisible by their product also. Example 3, 5 are co-prime numbers (Because HCF of 3 and 5 = 1) 14325 is divisible by 3 and 5. 3 × 5 = 15 Hence 14325 is divisible by 15 also
If a number is divisible by more than two pairwise co-prime numbers, then the number is divisible by their product also. Example1 : The numbers 3, 4, 5 are pairwise co-prime because HCF of any pair of numbers in this is 1 1440 is divisible by 3, 4 and 5. 3 × 4 × 5 = 60. Hence 1440 is also divisible by 60 Example2 The numbers 3, 4, 9 are not pairwise co-prime because HCF (3, 9 ) = 3 ≠ 1 1440 is divisible by 3, 4 and 9. 3 X 4 X 9 = 108. However 1440 is not divisible by 108 as 3, 4, 9 are not pairwise co-prime
XV.
I mp ortant P oints to N ote on L CM and H CF
Product of two numbers = Product of their HCF and LCM. Example LCM (8, 14) = 56 HCF (8, 14) = 2 LCM (8, 14) × HCF (8, 14) = 56 × 2 = 112 8 × 14 = 112 Hence LCM (8, 14) × HCF (8, 14) = 8 × 14
Important Formulas - Height and Distance 1. Trigonometric Basics
sin θ =
opposite side y = hypotenuse r
cos θ =
adjacent side x = hypotenuse r
tan θ =
opposite side y = adjacent side x
csc θ =
hypotenuse r = opposite side y
sec θ =
hypotenuse r = adjacent side x
cot θ =
adjacent side x = opposite side y
From Pythagorean theorem, triangle mentioned above
x2 + y 2 = r 2
for the right angled
2. Basic Trigonometric Values
θ
θ
in degrees in radians
sin θ
cos θ tan θ
0°
0
0
1
0
30°
π 6
1 2
√3 2
45°
π 4
1
1
√2
√2
60°
π 3
√3 2
1 2
√3
90°
π 2
1
0
Not defined
1 √3 1
3. Trigonometric Formulas Degrees to Radians and vice versa
360° = 2π radian
Trigonometry Quotient Formulas
tan θ =
sin θ cos θ
cot θ =
cos θ sin θ
Trigonometry - Reciprocal Formulas
csc θ =
1 sin θ
sec θ =
1 cos θ
cot θ =
1 tan θ
Trigonometry - Pythagorean Formulas
sin2 θ + cos2 θ = 1 sec2 θ − tan 2 θ = 1 csc2 θ − cot2 θ = 1
4. Angle of Elevation
Suppose a man from a point O looks up at an object P, placed above the level of his eye. Then, angle of elevation is the angle between the horizontal and the line from the object to the observer's eye (the line of sight). i.e., angle of elevation =
AOP
5. Angle of Depression
Suppose a man from a point O looks down at an object P, placed below the level of his eye. Then, angle of depression is the angle between the horizontal and the observer's line of sight i.e., angle of depression =
AOP
6. Angle Bisector Theorem
Consider a triangle ABC as shown above. Let the angle bisector of angle A intersect side BC at a point D. Then
BD
AB
DC
=
AC
(Note that an angle bisector divides the angle into two angles with equal measures. i.e.,
BAD =
CAD in the above diagram)
7. Few Important Values to memorise
√ 2 = 1.414 √ 3 = 1.732 √ 5 = 2.236
1. Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is: A. 300 m
B. 173 m
C. 273 m
D. 200 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Let BD be the lighthouse and A and C be the positions of the ships. Then, BD = 100 m,
BAD = 30° ,
BCD = 45°
BD BA 1 100 ⇒ = √ 3 BA
tan 30° =
⇒ BA = 100√ 3 BD BC 100 ⇒1= BC ⇒ BC = 100
tan 45° =
Distance between the two ships = AC = BA + BC
= 100√ 3 + 100 = 100(√ 3 + 1) = 100(1.73 + 1) = 100 × 2.73 = 273 m
2. A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 45º. What is the distance between the base of the tower and the point P?
3√ 3 units
A. 9 units
B.
C. Data inadequate
D. 12 units
Hide Answer | Notebook | Discuss
Here is the answer and explanation Answer : Option C Explanation :
SR QR SR SR tan 30° = = PR (PQ + QR)
tan 45° =
two equations and 3 variables. Hence we can not find the required value with the given data. (Note that if one of SR, PQ, QR is known, this becomes two equations and two variables and if that was the case, we could have found out the required value)
3. From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 200 m high, the distance of point P from the foot of the tower is: A. 346 m
B. 400 m
C. 312 m
D. 298 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
RQ PQ 1 200 = √ 3 PQ PQ = 200√ 3 = 200 × 1.73 = 346 m
tan 30° =
4. The angle of elevation of the sun, when the length of the shadow of a tree is equal to the height of the tree, is: A. None of these
B. 60°
C. 45°
D. 30°
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Consider the diagram shown above where QR represents the tree and PQ represents its shadow We have, QR = PQ Let
QPR = θ
tan θ =
QR =1 PQ
=> θ = 45°
(since QR = PQ)
i.e., required angle of elevation = 45°
5. An observer 2 m tall is 10√ 3 m away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is: A. None of these
B. 12 m
C. 14 m
D. 10 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
SR = PQ = 2 m
PS = QR = 10√ 3 m TS tan 30° = PS 1 TS = √ 3 10√ 3 10√ 3 TS = = 10 m √3 TR = TS + SR = 10 + 2 = 12 m
6. The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 12.4 m away from the wall. The length of the ladder is: A. 14.8 m
B. 6.2 m
C. 12.4 m
D. 24.8 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Consider the diagram shown above where PR represents the ladder and RQ represents the wall.
cos 60° =
PQ PR
1 12.4 = 2 PR PR = 2 × 12.4 = 24.8 m
7. A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower? A. 8 min 17 second
B. 10 min 57 second
C. 14 min 34 second
D. 12 min 23 second
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car Then,
ADC = 30° ,
ACB = 45°
Let AB = h, BC = x, CD = y
tan 45° =
AB h = BC x
h x => h = x ------(1) => 1 =
AB AB h = = BD (BC + CD) x+y 1 h => = √ 3 x+y => x + y = √ 3h => y = √ 3h - x => y = √ 3h − h (∵ Substituted the value of x from equation 1 ) => y = h(√ 3 − 1)
tan 30° =
Given that distance y is covered in 8 minutes
i.e, distance h(√ 3 − 1) is covered in 8 minutes
Time to travel distance x = Time to travel distance h (∵ Since x = h as per equation 1). Let distance h is covered in t minutes since distance is proportional to the time when the speed is constant, we have
h(√ 3 − 1) ∝ 8 h∝ t
---(A) --- (B)
(A) h(√ 3 − 1) 8 => = h t (B) 8 => (√ 3 − 1) = t 8 8 8 800 => t = = = = minutes .73 73 (1.73 − 1) (√ 3 − 1) 70 = 10 minutes ≈ 10 minutes 57 seconds 73
8. A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man's eye when at a distance of 100 metres from the tower. After 10 seconds, the angle of depression becomes 30°. What is the approximate speed of the boat, assuming that it is running in still water? A. 26.28 km/hr
B. 32.42 km/hr
C. 24.22 km/hr
D. 31.25 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Consider the diagram shown above. Let AB be the tower. Let C and D be the positions of the boat Then,
ACB = 45° ,
AB BC AB => 1 = 100 => AB = 100
ADC = 30° , BC = 100 m
tan 45° =
------(1)
AB BD 1 100 => = (∵ Substituted the value of AB from equation 1) √ 3 BD => BD = 100√ 3 CD = (BD - BC) = (100√ 3 − 100) = 100(√ 3 − 1)
tan 30° =
It is given that the distance CD is covered in 10 seconds.
i.e., the distance 100(√ 3 − 1) is covered in 10 seconds Required speed = = 7.3 ×
100(√ 3 − 1) Distance = = 10(1.73 − 1) = 7.3 meter/seconds Time 10
18 km/hr = 26.28 km/hr 5
9. The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole? A. 5 metres
B. 8 metres
C. 10 metres
D. 12 metres
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Consider the diagram shown above. AC represents the tower and DE represents the pole Given that AC = 15 m ,
ADB = 30°,
AEC = 60°
Let DE = h Then, BC = DE = h, AB = (15-h) (∵ AC=15 and BC = h), BD = CE
AC CE 15 => √ 3 = CE 15 => CE = √3
tan 60° =
--- (1)
AB BD 15 − h = BD 15 − h = 15 ( ) √3
tan 30° = 1 √3 1 => √3 =>
=> (15 − h) =
1
(∵ BD = CE and Substituted the value of CE from equation 1 )
×
15
√ 3 √3 => h = 15 − 5 = 10 m
=
15 =5 3
i.e., height of the electric pole = 10 m
10. The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 40 m towards the tower, the angle of elevation of the top of the tower increases by 15°. The height of the tower is: A. 64.2 m
B. 62.2 m
C. 52.2 m
D. 54.6 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Let DC be the tower and A and B be the positions of the observer such that AB = 40 m We have
DAC = 30°,
DBC = 45°
Let DC = h
DC AC 1 h => = √ 3 AC => AC = h√ 3
tan 30° =
DC BC h => 1 = BC => BC = h
------(1)
tan 45° =
------(2)
We know that, AB = (AC - BC) => 40 = (AC - BC)
=> 40 = (h√ 3 − h)
[∵ from (1) & (2)]
=> 40 = h(√ 3 − 1) ( √ 3 + 1) 40(√ 3 + 1) 40 40 = × = (3 − 1) (√ 3 − 1) (√ 3 − 1) ( √ 3 + 1) 40(√ 3 + 1) = = 20(√ 3 + 1) = 20(1.73 + 1) = 20 × 2.73 = 54.6 m 2 => h =
11. On the same side of a tower, two objects are located. Observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 600 m, the distance between the objects is approximately equal to : A. 272 m
B. 284 m
C. 288 m
D. 254 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Let DC be the tower and A and B be the objects as shown above. Given that DC = 600 m ,
DC BC 600 √3= BC 600 BC= √3
DAC = 45°,
DBC = 60°
tan 60° =
tan 45° = 600 AC AC= 600
------ (1)
DC AC
1=
------ (2)
Distance between the objects = AB = (AC - BC) 600 = 600 − [∵ from (1) and (2) ] √3 = 600 [1 −
1 √ 3−1 √ 3−1 √ 3 600√ 3(√ 3 − 1) ] = 600 [ ] = 600 [ ]× = 3 √3 √3 √3 √3
= 200√ 3(√ 3 − 1) = 200(3 − √ 3) = 200(3 − 1.73) = 254 m
12. A ladder 10 m long just reaches the top of a wall and makes an angle of 60° with the wall.Find the distance of the foot of the ladder from the wall (√ 3 = 1.73) A. 4.32 m
B. 17.3 m
C. 5 m
D. 8.65 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Let BA be the ladder and AC be the wall as shown above. Then the distance of the foot of the ladder from the wall = BC Given that BA = 10 m ,
BAC = 60°
BC BA √ 3 BC = 2 10 √3 BC = 10 × = 5 × 1.73 = 8.65 m 2
sin 60° =
13. From a tower of 80 m high, the angle of depression of a bus is 30°. How far is the bus from the tower? A. 40 m
B. 138.4 m
C. 46.24 m
D. 160 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Let AC be the tower and B be the position of the bus. Then BC = the distance of the bus from the foot of the tower. Given that height of the tower, AC = 80 m and the angle of depression, ABC =
tan 30° =
DAB = 30°
DAB = 30°
( Because DA || BC)
AC BC
80 BC 80 => BC = = tan 30° => tan 30° =
80 1 ( ) √3
= 80 × 1.73 = 138.4 m
i.e., Distance of the bus from the foot of the tower = 138.4 m
14. The angle of elevation of the top of a lighthouse 60 m high, from two points on the ground on its opposite sides are 45° and 60°. What is the distance between these two points? A. 45 m
B. 30 m
C. 103.8 m
D. 94.6 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Let BD be the lighthouse and A and C be the two points on ground. Then, BD, the height of the lighthouse = 60 m BAD = 45° ,
BCD = 60°
BD BA 60 ⇒1= BA ⇒ BA = 60 m
tan 45° =
------ (1)
BD BC 60 ⇒√3= BC 60 60 × √ 3 60√ 3 ⇒ BC = = = = 20 √ 3 = 20 × 1.73 = 34.6 m 3 √ 3 √3×√ 3
tan 60° =
------ (2)
Distance between the two points A and C = AC = BA + BC = 60 + 34.6 [∵ Substituted the value of BA and BC from (1) and (2)] = 94.6 m
15. From the top of a hill 100 m high, the angles of depression of the top and bottom of a pole are 30° and 60° respectively. What is the height of the pole? A. 52 m
B. 50 m
C. 66.67 m
D. 33.33 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Consider the diagram shown above. AC represents the hill and DE represents the pole Given that AC = 100 m XAD =
ADB = 30°
(∵ AX || BD )
XAE =
AEC = 60°
(∵ AX || CE)
Let DE = h Then, BC = DE = h, AB = (100-h) = CE
AC CE 100 => √ 3 = CE 100 => CE = √3
(∵ AC=100 and BC = h), BD
tan 60° =
AB BD 100 − h = BD 100 − h = 100 ( ) √3
--- (1)
tan 30° = 1 √3 1 => √3 =>
(∵ BD = CE and Substituted the value of CE from equation 1 )
1 100 100 × = = 33.33 3 √3 √3 => h = 100 − 33.33 = 66.67 m => (100 − h) =
i.e., the height of the pole = 66.67 m
16. A vertical tower stands on ground and is surmounted by a vertical flagpole of height 18 m. At a point on the ground, the angle of elevation of the bottom and the top of the flagpole are 30° and 60° respectively. What is the height of the tower? A. 9 m
B. 10.40 m
C. 15.57 m
D. 12 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation. Given that AD = 18 m,
ABC = 60°,
Let DC be h.
DC BC 1 h = √ 3 BC BC h= ------ (1) √3
tan 30° =
AC BC 18 + h √3= BC 18+h = BC × √ 3
tan 60° =
------ (2)
DBC = 30°
(
BC √3
)
(1) h => = 18 + h (2) (BC × √ 3)
=
1 3
=> 3h = 18 + h => 2h = 18 => h = 9 m i.e., the height of the tower = 9 m
17. A balloon leaves the earth at a point A and rises vertically at uniform speed. At the end of 2 minutes, John finds the angular elevation of the balloon as 60°. If the point at which John is standing is 150 m away from point A, what is the speed of the balloon? A. 0.63 meter/sec
B. 2.16 meter/sec
C. 3.87 meter/sec
D. 0.72 meter/sec
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Let C be the position of John. Let A be the position at which balloon leaves the earth and B be the position of the balloon after 2 minutes. Given that CA = 150 m,
BA CA BA √3= 150 BA = 150√ 3
tan 60° =
BCA = 60°
i.e, the distance travelled by the balloon = 150√ 3 meters time taken = 2 min = 2 × 60 = 120 seconds Distance 150√ 3 Speed = = = 1.25√ 3 = 1.25 × 1.73 = 2.16 meter/second Time 120
18. The angles of depression and elevation of the top of a wall 11 m high from top and bottom of a tree are 60° and 30° respectively. What is the height of the tree? A. 22 m
B. 44 m
C. 33 m
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Let DC be the wall, AB be the tree. Given that
DBC = 30°,
DC BC 1 11 = √ 3 BC BC = 11√ 3 m AE = BC =11√ 3 m
DAE = 60°, DC = 11 m
tan 30° =
------ (1)
ED AE ED √3= [∵ Substituted the value of AE from (1)] 11√ 3 ED =11√ 3 × √ 3 = 11 × 3 = 33
tan 60° =
Height of the tree = AB = EC = (ED + DC) = (33 + 11) = 44 m
19. Two vertical poles are 200 m apart and the height of one is double that of the other. From the middle point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary. Find the heights of the poles. A. 141 m and 282 m
B. 70.5 m and 141 m
C. 65 m and 130 m
D. 130 m and 260 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Let AB and CD be the poles with heights h and 2h respectively Given that distance between the poles, BD = 200 m Let E be the middle point of BD. Let AEB = θ and CED = (90-θ) elevations are complementary )
(∵ given that angular
Since E is the middle point of BD, we have BE = ED = 100 m From the right
ABE,
AB BE h tan θ = 100 h = 100 tan θ
------ (1)
From the right
EDC,
tan θ =
tan(90 − θ) =
CD ED
2h 100 2h = 100 cot θ
cot θ =
[∵ tan(90 − θ) = cot θ] ------ (2)
(1) × (2) => 2h 2 = 1002
[∵ tan θ × cot θ = tan θ ×
1 = 1] tan θ
=> √ 2h = 100 100 100 × √ 2 => h = = = 50√ 2 = 50 × 1.41 = 70.5 √2 √ 2×√2 2h = 2 × 70.5 = 141 i.e., the height of the poles are 70.5 m and 141 m.
20. To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and he is 5 m away from the wall, what is the length of the window? A. 8.65 m
B. 2 m
C. 2.5 m
D. 3.65 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Let AB be the man and CD be the window Given that the height of the man, AB = 180 cm, the distance between the man and the wall, BE = 5 m, DAF = 45° ,
CAF = 60°
From the diagram, AF = BE = 5 m
From the right
tan 45° = DF 5 DF = 5
AFD,
DF AF
1=
------ (1)
From the right
AFC,
CF AF CF √3= 5 CF = 5√ 3
------ (2)
tan 60° =
Length of the window = CD = (CF - DF)
= 5√ 3 − 5
[∵ Substitued the value of CF and DF from (1) and (2)]
= 5 (√ 3 − 1) = 5(1.73 − 1) = 5 × 0.73 = 3.65 m
21. The elevation of the summit of a mountain from its foot is 45°. After ascending 2 km towards the mountain upon an incline of 30°,the elevation changes to 60°. What is the approximate height of the mountain? A. 1.2 km
B. 0.6 km
C. 1.4 km
D. 2.7 km
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Let A be the foot and C be the summit of a mountain. Given that
CAB = 45°
From the diagram, CB is the height of the mountain. Let CB = x Let D be the point after ascending 2 km towards the mountain such that AD = 2 km and given that
DAY = 30°
It is also given that from the point D, the elevation is 60° i.e.,
CDE = 60°
From the right
CB AB x => 1 = AB => AB = x
ABC,
tan 45° =
From the right
[∵ CB = x (the height of the mountain)] ------ (eq:1) AYD,
DY AD 1 DY => = (∵ Given that AD = 2) 2 2 => DY = 1 ------ (eq:2) AY cos 30° = AD √ 3 AY => = (∵ Given that AD = 2) 2 2 => AY = √ 3 ------ (eq:3)
sin 30° =
From the right
tan 60° =
CED,
CE DE
(CB - EB) YB (CB - DY) => tan 60° = (AB - AY) (x - 1) => tan 60° = (x - √ 3) (x - 1) => √ 3 = (x - √ 3) => tan 60° =
∵ [CE = (CB - EB) and DE = YB)] [ ∵ EB = DY and YB = (AB - AY)] ∵ [CB = x, DY = 1(eq:2), AB=x (eq:1) and AY = √ 3(eq : 3)]
=> x√ 3 − 3 = x − 1 => x(√ 3 − 1) = 2 => 0.73x = 2 2 => x = = 2.7 0.73 i.e., the height of the mountain = 2.7 km
22. Two persons are on either sides of a tower of height 50 m. The persons observers the top of the tower at an angle of elevation of 30° and 60°. If a car crosses these two persons in 10 seconds, what is the speed of the car?
24√ 3 km/hr 24 C. km/hr √3 A.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
B. None of these D.
20√ 3 km/hr 3
Let BD be the tower and A and C be the positions of the persons. Given that BD = 50 m, From the right
BAD = 30° ,
BCD = 60°
ABD,
BD BA 1 50 ⇒ = √ 3 BA ⇒ BA = 50√ 3
tan 30° =
BD BC 50 ⇒√3= BC 50 50 × √ 3 50√ 3 ⇒ BC = = = 3 √ 3 √3×√ 3
tan 60° =
Distance between the two persons = AC = BA + BC
= 50√ 3 +
50√ 3 50 200√ 3 )= = √ 3 (50 + m 3 3 3
i.e., the distance travelled by the car in 10 seconds =
Speed of the car = =
Distance = Time
(
200√ 3 ) 3
20√ 3 18 × km/hr = 24 √ 3 km/hr 3 5
10
=
200√ 3 m 3
20√ 3 meter/second 3
23. Find the angle of elevation of the sun when the shadow of a pole of 18 m height is A. 30°
B. 60°
C. 45°
D. None of these
6√ 3 m long?
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Let RQ be the pole and PQ be the shadow
Given that RQ = 18 m and PQ = 6√ 3 m Let the angle of elevation, From the right
RPQ = θ
PQR,
RQ 18 3 3×√3 3√ 3 = = = = =√3 PQ 3 6√ 3 √ 3 √ 3 ×√ 3 θ = tan −1 (√ 3) = 60°
tan θ =
24. The angle of elevation of the top of the tower from a point on the ground is
3 sin −1 ( ) . If the point 5
of observation is 20 meters away from the foot of the tower, what is the height of the tower? A. 9 m
B. 18 m
C. 15 m
D. 12 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : ------------------------------------------------------------------------------------Solution 1 ------------------------------------------------------------------------------------Consider a right-angled triangle PQR as shown below.
Let QR = 3 and PR = 5 such that sin θ =
3 3 [i.e., θ = sin −1 ( )] 5 5
−−−−−−−−− => PQ = √ PR 2 − QR2 (∵ Pythagorean theorem) −−−−−− = √ 52 − 32 = 4 3 i.e., when θ = sin −1 ( ) , PQ : QR = 4 : 3 ------(eq:1) 5
Now Let's solve the question. Let P be the point of observation and QR be the tower as shown the below diagram.
3 Given that θ = sin −1 ( ) and PQ = 20 m 5 We know that PQ : QR = 4 : 3
(From eq:1)
i.e., 20 : QR = 4 : 3 => 20 × 3 = QR × 4 => QR = 15 m i.e., height of the tower = 15 m ------------------------------------------------------------------------------------Solution 2 -------------------------------------------------------------
-------------------------
Let P be the point of observation and QR be the tower.
3 Given that θ = sin −1 ( ) and PQ = 20 m 5 Let the height of the tower, QR = h and PR = x From the right
sin θ =
PQR,
QR PR
3 h => sin [sin −1 ( )] = 5 x 3 h => = 5 x 5h => x = ------ (eq:1) 3 From Pythagorean theorem, we have
PQ 2 + QR2 = PR 2 202 + h 2 = x2 5h 202 + h 2 = ( ) 3
2
(∵ Substituted the value of x from eq:1)
25h 2 20 + h = 9 2 16h = 202 9 4h = 20 3 3 × 20 h= = 15 m 4 2
2
i.e., height of the tower = 15 m
25. A person, standing exactly midway between two towers, observes the top of the two towers at angle of elevation of 22.5° and 67.5°. What is the ratio of the height of the taller tower to the height of the shorter tower? (Given that tan 22.5° = √
2 − 1)
1 − 2√ 2 : 1 C. 3 + 2√ 2 : 1
1 + 2√ 2 : 1 D. 3 − 2√ 2 : 1
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Let ED be the taller tower and AB be the shorter tower. Let C be the point of observation. Given that
ACB = 22.5° and
DCE = 67.5°
Given that C is the midpoint of BD. Hence, BC = CD From the right
tan 22.5° =
AB BC
From the right
tan 67.5° =
ED CD
ABC,
------ (eq:1) CDE,
------ (eq:2)
(eq:2) (eq:1)
=>
tan 67.5° = tan 22.5°
=>
tan(90° − 22.5°) tan 22.5°
=>
cot 22.5° ED = tan 22.5° AB
1 ) tan 22.5° => tan 22.5° (
=
(
ED ) CD
AB ( ) BC
=
ED AB
(∵ CD = BC )
ED AB
=
[ ∵ tan(90-θ)= cot θ ]
ED AB
ED 1 => i.e., = AB (tan 22.5°) (√ 2 + 1) ] =[ (√ 2 − 1)(√ 2 + 1)
2
1 ] tan θ
[ ∵ cot θ =
2
=
1 (√ 2 − 1)
(√ 2 + 1) ] =[ (2 − 1)
= (2 + 2√ 2 + 1) = (3 + 2√ 2) Required Ratio = ED : AB = (3 + 2√ 2) : 1
2
2
1 ) =( √ 2−1 (√ 2 + 1) ] =[ 1
2
2
= (√ 2 + 1)
2
26. A flagstaff is placed on top of a building. The flagstaff and building subtend equal angles at a point on level ground which is 200 m away from the foot of the building. If the height of the flagstaff is 50 m and the height of the building is h, which of the following is true? A. h3 - 50h2 + (200)2h + (200)250 = 0
B. None of these
C. h3 + 50h2 + (200)2h - (200)250 = 0
D. h3 - 50h2 - (200)2h + (200)250 = 0
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Let AD be the flagstaff and CD be the building. Assume that the flagstaff and building subtend equal angles at point B. Given that AD = 50 m, CD = h and BC = 200 m Let ABD = θ, DBC = θ (∵ flagstaff and building subtend equal angles at a point on level ground) Then,
ABC = 2θ
From the right
tan θ =
BCD,
DC h = BC 200
From the right
BCA,
------ (eq: 1)
tan 2θ = =>
AC AD + DC 50 + h = = BC 200 200
2 tan θ 50 + h = 200 1 − tan 2 θ 2(
=>
h ) 200
1−
h
2
[∵ tan(2θ) =
50 + h 200
=
2 tan θ ] 1 − tan 2 θ
[∵ substituted the value of tan θ from eq:1 ]
2002
=> 2h = (1 −
h2 2002
=> 2h = 50 + h −
) (50 + h)
50h 2 2002
−
h3 2002
=> 2 (2002 ) h = 50(200) 2 + h(200) 2 − 50 h2 − h 3
(∵ Multiplied LHS and RHS by 2002 )
=> h 3 + 50h 2 + (200) 2 h − 50(200) 2 = 0
27. From the foot and the top of a building of height 230 m, a person observes the top of a tower with angles of elevation of b and a respectively. What is the distance between the top of these buildings if tan a = 5/12 and tan b = 4/5 A. 400 m
B. 250 m
C. 600 m
D. 650 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Let ED be the building and AC be the tower. Given that ED = 230 m,
ADC = b,
AEB = a
Also given that tan a = 5/12 and tan b = 4/5 Let AC = h Required Distance buildings = AE From the right
= Distance between the
top of these
ABE,
AB BE (h-230) 5 => = [ ∵ tan a = 5/12 (given), AB = (AC-BC) = (AC-ED) = (h-230) ] 12 BE 12(h-230) => BE = ------ (eq: 1) 5
tan (a) =
From the right
AC CD 4 h => = 5 CD 5h => CD = 4
ACD,
tan (b) =
[ ∵ tan b = 4/5 (given), AC = h) ] ------ (eq: 2)
From the diagram, BE = CD
12(h-230) 5h = (from eq: 1 and eq:2) 5 4 => 48h − (4 × 12 × 230) = 25h => 23h = (4 × 12 × 230) (4 × 12 × 230) => h = = 480 m ------ (eq: 3) 23
=>
Consider the right
ABE
AB = (AC - BC) = (480 - 230) [ ∵ Since AC = h = 480(from eq:3) and BC = ED = 230 m(given) ] = 250 m In this triangle, tan(a) = 5/12. Let's figure out the value of sin(a) now.
Consider a triangle with opposite side = 5 and adjacent side = 12 such that tan(a) = 5/12
−−−−−−− hypotenuse = √ 52 + 122 = 13 opposite side 5 i.e., sin(a) = = hypotenuse 13 We have seen that sin(a) = =>
AB 5 = 13 AE
=> AE = AB ×
5 13
13 13 = 250 × = 650 m 5 5
i.e., Distance between the top of the buildings = 650 m
28. A vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 15 m away from the base of the pole, what is the height of the pole?
60√ 5 m C. 15√ 3 m A.
15√ 5 m D. 60√ 3 m B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
(Reference : Angle Bisector Theorem) Let CB be the pole and point D divides it such that BD : DC = 1 :9 Given that AB = 15 m Let the the two parts subtend equal angles at point A such that
CAD =
BAD = θ
From Angle Bisector Theorem, we have
BD AB = DC AC 1 15 => = [ ∵ BD : DC = 1 : 9 and AB = 15(given) ] 9 AC AC = 15 × 9 m ------ (eq: 1) From the right
ABC,
−−−−−−−−− CB = √ AC 2 − AB2 ( ∵ Pythagorean theorem) −−−−−−−−−−− − 2 = √ (15 × 9) − 152 [ ∵ AC = 15 × 9 (eq:1) and AB = 15 m (given) ] −−−−−−−−−− −−−−−−−−−−− −−−−−−− 2 2 2− √ = 15 × 9 − 15 = √ 152 (92 − 1) = √ 152 × 80 −−−−−−−−−− = √ 152 × 16 × 5 = 15 × 4 × √ 5 = 60√ 5 m
29. An aeroplane when 900 m high passes vertically above another aeroplane at an instant when their angles of elevation at same observing point are 60° and 45° respectively. Approximately, how many meters higher is the one than the other? A. 381 m
B. 169 m
C. 254 m
D. 211 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Let C and D be the position of the aeroplanes. Given that CB = 900 m, From the right
ABC,
CAB = 60°,
DAB = 45°
CB AB 900 √3= AB 900 900 × √ 3 900√ 3 AB = = = = 300√ 3 3 √ 3×√3 √3
tan 60° =
From the right
tan 45° =
ABD,
DB AB
DB AB DB = AB = 300√ 3 1=
Required height = CD = (CB-DB) = (900 − 300√ 3) = (900 − 300 × 1.73) = (900 − 519) = 381 m
30. When the sun's altitude changes from 30° to 60°, the length of the shadow of a tower decreases by 70m. What is the height of the tower? A. 35 m
B. 140 m
C. 60.6 m
D. 20.2 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Let AD be the tower, BD be the initial shadow and CD be the final shadow. Given that BC = 70 m,
ABD = 30°,
ACD = 60°,
Let CD = x, AD = h From the right
tan 60° = √3=
h x
CDA,
AD CD ------ (eq :1)
From the right
AD BD 1 h = √ 3 70 + x
BDA,
tan 30° =
eq :1 => eq :2
√3 (
1 ) √3
------ (eq :2) h ( ) x = h ( ) 70 + x
70 + x x => 2x = 70 => x = 35 => 3 =
Substituting this value of x in eq :1, we have
h 35 => h = 35√ 3 = 35 × 1.73 = 60.55 ≈ 60.6
√3=
I mportant Concepts and Formulas - A lgebra I.
Basic A lgebraic Formulas
1.
a(b + c) = ab + ac
2.
(a + b) = a 2 + 2ab + b2
3.
(a − b) = a 2 − 2ab + b2
4.
(a + b) + (a − b) = 2(a 2 + b2 )
5.
(a + b) = a 3 + 3a 2 b + 3a b + b = a 3 + 3ab(a + b) + b
6.
(a − b) = a 3 − 3a 2 b + 3a b2 − b3 = a 3 − 3ab(a − b) − b3
7.
a 2 − b2 = (a − b)(a + b)
8.
a 3 + b3 = (a + b)(a 2 − ab + b2 )
9.
a 3 − b3 = (a − b)(a 2 + ab + b2 )
(Distributive Law)
2
2
2
2
3
2
3
3
3
10.
a n − bn = (a − b)(a n−1 + a n−2 b + a n−3 b2 + ... + bn−1 )
11.
2 2 (a + b + c) = a 2 + b + c2 + 2(ab + bc + ca)
12.
(a + b)(a + c) = a 2 + (b + c)a + bc
1.
a n = a. a. a ... (n times)
2.
a m . a n . . . a p = a m+n+...+p
3.
a m . a n = a m+n
4.
am = a m−n an
5.
(a m ) = amn = (a n )
6.
(ab) n = an bn
7.
a ( ) b
8.
a −n =
9.
an =
n
n
=
m
an bn
1 an
1 a −n
10.
a 0 = 1 where a ∈ R, a ≠ 0
11.
q p a p/q = √ −− a
12.
if a m = a n where a ≠ 0 and a ≠ ±1, then m = n
13.
if a n = bn where n ≠ 0, then a = ±b
2
1.
(x + a ) = x 2 + 2ax + a 2
2.
(x − a ) = x 2 − 2ax + a 2
3.
(x + a)(x + b) = x 2 + (a + b)x + ab
4.
x 2 − a 2 = (x − a)(x + a)
5.
(x n + 1) is completely divisible by (x + 1) when n is odd
6.
(x n + a n ) is completely divisible by (x + a) when n is odd
7.
(x n − a n ) is completely divisible by (x − a) for every natural number n
8.
(x n − a n ) is completely divisible by (x + a) when n is even
2
Binom ial Theorem
If a and b are any real numbers and n is a positive integer,
(a + b)
n
= a n + na n−1 b +
n(n − 1) n−2 2 n(n − 1)(n − 2) a b + 2! 3!
a n−3 b3 + ⋯ +
n(n − 1)(n − 2) ⋯ (n − r + 1) r!
an−r br + ⋯ + bn
n n n n n n = ( ) a n + ( ) a n−1 b + ( ) a n−2 b2 + ( ) an−3 b3 + ⋯ + ( ) a n−r br + ⋯ + ( ) bn 0 1 2 3 r n n n n−r r =∑ ( )a b r r=0
n where ( ) is known as Binomial Coefficient and is defined by r n n! ( )= r (r!)(n − r)!
=
n(n − 1)(n − 2) ⋯ (n − r + 1) r!
n where r = 1,2,...,n and ( ) = 1 0
Binomial Coefficient also occurs in many other mathematical areas than algebra, especially in combinatorics where
( nr ) represents the number of combinations of n distinct things taking r at a time and is denoted by nCr or C(n,r). The properties of binomial coefficients have led to extending its meaning beyond the basic case where n and r are nonnegative integers with r ≤ n; such expressions are also called binomial coefficients.
Binom ial Ex pansions 0
(a + b) = 1 1
(a + b) = a + b 2
(a + b) = a 2 + 2ab + b2 3
(a + b) = a 3 + 3a 2 b + 3a b2 + b3 4
2
3
4
(a + b) = a 4 + 4a 3 b + 6 a2 b + 4a b + b ... and so on
Binom ial Series One of Newton's achievement was to extend the Binomial Theorem to the case where n can be any real number. If n is any real number and -1 < x < 1, then
(1 + x)
n
= 1 + nx +
n(n − 1) n(n − 1)(n − 2) x2 + 2! 3!
x3 + ⋯
n n n n = ( ) x 0 + ( ) x 1 + ( ) x2 + ( ) x 3 + ⋯ 0 1 2 3 ∞ n = ∑ ( ) xr r r=0
n(n − 1)(n − 2) ⋯ (n − r + 1) n where ( ) = r r!
II.
n where r ≥ 1 and ( ) = 1 0
Surds and I mportant P roperties i.
Surds Let
a be a rational number and n be a positive integer such that √n a is irrational. Then √n a is called a surd of order n.
Examples :
√ 3 is a surd of order 2 √ 7 is a surd of order 3 3
4√ 3 is a surd of order 2 Please note that numbers like
3 −− √ 9 , √ 27 etc are not surds because they are not irrational numbers
Every surd is an irrational number. But every irrational number is not a surd. (eg : are irrational numbers.)
ii.
Quadratic Surds
π , e etc are not surds though they
A surd of order 2 is called a quadratic surd Examples :
iii.
√ 2, √ 3, (3 + √ 5), etc.
Rules of Surds
1.
2.
−− − −−− √ m × √ n = √ mn − −− − m √− m =√ − n √n
3.
1 −−−−−−−−−−−−−−− 1− n −−−−−−−−−−−− −−−−−−−−−− √ a. √ a. √ a ⋯ n times = a 2
4.
−−−−−−−−−−− − −−−−−−−− −−−−−− √ a. √ a. √ a ⋯ ∞ = a
5.
if
1.
The conjugate surd of
2.
if √ b and √ d are quadratic surds and if then a = c and b = d
−−−−−−−−−−−−−− −−−−−−−−− −− a−−−− ⋯ ∞− = p, then p(p − 1) = a √ a +√ a +√−
√ a + √ b = ±(√ a − √ b) a +√ b= c+√ d
,
i.e., rational part on the left side = rational part on the right side. and irrational part on the left side = irrational part on the right side.
3.
III.
if √ b and √ d are quadratic surds and if then a = 0 and b = d
a +√ b= √ d ,
Quadratic Equations and How to Solve Quadratic Equations A . Quadratic Equations A quadratic equation is a second degree univariate polynomial equation.
A quadratic equation can be written as (general form or standard form of quadratic equation)
a x 2 + bx + c = 0
Example :
where x is a variable, a, b and c are constants and
x 2 + 5x + 6 = 0
(Please note that if a=0, equation becomes a linear equation) The solutions of a quadratic equation are called its roots.
B. How to Solve Quadratic Equations
a≠0
There are many methods to solve Quadratic equations. Quadratic equations can be solved by factoring, completing the square, graphing, Newton's method, and using the quadratic formula. We can go through some of the popular methods for solving quadratic equations here. 1.
How to Solve Quadratic Equations By Factoring
Step 1: Write the quadratic equation in standard form. i.e. , bring all terms on the left side of the equal sign and 0 on the right side of the equal sign. Step 2: Factor the left side of the equation Step 3: Equate each factor to 0 and solve the equations
Factoring is one of the fastest ways for solving quadratic equations. The concept will be clear from the following examples.
Example1 : Solve the Quadratic Equation
x 2 + 4x − 11 = x − 1
Step 1 : Write the quadratic equation in standard form.
x 2 + 4x − 11 = x − 1 ⇒ x 2 + 4x − 11 − x + 1 = 0 ⇒ x 2 + 3x − 10 = 0 Step 2 : Factor the left side of the equation. We know that Assume
x 2 + (a + b)x + ab = (x + a)(x + b)
. We will use the same concept for factoring.
x 2 + 3x − 10 = x 2 + (a + b)x + ab
We need to find out a and b such that a + b = 3 and ab = -10 a = +5 and b = -2 satisfies the above condition. Hence
x 2 + 3x − 10 = (x + 5)(x − 2)
x 2 + 3x − 10 = 0 ⇒ (x + 5)(x − 2) = 0 Step 3 :Equate each factor to 0 and solve the equations (x + 5)(x - 2) = 0 => (x + 5) = 0 or (x - 2) = 0 => x = -5 or 2 Hence, the solutions of the quadratic equation
x 2 + 4x − 11 = x − 1
are x = -5 and x = 2.
(In other words, x = -5 and x = 2 are the roots of the quadratic equation
Example2 : Solve the Quadratic Equation
x 2 − 7x + 10 = 0
This equation is already in the standard form. Hence let's go to step 2
x 2 + 4x − 11 = x − 1
Step 2 : Factor the left side of the equation. Here we need to find out a and b such that a + b = -7 and ab = +10 a = -5 and b = -2 satisfies the above condition. Hence
Hence,
x 2 − 7x + 10 = (x − 5)(x − 2) x 2 + 3x − 10 = 0
⇒ (x − 5)(x − 2) = 0 Step 3 :Equate each factor to 0 and solve the equations (x - 5)(x - 2) = 0 => (x - 5) = 0 or (x - 2) = 0 => x = 5 or 2 Hence, the solutions of the quadratic equation
x 2 − 7x + 10 = 0
are x = 5 and x = 2.
(In other words, x = 5 and x = 2 are the roots of the quadratic equation
Example3 : Solve the Quadratic Equation
x 2 − 7x + 10 = 0
)
3x 2 − 14x + 8 = 0
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation. Here we need to follow a slightly different approach for factoring because here the coefficient of x2 ≠ 1 (whereas In example 1 and example 2, the coefficient of x2 was 1) Hence, to factor
3x 2 − 14x + 8 , we need to do the followings
1. Product of the second degree term and the constant. i.e.,
3x 2 × 8 = 24x 2
2. We got product as 24x2 and have middle term as -14x. From this , it is clear that we can take -12x and -2x such that their sum is -14x and product is 24x2 Hence,
3x 2 − 14x + 8 = 0
Hence,
3x 2 − 14x + 8 = 0
= 3x
2
− 12x − 2x + 8 = 3x(x − 4) − 2(x − 4) = (x − 4)(3x − 2)
⇒ (x − 4)(3x − 2) = 0 Step 3 :Equate each factor to 0 and solve the equations (x - 5)(x - 2) = 0 => (x - 4) = 0 or (3x - 2) = 0 => x = 4 or
2 . 3
Hence, the solutions of the quadratic equation (In other words, x = 4 and x =
3x 2 − 14x + 8 = 0
are x = 4 and x =
2 3
2 2 are the roots of the quadratic equation 3x − 14x + 8 = 0 3
Example4 : Solve the Quadratic Equation
6x 2 − 17x + 12 = 0
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation. coefficient of x2 ≠ 1 Hence, to factor
6x 2 − 17x + 12 , we need to do the followings
1. Product of the second degree term and the constant. i.e.,
6x 2 × 12 = 72x 2
2. We got product as 72x2 and have middle term as -17x. From this , it is clear that we can take -8x and -9x such that their sum is -17x and product is 72x2 Hence,
6x 2 − 17x + 12 = 6x 2 − 8x − 9x + 12 = 2x(3x − 4) − 3(3x − 4) = (3x − 4)(2x − 3)
Hence,
6x 2 − 17x + 12 = 0
⇒ (3x − 4)(2x − 3) = 0 Step 3 :Equate each factor to 0 and solve the equations (3x - 4)(2x - 3) = 0 => (3x - 4) = 0 or (2x - 3) = 0 => x =
4 3 or . 3 2
Hence, the solutions of the quadratic equation (In other words, x =
6x 2 − 17x + 12 = 0
are x =
4 3 and x = . 3 2
4 3 2 and x = are the roots of the quadratic equation 6x − 17x + 12 = 0 3 2
(Special Factorization Examples) Example5 : Solve the Quadratic Equation
x2 − 9 = 0
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation. Here,x
2
− 9 is in the form a 2 − b2 where a = x and b = 3
We know that
a 2 − b2 = (a − b)(a + b)
We will use the same concept for factoring this.
x 2 − 9 = x 2 − 32 = (x − 3)(x + 3) Hence,
x2 − 9 = 0
⇒ (x − 3)(x + 3) = 0 Step 3 :Equate each factor to 0 and solve the equations (x - 3)(x + 3) = 0
=> x = 3 or -3 Hence, the solutions of the quadratic equation
x2 − 9 = 0
are x = 3 and x = -3
(In other words, x = 3 and x = -3 are the roots of the quadratic equation
Example6 : Solve the Quadratic Equation
x2 − 9 = 0
25x 2 − 16 = 0
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation. 2
25x 2 − 16 = (5x) − 42 = (5x − 4)(5x + 4) Hence,
25x 2 − 16 = 0
⇒ (5x − 4)(5x + 4) = 0 Step 3 : Equate each factor to 0 and solve the equations (5x - 4)(5x + 4) = 0 =>5x = 4 or -4
x=
4 −4 . or 5 5
Hence, the solutions of the quadratic equation (In other words, x =
25x 2 − 16 = 0
are x =
4 −4 and x = 5 5
4 −4 2 and x = are the roots of the quadratic equation 25x − 16 = 0 5 5
Example7 : Solve the Quadratic Equation
x 2 + 6x + 9 = 0
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation. Here,x 2
+ 6x + 9
We know that
is in the form a2
+ 2ab + b2
a 2 + 2ab + b2 = (a + b)
2
where a = x and b = 3
We will use the same concept for factoring this.
x 2 + 6x + 9 = x 2 + 2 × x × 3 + 32 = (x + 3 )2 Hence,
x 2 + 6x + 9 = 0
⇒ (x + 3) 2 = 0 Step 3 :
(x + 3) 2 = 0 => x + 3 = 0 => x = -3 Hence, the solution of the quadratic equation
x 2 + 6x + 9 = 0
is x = -3
(In other words, x = -3 is the root of the quadratic equation
Example8 : Solve the Quadratic Equation
x 2 + 6x + 9 = 0
x 2 − 6x + 9 = 0
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation. Here,x
2
− 6x + 9
We know that
2
is in the form a
− 2ab + b2
a 2 − 2ab + b2 = (a − b) 2
where a = x and b = 3
We will use the same concept for factoring this.
x 2 − 6x + 9 = x 2 − 2 × x × 3 + 32 = (x − 3 )2 Hence,
x 2 + 6x + 9 = 0 2
⇒ (x − 3) = 0 Step 3 : 2
(x − 3) = 0 => x - 3 = 0 => x = 3 Hence, the solution of the quadratic equation
x 2 − 6x + 9 = 0
(In other words, x = 3 is the root of the quadratic equation
2.
is x = 3
x 2 − 6x + 9 = 0
How to Solve Quadratic Equations By Quadratic Formula
Quadratic F orm ula Consider a quadratic equation Its solutions can be given by
a x 2 + bx + c = 0
wherea
≠0
−−−−−−− −b ± √ b2 − 4ac x= 2a Example1 : Solve the Quadratic Equation
x 2 + 4x − 11 = x − 1 ⇒ x 2 + 4x − 11 − x + 1 = 0 ⇒ x 2 + 3x − 10 = 0
x 2 + 4x − 11 = x − 1
−−−−−−− −b ± √ b2 − 4ac x= 2a =
−3 ± 7 2
=
4 −10 or 2 2
=
−−−−−−−−−−−−−−− −3 ± √ 32 − 4 × 1 × (−10)
=
2×1
9−−− + 40− −3 ± √ − 2
= 2 or − 5 Hence, the solutions of the quadratic equation
x 2 + 4x − 11 = x − 1
are x = 2 and x = -5
(In other words, x = 2 and x = -5 are the roots of the quadratic equation
Example2 : Solve the Quadratic Equation
−−−−−−− −b ± √ b2 − 4ac x= 2a =
7±√ 9 7±3 = 2 2
=
10 4 or 2 2
=
x 2 + 4x − 11 = x − 1
x 2 − 7x + 10 = 0
−−−−−−−−−−−−−− − 2 7 ± √ (−7) − 4 × 1 × 10
=
2×1
−−−−−− 7 ± √ 49 − 40 2
= 5 or 2 Hence, the solutions of the quadratic equation
x 2 − 7x + 10 = 0
are x = 5 and x = 2.
(In other words, x = 5 and x = 2 are the roots of the quadratic equation
Example3 : Solve the Quadratic Equation
−−−−−− − 2 −b ± √ b − 4ac x= 2a
=
=
−−− 14 ± √ 100 14 ± 10 = 6 6
=
24 4 or 6 6
= 4 or
3x 2 − 14x + 8 = 0
−−−−−−−−−−−−−− − 2 14 ± √ (−14) − 4 × 3 × 8 2×3
x 2 − 7x + 10 = 0)
=
−−−−−−− 14 ± √ 196 − 96 6
2 3
Hence, the solutions of the quadratic equation (In other words, x = 4 and x =
3x 2 − 14x + 8 = 0
are x = 4 and x =
2 3
2 are the roots of the quadratic equation 3x 2 − 14x + 8 = 0 3
Example4 : Solve the Quadratic Equation
6x 2 − 17x + 12 = 0
−−−−−−− −b ± √ b2 − 4ac x= 2a =
17 ± √ 1 17 ± 1 = 12 12
=
18 16 or 12 12
=
3 4 or 2 3
=
−−−−−−−−−−−−−−− − 2 17 ± √ (−17) − 4 × 6 × 12 2×6
Hence, the solutions of the quadratic equation (In other words, x =
=
−− − −6 ± √ −4 2
=
−−−−−−−− 17 ± √ 289 − 288 12
6x 2 − 17x + 12 = 0
are x =
4 3 and x = . 3 2
4 3 2 and x = are the roots of the quadratic equation 6x − 17x + 12 = 0 3 2
Example5 : Solve the Quadratic Equation
−−−−−−− −b ± √ b2 − 4ac x= 2a
=
=
−−−−−−−−−−−−− − 2 −6 ± √ (6) − 4 × 1 × 10 2×1
x 2 + 6x + 10 = 0 −−−−−− −6 ± √ 36 − 40 = 2
−6 ± 2i 2
= −3 ± i = (−3 + i) or (−3 − i) Hence, the solutions of the quadratic equation
x 2 + 6x + 10 = 0
are x =(-3 + i) and x = (-3 - i)
(In other words, x =(-3 + i) and x = (-3 - i) are the roots of the quadratic equation
Example5 : Solve the Quadratic Equation
−−−−−−− −b ± √ b2 − 4ac x= 2a =
=
−−−−−−−−−−−−− − 2 −6 ± √ (−6) − 4 × 1 × 9 2×1
x 2 + 6x + 9 = 0 =
−−−−−−− −6 ± √ (36 − 36 2×1
−6 ± √ 0 −6 = 2 2
= −3 Hence, the solutions of the quadratic equation (In other words,
x 2 + 6x + 9 = 0
is
x = −3
x = −3 is the roots of the quadratic equation x 2 + 6x + 9 = 0
Example6 : Solve the Quadratic Equation
x2 − 9 = 0
x 2 + 6x + 10 = 0
−−−−−−− −b ± √ b2 − 4ac x= 2a =±
=
−−−−−−−−−−−−−−− 0 ± √ (0) 2 − 4 × 1 × (−9) 2×1
=
±√ −− 36 2
6 2
= 3 or − 3 Hence, the solutions of the quadratic equation
x2 − 9 = 0
are x = 3 and x = -3
(In other words, x = 3 and x = -3 are the roots of the quadratic equation
x2 − 9 = 0
C. How to find out the number of solutions(roots) of a quadratic equation quickly?
To find out the number of solutions (roots) of a quadratic equation 2
discriminant, D. The discriminant, D = b
– 4ac
a x 2 + bx + c = 0
quickly, we need to find out the
If the discriminant is positive, there will be two real solutions for the quadratic equation. If the discriminant is 0, there will be one real solution (it repeats itself) for the quadratic equation. If the discriminant is negative, there will be two complex solutions for the quadratic equation.
Ex am ple1 : Find out the number of roots for 2
Discriminant, D = b D > 0. Hence
x 2 − 7x + 12 = 0
– 4ac = (−72 ) − (4 × 1 × 12) = 49 − 48 = 1.
x 2 − 7x + 12 = 0
will have two real roots.
Let's verify if this is correct by solving the quadratic equation.
x 2 − 7x + 12 = 0 ⇒ (x − 3)(x − 4) = 0 ⇒ x = 3 or 4. Yes, two real roots we got.
Ex am ple2 : Find out the number of roots for 2
Discriminant, D = b D = 0. Hence
x 2 − 4x + 4 = 0
– 4ac = (−42 ) − (4 × 1 × 4) = 16 − 16 = 0.
x 2 − 4x + 4 = 0
will have one real root (it repeats itself)
Let's verify if this is correct by solving the quadratic equation.
x 2 − 4x + 4 = 0 ⇒ (x − 2) 2 = 0 ⇒ x = 2 Yes, real root (it repeats itself).
Ex am ple3 : Find out the number of roots for 2
Discriminant, D = b D < 0. Hence
x2 + 1 = 0
– 4ac = (02 ) − (4 × 1 × 1) = 0 − 4 = −4
x2 + 1 = 0
will have two complex roots
Let's verify if this is correct by solving the quadratic equation.
x2 + 1 = 0
−−−−−−− −b ± √ b2 − 4ac x= 2a
−−−−−−−−−−− −−−− − 0 ± √ 02 − 4 × 1 × 1 √ −− −4 ±2 i √ 0−4 = = = = =±i 2×1 2 2 2
Yes, we got two complex roots, +i and -i
D. Sum and P roducts of the roots of a quadratic equation
Let
x 1 and x 2 are the roots of a quadratic equation a x 2 + bx + c = 0 where a ≠ 0
Sum of the roots of the quadratic equation,
x1 + x2 =
Product of the roots of the quadratic equation,
−b a
x1 × x2 =
c a
Ex am ple 1: Find the sum and product of the roots of the quadratic equation sum of the roots=
−b 7 = =7 a 1
product of the roots=
c 10 = = 10 a 1
Ex am ple 2: Find the sum and product of the roots of the quadratic equation sum of the roots=
x 2 − 7x + 10 = 0
−b −3 = a 2
product of the roots=
c 1 = a 2
2x 2 + 3x + 1 = 0
I mportant Concepts and Formulas - Complex Numbers I.
Basics of Complex Numbers
Definition of i 1.
−− − i = √ −1
i 2 = −1 i 3 = −i 2.
i4 = 1
A complex number is any number which can be written as −− − and i = √ −1
a + ib where a and b are real numbers
a is the real part of the complex number and b is the imaginary part of the complex number. 3.
Example for a complex number : 9 + i2
if z = of z. 4.
a + ib
is a complex number, a is called the real part of z and b is called the imaginary part
It can be represented as Re(z) = a and Im(z) = b
Conjugate of the complex number
¯4¯¯+ ¯¯¯i2 ¯¯¯ = 4 − i2
z = x + iy
¯¯¯i2 ¯¯¯ = 4¯¯− and ¯
can be defined as
4 + 2i
5.
Example :
6.
if the complex number
a + ib = 0 , then a = b = 0
7.
if the complex number
a + ib = x + iy
, then
a = x and b = y
z¯ = x − iy
−−−−− 2 2−
if x + iy is a complex numer, then the non-negative real number √ x + y is the modulus (or absolute value or magnitude) of the complex number x + iy . It can be denoted as 8.
−−−−−− | x + iy | = √ x 2 + y 2
9.
eiθ = cos θ + i sin θ
(Note that modulus is a non-negative real number )
(Euler Formula)
Rectangular (Cartesian) F orm of Com plex Num bers 10.
A complex number when written in the form a + ib , it is in the rectangular (Cartesian) form
11.
eiθ = cos θ + i sin θ
(Euler Formula)
Cube Roots of Unity = (1)
= 1, 12.
1/3
−1 + i√ 3 −1 − i√ 3 , 2 2
= 1, w, w2 where w =
−1 + i√ 3 2
Properties of Cube Roots of Unity i. Cube Roots of Unity are in G.P. ii. Each complex cube root of unity is the square of the other complex cube root of unity. Example :
iii.
w=
−1 + √ 3i −1 − √ 3i , w2 = 2 2
1 + w + w2 = 0
iv. Product of all cube roots of unity = 1. i.e., 13.
v.
1 1 = w2 and 2 = w w w
w3 = 1
14.
II.
Fourth Roots of Unity ,
(1)
1/4
are +1, -1, +i, -i
P olar and Exponential Forms of Complex Numbers Polar and Exponential Forms are very useful in dealing with the multiplication, division, power etc. of complex numbers.
Polar F orm of a Com plex Num ber
A complex number
z = x + iy
can be expressed in polar form as
z = r∠θ = r cisθ = r(cos θ + i sin θ)
(Please not that θ can be in degrees or radians)
−−−−−−
where r = |z| = √ x 2 + y 2 (note that r ≥ 0 and and r = modulus or absolute value or magnitude of the complex number)
y θ = arg z = tan −1 ( ) x
(θ denotes the angle measured counterclockwise from the positive
real axis.) θ is called the argument of z. it should be noted that 2π n + θ is also an argument of z where n = ⋯ − 3, −2, −1, 0, 1, 2, 3, ⋯ . Note that while there can be many values for the argument, we will normally select the smallest positive value) Please note that we need to make sure that θ is in the correct quadrant. i.e., θ should be in the same quadrant where the complex number is located in the complex plane. This will be clear from the next topic where we will go through various examples to convert complex numbers between polar form and rectangular form. It is strongly recommended to go through those examples to get the concept clear.
x = r cos θ y = r sin θ If −π < radian)
θ ≤ π,
θ
is called as principal argument of z
(In this statement, θ is expressed in
Ex ponential form of a Com plex Num ber We
have
already
z = r(cos θ + i sin θ)
seen that
in
polar form ,
. By Euler's Formula, we have
iθ
a
complex number can be
expressed as
e = cos θ + i sin θ
Hence, we can express a complex number in Exponential form as radians)
z = r eiθ
(Note that θ is in
While there may be many values of θ satisfying this, we will normally select the smallest positive value.
From Cartesian form, complex form and exponential form, a complex number can be expressed as
z = r cis θ = r∠θ = r(cos θ + i sin θ) = r eiθ
Note that radians and degrees are two units for measuring angles.
360° = 2π radian
III.
How to Convert Complex Numbers from R ectangular Form to P olar Form and Exponential Form
1.
Example 1 : Convert
z = 1 + i√ 3
to polar form
−−−−−−−−2− −−−−−− r = |z| = √ x 2 + y 2 = √ 12 + (√ 3) = √ − 1−− + 3− = √ 4 = 2 y π √3 ) = tan −1 (√ 3) = arg z = θ = tan −1 ( ) = tan −1 ( x 1 3 Here the complex number is in first quadrant in the complex plane. The angle we got, the first quadrant .Hence we select this value. Hence, the polar form is
π is also in 3
π π π z = 2∠( ) = 2 [cos ( ) + i sin ( )] 3 3 3
iπ . 3 Similarly we can write the com plex num ber in ex ponential form as z = r eiθ = 2e .
(Please
note
that
all
π 2π n + where n = 0, ±1, ±2, ⋯ 3 exponential forms also)
possible
values
of
the
argument,
arg
z
are
. Accordingly we can get other possible polar forms and
2.
Example 2 : Convert
z = −1 + i√ 3
to polar form
−−−−−−−−−− − −−−−−− 2 2 1−− + 3− = √ 4 = 2 r = |z| = √ x 2 + y 2 = √ (−1) + (√ 3) = √ − y √3 ) = tan −1 (−√ 3) arg z = θ = tan −1 ( ) = tan −1 ( x −1 Here
−
π 3
is one value of θ which meets the condition
θ = tan −1 (− √ 3) . But it is in fourth
quadrant. We know that θ should be in second quadrant because the complex number is in second quadrant in the complex plane.
π 2π +π = which is in second quadrant and also meets the condition 3 3 (− √ 3) . Hence we take that value.
θ=−
Hence
θ = tan −1
z = 2∠(2
Hence, the polar form is
π 2π π ) = 2 [cos ( ) + i sin (2 )] 3 3 3
i 2π iθ 3 Similarly we can write the com plex num ber in ex ponential form as z = r e = 2e (Please
note
that
all
possible
2π 2π n + where n = 0, ±1, ±2, ⋯ 3
values
of
the
argument,
arg
z
are
. Accordingly we can get other possible polar forms and
exponential forms also)
3.
Example 3 : Convert
z = −1 − i√ 3
to polar form
−−−−−−−−−−−− − −−−−−− 2 2 −−−− r = |z| = √ x 2 + y 2 = √ (−1) + (−√ 3) = √ 1+3 =√ 4 = 2 y −√ 3 ) = tan −1 (√ 3) arg z = θ = tan −1 ( ) = tan −1 ( x −1 Here
π is one value of θ which meets the condition θ = tan −1 (√ 3) . But it is in first quadrant. 3
We know that θ should be in third quadrant because the complex number is in third quadrant in the complex plane. Hence
θ=
π 4π +π = 3 3
which is in third quadrant and also meets the condition
θ = tan −1 (√ 3) .
Hence we take that value. Hence, the polar form is
z = 2∠(4
π 4π π ) = 2 [cos ( ) + i sin (4 )] 3 3 3
i 4π 3 Similarly we can write the com plex num ber in ex ponential form as z = r eiθ = 2e
(Please
note
that
all
possible
4π 2π n + where n = 0, ±1, ±2, ⋯ 3
values
of
the
argument,
arg
z
are
. Accordingly we can get other possible polar forms and
exponential forms also)
4.
Example 4 : Convert
z = 1 − i√ 3
to polar form
−−−−−−−−−− − −−−−− 2 2 2 + y 2− √ x (1 ) + (− √ 3 ) 1−− + 3− = √ 4 = 2 r = |z| = √ = =√− y −√ 3 ) = tan −1 (−√ 3) arg z = θ = tan −1 ( ) = tan −1 ( x 1 Here
−
π is one value of θ which meets the condition and also in the fourth quadrant. The 3
complex number is also in fourth quadrant. However we will normally select the smallest positive value for θ. Hence
θ=−
π 5π + 2π = 3 3
which meets the condition
θ = tan −1 (√ 3)
and also is in the fourth
quadrant. Hence we take that value. Hence, the polar form is
z = 2∠(
5π 5π 5π ) = 2 [cos ( ) + i sin ( )] 3 3 3
i 5π iθ 3 Similarly we can write the com plex num ber in ex ponential form as z = r e = 2e (Please
note
that
all
possible
5π 2π n + where n = 0, ±1, ±2, ⋯ 3
values
of
the
argument,
arg
z
are
. Accordingly we can get other possible polar forms and
exponential forms also)
5.
Example 5 : Convert
z= 8
to polar form
−−−−−−−− − −−−2 −−−−− 2 2 2 2− √ √ x y ) ) r = |z| = √ + = (8 + (0 = (8) = 8 Here the complex number lies in the positive real axis . Hence
θ=0 .
y 0 arg z = θ = tan −1 ( ) = tan −1 ( ) = tan −1 0 x 8 Hence, the polar form is
z = 8∠0 = 8 (cos 0 + i sin 0)
Similarly we can write the com plex num ber in ex ponential form as
(Please
note
that
all
possible
2π n + 0 = 2π n where n = 0, ±1, ±2, ⋯
z = r eiθ = 8e0i
values of the argument, arg z are . Accordingly we can get other possible polar forms and
exponential forms also)
6.
Example 6 : Convert
z = −8
to polar form
−−−−−−−−− − −−−−2− −−−−−− 2 2 r = |z| = √ x 2 + y 2 = √ (−8) + (0) = √ (−8) = 8 Here the complex number lies in the negative real axis . Hence Hence, the polar form is
θ=π .
z = 8∠π = 8 (cos π + i sin π)
Similarly we can write the com plex num ber in ex ponential form as
(Please
note
that
all
2π n + π where n = 0, ±1, ±2, ⋯
z = r eiθ = 8eiπ
possible values of the argument, arg z are . Accordingly we can get other possible polar forms and
exponential forms also)
7.
Example 7 : Convert
z = i8
to polar form
−−−−−−−− − −−−2 −−−−−− 2 2 r = |z| = √ x 2 + y 2 = √ (0) + (8) = √ (8) = 8 Here the complex number lies in the positive imaginary axis . Hence Hence, the polar form is
z = 8∠
θ=
π . 2
π π π = 8 [cos ( ) + i sin ( )] 2 2 2
iπ 2 Similarly we can write the com plex num ber in ex ponential form as z = r eiθ = 8e (Please
note
that
all
π 2π n + where n = 0, ±1, ±2, ⋯ 2
possible
values
of
the
argument,
arg
z
. Accordingly we can get other possible polar forms and
exponential forms also)
8.
Example 8 : Convert
z = −i8
to polar form
−−−−−−−−− − −−−−2− −−−−−− 2 2 r = |z| = √ x 2 + y 2 = √ (0) + (−8) = √ (−8) = 8 Here the complex number lies in the negavive imaginary axis . Hence
θ=−
However we will normally select the smallest positive value for θ. Hence Hence, the polar form is
z = 8∠
are
3π 3π 3π = 8 [cos ( ) + i sin ( )] 2 2 2
π . 2
θ=−
π 3π + 2π = 2 2
i3π 2 iθ Similarly we can write the com plex num ber in ex ponential form as z = r e = 8e (Please
note
that
all
possible
3π 2π n + where n = 0, ±1, ±2, ⋯ 2
values
of
the
argument,
arg
z
are
. Accordingly we can get other possible polar forms and
exponential forms also)
IV.
How to Convert Complex Numbers from P olar Form to R ectangular (Cartesian) Form Examples 1.
Convert the complex number
x = r cos θ = 2 cos y = r sin θ = 2 sin
2 [cos (
5π 5π ) + i sin ( )] 3 3
to rectangular (Cartesian) form
5π 1 = 2× = 1 3 2
5π √3 ) = −√ 3 = 2 × (− 3 2
z = x + iy = 1 − i√ 3
2.
Convert the complex number
x = r cos θ = 8 cos y = r sin θ = 8 sin
π π 8 [cos ( ) + i sin ( )] 2 2
to rectangular (Cartesian) form
π = 2×0 = 0 2
π = 8×1 = 8 2
z = x + iy = 0 + 8i = 8i V.
How to Convert Complex Numbers from Exponential Form to R ectangular (Cartesian) Form Examples
1.
i2π 3 Convert the complex number 2e x = r cos θ = 2 cos y = r sin θ = 2 sin
2π 1 = 2 × (− ) = −1 3 2
2π √3 = 2× = √3 3 2
to rectangular (Cartesian) form
z = x + iy = −1 + i√ 3
2.
iπ 3 Convert the complex number 2e x = r cos θ = 2 cos y = r sin θ = 2 sin
to rectangular (Cartesian) form
π 1 = 2× = 1 3 2
π √3 = 2× =√3 3 2
z = x + iy = 1 + i√ 3 VI.
A rithmetical Operations of Complex Numbers Addition, subtraction, multiplication and division can be carried out on complex numbers in either rectangular form or polar form. Addition and subtraction of complex numbers is easy in rectangular form. Multiplication and division of complex numbers is easy in polar form.
1.
A ddition of Complex Numbers To add complex numbers, add their real parts and add their imaginary parts.
(a + ib) + (c + id) = (a + c) + i(b + d) Example : Find (9 + i2) + (8 + i6) (9 + i2) + (8 + i6) = (9 + 8) + i(2 + 6) = 17 + i8
2.
Subtraction of Complex Numbers To subtract complex numbers, subtract their real parts and subtract their imaginary parts.
(a + ib) - (c + id) = (a - c) + i(b - d) Example : Find (9 + 2i) - (8 + 6i) (9 + 2i) - (8 + 6i) = (9 - 8) + i(2 - 6) = 1 - i4
3.
Multiplication of Complex Numbers A. Multiplication of Com plex Num bers in Rectangular F orm
(a + ib)(c + id) = ac + iad + ibc + i2bd = ac + iad + ibc - bd = (ac - bd) + i(ad + bc) Example : Find (9 + 2i)(8 - 6i)
(9 + i2)(8 − i6) = 72 − i54 + i16 − i 2 12 = 72 − i(54 − 16) + 12 = 84 − i38
B. Multiplication of Com plex Num bers in Polar F orm
r 1 ∠θ1 × r 2 ∠ θ2 = r 1 r 2 ∠ (θ1 + θ2 ) Example : Find
3∠30° × 4∠40°
3∠30° × 4∠40° = (3 × 4) ∠ (30° + 40°) = 12∠70° 4.
Division of Complex Numbers A. Div ision of Com plex Num bers in Rectangular F orm
(a + ib) (a + ib) (c − id) (ac + bd) − i(ad − bc) = × = (c + id) (c + id) (c − id) c2 + d 2 (We multiplied denominator and numerator with the conjugate of the denominator to proceed)
Example : Find
(9 + 2i) (8 − 6i)
(9 + 2i) (9 + 2i)(8 + 6i) = (8 − 6i) (8 − 6i)(8 + 6i)
=
72 + 54i + 16i − 12 64 + 36
B. Div ision of Com plex Num bers in Polar F orm
r 1 ∠ θ1 r1 = ∠ (θ1 − θ2 ) r 2 ∠ θ2 r2
Example : Find
5∠135° 4∠75°
5∠135° 5 5 = ∠ (135° − 75°) = ∠60° 4 4 4∠75°
=
60 + 70i = .6 + .7i 100
VII.
P ow ers and R oots of Complex Numbers
De'Moiv re's Theorem If n is any integer, then
n
(cos θ + i sin θ) = cos nθ + i sin nθ
From De'Moivre's formula, it is clear that for any complex number
z = r cis θ = r∠θ = r(cos θ + i sin θ) = r eiθ
,
z n = r n ∠ (θ × n) = r n (cos nθ + i sin nθ) = r n einθ A . Examples to find out P ow ers of Complex Numbers
1.
Find
(−1 + √ 3 i)
12
−−−−−−−−−−− 2− 2 r = √ (−1) + (√ 3) = √ − 1−− + 3− = √ 4 = 2 θ = tan −1 (
2π √3 ) = tan −1 (−√ 3) = tan −1 ( ) −1 3
(∵The
complex
number
is
in
second
quadrant) Hence,
−1 + √ 3 i = 2 [cos (
⇒ (−1 + √ 3 i)
12
= [2 (cos
2π π ) + i sin (2 )] 3 3 2π π + i sin 2 )] 3 3
12
= 212 [cos (
2π π × 12) + i sin (2 × 12)] 3 3
= 4096 (cos 8π + i sin 8π) = 4096 (cos 0 + i sin 0) = 4096(1 + 0) = 4096
(Note
that
θ
is
mentioned in radians)
2.
Find
(2∠135°)
5
(2∠135°) 5 = 25 (∠135° × 5) = 32∠675° = 32∠ − 45° = 32 [cos(−45°) + i sin(−45°)] = =
32 (cos(−45°) + i sin(−45°))
32 [cos(45°) − i sin(45°)] = 32 (
1 1 32 )= −i (1 − i) √2 √2 √2
degrees)
3.
Find
[4 (cos 30° + i sin 30°)]
6
(Note that θ is mentioned in
[4 (cos 30° + i sin 30°)]
6
= 46 [cos (30° × 6) + i sin (30° × 6)]
= 4096 (cos 180° + i sin 180°) = mentioned in degrees)
4.
Find
(2e0.3i )
4096 (−1 + i × 0) = 4096 × (−1) = −4096
(Note that θ is
8
(2e0.3i ) 8 = 28 e(0.3i×8) = 256e2.4i = 256 (cos 2.4 + i sin 2.4)
(Note
that
θ
is
radians)
R oots of Complex Numbers
nth roots of a complex number
wk = r 1/n [cos (
z = r(cos θ + i sin θ)
θ + 2π k θ + 2π k ) + i sin ( )] n n
(If θ is in degrees, substitute 360° for
can be given by where k = 0, 1, 2, . . . (n-1)
2π )
Examples to find out R oots of Complex Numbers
1.
Find 5 th roots of 32i
32i = 32 (cos
π π + i sin ) 2 2
(Converted to polar form, reference : click here )
The 5th roots of 32i can be given by
wk = r 1/n [cos (
1/5
= 32
=2
cos
cos
w0 = 2 cos
θ + 2π k θ + 2π k ) + i sin ( )] n n π + 2π k 2 5
π + 2π k 2 5
π +0 2 5
+ i sin
+ i sin
+ i sin
π + 2π k 2 5
π + 2π k 2 5
π +0 2 5
where k = 0, 1, 2,3 and 4
= 2 (cos
π π + i sin ) 10 10
mentioned
in
π + 2π 2 5
w1 = 2 cos
π + 4π 2 5
w2 = 2 cos
+ i sin
π + 6π 2 5
w3 = 2 cos
+ i sin
π + 8π 2 5
w4 = 2 cos
2.
+ i sin
Find cube roots of
+ i sin
π + 2π 2 5 π + 4π 2 5 π + 6π 2 5 π + 8π 2 5
= 2 (cos
π π + i sin ) = 2i 2 2
= 2 (cos
9π 9π ) + i sin 10 10
= 2 (cos
13π 13π ) + i sin 10 10
= 2 (cos
17π 17π ) + i sin 10 10
−4 − 4√ 3 i
−4 − 4√ 3i = 8 (cos 240° + i sin 240°)
(Converted to polar form, reference : click here. Here we took the angle in degrees. Remember that we can use radians or degrees ) The cube roots of
wk = r 1/n [cos ( = 81/3
[cos (
= 2 [cos (
3.
−4 − 4√ 3i can be given by
θ + 360° k θ + 360° k ) + i sin ( )] n n
240° + 360° k 240° + 360° k ) + i sin ( )] 3 3
240° + 360° k 240° + 360° k ) + i sin ( )] 3 3
where k = 0, 1 and 2
w0 = 2 [cos (
240° + 0 240° + 0 ) + i sin ( )] = 2 (cos 80° + i sin 80°) 3 3
w1 = 2 [cos (
240° + 360° 240° + 360° ) + i sin ( )] = 2 (cos 200° + i sin 200°) 3 3
w3 = 2 [cos (
240° + 720° 240° + 720° ) + i sin ( )] = 2 (cos 320° + i sin 320°) 3 3
Find cube roots of 1
1 = 8 (cos 0 + i sin 0)
(Converted to polar form,reference : click here. Here we took the angle in degrees. Remember that we can use radians or degrees ) The cube roots of 1 can be given by
wk = r 1/n [cos ( = 11/3
[cos (
θ + 360° k θ + 360° k ) + i sin ( )] n n
0° + 360° k 0° + 360° k ) + i sin ( )] 3 3
= cos (120° k) + i sin (120° k)
where k = 0, 1 and 2
w0 = cos (120° × 0) + i sin (120° × 0) = cos 0 + i sin 0 = 1 w1 = cos (120° × 1) + i sin (120° × 1) = cos 120° + i sin 120° =−
1 √ 3 −1 + i√ 3 +i = 2 2 2
w2 = cos (120° × 2) + i sin (120° × 2) = cos 240° + i sin 240° =−
1 √ 3 −1 − i√ 3 −i = 2 2 2
I mportant Concepts and Formulas - Logarithms I.
L ogarithm Basics y
if y = log b x , then b = x where log b x = y = log to the base b of x Please note that b(base) is a positive positive real number, other than 1.
if x = by , then log b x = y where log b x = y = log to the b b of x Please note that b(base) is a positive positive real number, other than 1. Ex am ple
16 = 24 (in this expression, 4 is the power or the exponent or the index and 2 is the base) Hence we can say that In other words, both
log 2 16 = 4 (i.e., log to the base 2 of 16 = 4)
16 = 24 and log2 16 = 4 are equivalent expressions.
If base = 10, then we can write log x instead of log x is called as the common logarithm of x If base = e, then we can write ln x instead of ln x is called as the natural logarithm of x
log 10 x
loge x
Please note that e is a mathematical constant which is the base of the natural logarithm. It is known as Euler's number. It is also called as Napier's constant
e= 1+
1 1 1 1 + + + + ⋯ ≈ 2.71828 1 1.2 1.2.3 1.2.3.4
ex = 1 + x +
II.
x 2 x 3 x4 + + +⋯ 2! 3! 4!
L ogarithms - I mportant P roperties
III.
1.
log b1 = 0
(∵
b0 = 1 )
2.
log bb = 1
(∵
b1 = b )
3.
y = ln x
⇒ ey = x
4.
x = ey
⇒ ln x = y
5.
x = ln ex = eln x
6.
blog b x = x
7.
log b by = y
L aw s of L ogarithms
1.
log b MN = logb M + log b N
2.
log b
3.
log b M c = c log b M
4.
log b M =
M = logb M − log b N N
1)
5.
log b a =
6. Let IV.
(where b, M, N are positive real numbers and b ≠ 1)
(where b, M, N are positive real numbers and b ≠ 1)
(where b and M are positive real numbers , b ≠ 1, c is any real number)
logk M log M ln M = = log b ln b logk b
1 log a b
(where b, k and M are positive real numbers, b ≠ 1, k ≠
(where a and are positive real numbers, a ≠ 1, b ≠ 1)
log b M = logb N . Then M = N
Mantissa and Characteristic
where b, M and N are positive real numbers and b ≠ 1.
The logarithm of a number has two parts, known as characteristic and mantissa. 1. Characteristic The internal part of the logarithm of a number is called its characteristic. Case I: When the number is greater than 1. In this case, the characteristic is one less than the number of digits in the left of the decimal point in the given number. Case II: When the number is less than 1. In this case, the characteristic is one more than the number of zeros between the decimal point and the first significant digit of the number and it is negative. Instead of -1, -2 etc. we write
¯ 1 (one bar), ¯2 (two bar), etc.
Exam ples Num ber
Characteristic
612.25
2
16.291
1
2.1854
0
0.9413
¯ 1
0.03754
¯ 2
0.00235
¯ 3
2. Mantissa The decimal part of the logarithm of a number is known is its mantissa. We normally find mantissa from the log table.
Impor tant Concepts and For mulas - Number s I.
Number Sets and P roperties of Numbers a.
Counting Num bers (Natural num bers) : 1, 2, 3 ...
b.
Whole Num bers : 0, 1, 2, 3 ...
c.
Integers : -3, -2, -1, 0, 1, 2, 3 ...
d.
Rational Num bers Rational numbers can be expressed as
11 4 −8 , , 0, etc. 2 2 11
Examples:
a where a and b are integers and b ≠ 0 b
All integers, fractions and terminating or recurring decimals are rational numbers.
e.
Irrational Num bers Any number which is not a rational number is an irrational number. In other words, an irrational number is a number which cannot be expressed as integers.
a where a and b are b
For instance, numbers whose decimals do not terminate and do not repeat cannot be written as a fraction and hence they are irrational numbers. Example :
3 π , √ 2 , (3 + √ 5) , 4√ 3 (meaning 4 × √ 3 ), √ 6 etc
Please note that the value of 70679... We cannot
f.
π = 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59230 78164 06286 20899 86280 34825 34211
π as a simple fraction (The fraction 22/7 = 3.14.... is just an approximate value of π )
Real Num bers Real numbers include counting numbers, whole numbers, integers, rational numbers and irrational numbers.
g.
Surds
a be any rational number and n be any positive integer such that √n a is irrational. Then √n a is a surd. 6 −− Example : √ 3 , √ 10 , 4√ 3 etc √ 9 √3 −− 27 Let
Please note that numbers like
,
etc are not surds because they are not irrational numbers
Every surd is an irrational number. But every irrational number is not a surd. (eg :
II.
π , e etc are not surds though they are irrational numbers.)
A ddition, Subtraction and Multiplication R ules for Even and Odd Numbers
Addition Rules for Ev en and Odd Num bers 1. The sum of any number of even numbers is always even 2. The sum of even number of odd numbers is always even 3. The sum of odd number of odd numbers is always odd
Subtraction Rules for Ev en and Odd Num bers 1. The difference of two even numbers is always even 2. The difference of two odd numbers is always even
Multiplication Rules for Ev en and Odd Num bers 1. The product of even numbers is always even 2. The product of odd numbers is always odd 3. If there is at least one even number multiplied by any number of odd numbers, the product is always even III.
Divisibility a.
Divisible By One whole number is divisible by another if the remainder we get after the division is zero. Examples
b.
i.
36 is divisible by 4 because 36 ÷ 4 = 9 with a remainder of 0.
ii.
36 is divisible by 6 because 36 ÷ 6 = 6 with a remainder of 0.
iii.
36 is not divisible by 5 because 36 ÷ 5 = 7 with a remainder of 1.
Divisibility Rules
By using divisibility rules we can easily find out whether a given number is divisible by another number without actually performing the division. This helps to save time especially when working with numbers.
Divisibility Rule
Description
Ex am ples
Example1: Check if 64 is divisible by 2.
Divisibility by 2
A number is divisible by 2 if the last digit is even. i.e., if the last digit is 0 or 2 or 4 or 6 or 8
The last digit of 64 is 4 (even). Hence 64 is divisible by 2 Example2: Check if 69 is divisible by 2. The last digit of 69 is 9 (not even). Hence 69 is not divisible by 2
A number is divisible by 3 if the sum of the digits is divisible by 3 Divisibility by 3
(Please note that we can apply this rule to the answer again and again if we need)
Example1: Check if 387 is divisible by 3. 3 + 8 + 7 = 18. 18 is divisible by 3. Hence 387 is divisible by 3 Example2: Check if 421 is divisible by 3. 4 + 2 + 1 = 7. 7 is not divisible by 3. Hence 421 is not divisible by 3 Example1: Check if 416 is divisible by 4.
Divisibility by 4
A number is divisible by 4 if the number formed by the last two digits is divisible by 4.
Number formed by the last two digits = 16. 16 is divisible by 4. Hence 416 is divisible by 4 Example2: Check if 481 is divisible by 4. Number formed by the last two digits = 81. 81 is not divisible by 4. Hence 481 is not divisible by 4 Example1: Check if 305 is divisible by 5. Last digit is 5. Hence 305 is divisible by 5.
Divisibility
A number is divisible by 5 if
Example2: Check if 420 is divisible by 5. Last digit is 0. Hence 420 is divisible by 5.
by 5
the last digit is either 0 or 5.
Example3: Check if 312 is divisible by 5. Last digit is 2. Hence 312 is not divisible by 5. Example1: Check if 546 is divisible by 6. 546 is divisible by 2. 546 is also divisible by 3. (Check the divisibility rule of 2 and3 to find out this) Hence 546 is divisible by 6 Example2: Check if 633 is divisible by 6. 633 is not divisible by 2 though 633 is divisible by 3. (Check the divisibility rule of 2 and 3 to find out this)
Divisibility by 6
A number is divisible by 6 if it is divisible by both 2 and 3.
Hence 633 is not divisible by 6 Example3: Check if 635 is divisible by 6. 635 is not divisible by 2. 635 is also not divisible by 3. (Check the divisibility rule of 2 and 3 to find out this) Hence 635 is not divisible by 6 Example4: Check if 428 is divisible by 6. 428 is divisible by 2 but 428 is not divisible by 3.(Check the divisibility rule of 2 and 3 to find out this) Hence 428 is not divisible by 6 Example1: Check if 349 is divisible by 7.
To find out if a number is divisible by 7, double the last digit and subtact it from the number formed by the remaining digits. Divisibility by 7
Repeat this process until we get at a smaller number whose divisibility we know. If this smaller number is 0 or divisible by 7, the original number is also divisible by 7.
Given number = 349 34 - (9 × 2) = 34 - 18 = 16 16 is not divisible by 7. Hence 349 is not divisible by 7 Example2: Check if 364 is divisible by 7. Given number = 364 36 - (4 × 2) = 36 - 8 = 28 28 is divisible by 7. Hence 364 is also divisible by 7 Example3: Check if 3374 is divisible by 7. Given number = 3374 337 - (4 × 2) = 337 - 8 = 329 32 - (9 × 2) = 32 - 18 = 14 14 is divisible by 7. Hence 329 is also divisible by 7. Hence 3374 is also divisible by 7. Example1: Check if 7624 is divisible by 8. The number formed by the last three digits of 7624 = 624. 624 is divisible by 8. Hence 7624 is also divisible by 8. Example2: Check if 129437464 is divisible by 8.
Divisibility by 8
A number is divisible by 8 if the number formed by the last three digits is divisible by 8.
The number formed by the last three digits of 129437464 = 464. 464 is divisible by 8. Hence 129437464 is also divisible by 8.
Example3: Check if 737460 is divisible by 8. The number formed by the last three digits of 737460 = 460. 460 is not divisible by 8. Hence 737460 is also not divisible by 8. Example1: Check if 367821 is divisible by 9. 3 + 6 + 7 + 8 + 2 + 1 = 27 27 is divisible by 9. Hence 367821 is also divisible by 9.
Divisibility by 9
A number is divisible by 9 if the sum of its digits is divisible by 9. (Please note that we can apply this rule to the answer again and again if we need)
Example2: Check if 47128 is divisible by 9. 4 + 7 + 1 + 2 + 8 = 22 22 is not divisible by 9. Hence 47128 is not divisible by 9. Example3: Check if 4975291989 is divisible by 9. 4 + 9+ 7 + 5 + 2 + 9 + 1 + 9 + 8 + 9= 63 Since 63 is big, we can use the same method to see if it is divisible by 9. 6+3=9 9 is divisible by 9. Hence 63 is also divisible by 9. Hence 4975291989 is also divisible by 9. Example1: Check if 2570 is divisible by 10.
Divisibility by 10
A number is divisible by 10 if the last digit is 0.
Last digit is 0. Hence 2570 is divisible by 10. Example2: Check if 5462 is divisible by 10. Last digit is not 0. Hence 5462 is not divisible by 10 Example1: Check if 85136 is divisible by 11. 8 + 1 + 6 = 15 5+3=8 15 - 8 = 7 7 is not divisible by 11. Hence 85136 is not divisible by 11.
To find out if a number is divisible by 11, find the sum of the odd numbered digits and the sum of the even numbered digits. Divisibility by 11
Now substract the lower number obtained from the bigger number obtained. If the number we get is 0 or divisible by 11, the original number is also divisible by 11.
Example2: Check if 2737152 is divisible by 11. 2+3+1+2=8 7 + 7 + 5 = 19 19 - 8 = 11 11 is divisible by 11. Hence 2737152 is also divisible by 11. Example3: Check if 957 is divisible by 11. 9 + 7 = 16 5=5 16 - 5 = 11 11 is divisible by 11. Hence 957 is also divisible by 11. Example4: Check if 9548 is divisible by 11. 9 + 4 = 13 5 + 8 = 13 13 - 13 = 0 We got the difference as 0. Hence 9548 is divisible by 11.
Example1: Check if 720 is divisible by 12. 720 is divisible by 3 and 720 is also divisible by 4. (Check the divisibility rules of 3 and 4 to find out this) Hence 720 is also divisible by 12 Example2: Check if 916 is divisible by 12. 916 is not divisible by 3 , though 916 is divisible by 4.(Check the divisibility rules of 3 and 4 to find out this) Divisibility by 12
A number is divisible by 12 if the number is divisible by both 3 and 4
Hence 916 is not divisible by 12 Example3: Check if 921 is divisible by 12. 921 is divisible by 3. But 921 is not divisible by 4.(Check the divisibility rules of 3 and 4 to find out this) Hence 921 is not divisible by 12 Example4: Check if 827 is divisible by 12. 827 is not divisible by 3. 827 is also not divisible by 4.(Check the divisibility rules of 3 and 4 to find out this) Hence 827 is not divisible by 12 Example1: Check if 349 is divisible by 13. Given number = 349 34 + (9 × 4) = 34 + 36 = 70 70 is not divisible by 13. Hence 349 is not divisible by 349 Example2: Check if 572 is divisible by 13.
To find out if a number is divisible by 13, multiply the last digit by 4 and add it to the number formed by the remaining digits. Divisibility by 13
Repeat this process until we get at a smaller number whose divisibility we know. If this smaller number is divisible by 13, the original number is also divisible by 13.
Given number = 572 57 + (2 × 4) = 57 + 8 = 65 65 is divisible by 13. Hence 572 is also divisible by 13 Example3: Check if 68172 is divisible by 13. Given number = 68172 6817 + (2 × 4) = 6817 + 8 = 6825 682 + (5 × 4) = 682 + 20 = 702 70 + (2 × 4) = 70 + 8 = 78 78 is divisible by 13. Hence 68172 is also divisible by 13. Example4: Check if 651 is divisible by 13. Given number = 651 65 + (1 × 4) = 65 + 4 = 69 69 is not divisible by 13. Hence 651 is not divisible by 13 Example1: Check if 238 is divisible by 14 238 is divisible by 2 . 238 is also divisible by 7. (Please check the divisibility rule of 2 and 7 to find out this)
Hence 238 is also divisible by 14 Example2: Check if 336 is divisible by 14 336 is divisible by 2 . 336 is also divisible by 7. (Please check the divisibility rule of 2 and 7 to find out this) Hence 336 is also divisible by 14
Divisibility by 14
A number is divisible by 14 if it is divisible by both 2 and 7.
Example3: Check if 342 is divisible by 14. 342 is divisible by 2 , but 342 is not divisible by 7.(Please check the divisibility rule of 2 and 7 to find out this) Hence 342 is not divisible by 12 Example4: Check if 175 is divisible by 14. 175 is not divisible by 2 , though it is divisible by 7.(Please check the divisibility rule of 2 and 7 to find out this) Hence 175 is not divisible by 14 Example5: Check if 337 is divisible by 14. 337 is not divisible by 2 and also by 7 (Please check the divisibility rule of 2 and7 to find out this) Hence 337 is not divisible by 14 Example1: Check if 435 is divisible by 15 435 is divisible by 3 . 435 is also divisible by 5. (Please check the divisibility rule of 3 and 5 to find out this) Hence 435 is also divisible by 15 Example2: Check if 555 is divisible by 15 555 is divisible by 3 . 555 is also divisible by 5. (Please check the divisibility rule of 3 and 5 to find out this) Hence 555 is also divisible by 15 Example3: Check if 483 is divisible by 15.
Divisibility by 15
A number is divisible by 15 If it is divisible by both 3 and 5.
483 is divisible by 3 , but 483 is not divisible by 5. (Please check the divisibility rule of 3 and 5 to find out this) Hence 483 is not divisible by 15 Example4: Check if 485 is divisible by 15. 485 is not divisible by 3 , though it is divisible by 5. (Please check the divisibility rule of 3 and 5 to find out this) Hence 485 is not divisible by 15 Example5: Check if 487 is divisible by 15. 487 is not divisible by 3 . It is also not divisible by 5 (Please check the divisibility rule of 3 and 5 to find out this) Hence 487 is not divisible by 15 Example1: Check if 5696512 is divisible by 16. The number formed by the last four digits
of 5696512 = 6512 6512 is divisible by 16. Hence 5696512 is also divisible by 16.
Divisibility by 16
A number is divisible by 16 if the number formed by the last four digits is divisible by 16.
Example2: Check if 3326976 is divisible by 16. The number formed by the last four digits of 3326976 = 6976 6976 is divisible by 16. Hence 3326976 is also divisible by 16. Example3: Check if 732374360 is divisible by 16. The number formed by the last three digits of 732374360 = 4360 4360 is not divisible by 16. Hence 732374360 is also not divisible by 16.
To find out if a number is divisible by 17, multiply the last digit by 5 and subtract it from the number formed by the remaining digits. Divisibility by 17
Repeat this process until you arrive at a smaller number whose divisibility you know. If this smaller number is divisible by 17, the original number is also divisible by 17.
Example1: Check if 500327 is divisible by 17. Given Number = 500327 50032 - (7 × 5 )= 50032 - 35 = 49997 4999 - (7 × 5 ) = 4999 - 35 = 4964 496 - (4 × 5 ) = 496 - 20 = 476 47 - (6 × 5 ) = 47 - 30 = 17 17 is divisible by 17. Hence 500327 is also divisible by 17 Example2: Check if 521461 is divisible by 17. Given Number = 521461 52146 - (1 × 5 )= 52146 -5 = 52141 5214 - (1 × 5 ) = 5214 - 5 = 5209 520 - (9 × 5 ) = 520 - 45 = 475 47 - (5 × 5 ) = 47 - 25 = 22 22 is not divisible by 17. Hence 521461 is not divisible by 17 Example1: Check if 31104 is divisible by 18. 31104 is divisible by 2. 31104 is also divisible by 9. (Please check the divisibility rule of 2 and 9 to find out this) Hence 31104 is divisible by 18 Example2: Check if 1170 is divisible by 18.
Divisibility by 18
A number is divisible by 18 if it is divisible by both 2 and 9.
1170 is divisible by 2. 1170 is also divisible by 9. (Please check the divisibility rule of 2 and 9 to find out this) Hence 1170 is divisible by 18 Example3: Check if 1182 is divisible by 18. 1182 is divisible by 2 , but 1182 is not divisible by 9. (Please check the divisibility rule of 2 and 9 to find out this) Hence 1182 is not divisible by 18 Example4: Check if 1287 is divisible by 18. 1287 is not divisible by 2 though it is divisible by 9. (Please check the divisibility rule of 2 and 9 to find out this) Hence 1287 is not divisible by 18
To find out if a number is divisible by 19, multiply
Example1: Check if 74689 is divisible by 19.
the last digit by 2 and add it to the number formed by the remaining digits. Divisibility by 19
Repeat this process until you arrive at a smaller number whose divisibility you know. If this smaller number is divisible by 19, the original number is also divisible by 19.
Given Number = 74689 7468 + (9 × 2 )= 7468 + 18 = 7486 748 + (6 × 2 ) = 748 + 12 = 760 76 + (0 × 2 ) = 76 + 0 = 76 76 is divisible by 19. Hence 74689 is also divisible by 19 Example2: Check if 71234 is divisible by 19. Given Number = 71234 7123 + (4 × 2 )= 7123 + 8 = 7131 713 + (1 × 2 )= 713 + 2 = 715 71 + (5 × 2 )= 71 + 10 = 81 81 is not divisible by 19. Hence 71234 is not divisible by 19 Example1: Check if 720 is divisible by 20 720 is divisible by 10. (Please check thedivisibility rule of 10 to find out this). The tens digit = 2 = even digit. Hence 720 is also divisible by 20 Example2: Check if 1340 is divisible by 20
A number is divisible by 20 if it is divisible by 10 and the tens digit is even. Divisibility by 20
(There is one more rule to see if a number is divisible by 20 which is given below. A number is divisible by 20 if the number is divisible by both 4 and 5)
1340 is divisible by 10. (Please check thedivisibility rule of 10 to find out this). The tens digit = 2 = even digit. Hence 1340 is divisible by 20 Example3: Check if 1350 is divisible by 20 1350 is divisible by 10. (Please check thedivisibility rule of 10 to find out this). But the tens digit = 5 = not an even digit. Hence 1350 is not divisible by 20 Example4: Check if 1325 is divisible by 20 1325 is not divisible by 10 (Please check the divisibility rule of 10 to find out this) though the tens digit = 2 = even digit. Hence 1325 is not divisible by 20
IV.
What are Factors of a Number and how to find it out? a.
F actors of a num ber If one number is divisible by a second number, the second number is a factor of the first number The lowest factor of any positive number = 1 The highest factor of any positive number = the number itself Exam ple The factors of 36 are 1, 2, 3, 4, 6, 9 12, 18, 36 because each of these numbers divides 36 with a remainder of 0
b.
How to find out factors of a num ber i.
Write down 1 and the number itself (lowest and highest factors).
ii.
Check if the given number is divisible by 2 (Reference: Divisibility by 2 rule) If the number is divisible by 2, write down 2 as the second lowest factor and divide the given number by 2 to get the second highest factor
iii.
Check for divisibility by 3, 4,5, and so on. till the beginning of the list reaches the end
Exam ple1: F ind out the factors of 72 i.
Write down 1 and the number itself (72) as lowest and highest factors. 1 . . . 72
ii.
72 is divisible by 2 (Reference: Divisibility by 2 Rule). 72 ÷ 2 = 36. Hence 2nd lowest factor = 2 and 2ndhighest factor = 36. So we can write as 1, 2 . . . 36, 72
iii.
72 is divisible by 3 (Reference: Divisibility by 3 Rule). 72 ÷ 3 = 24 . Hence 3rd lowest factor = 3 and 3rdhighest factor = 24. So we can write as 1, 2, 3, . . . 24, 36, 72
iv.
72 is divisible by 4 (Reference: Divisibility by 4 Rule). 72 ÷ 4 = 18. Hence 4th lowest factor = 4 and 4thhighest factor = 18. So we can write as 1, 2, 3, 4, . . . 18, 24, 36, 72
v.
vi.
72 is not divisible by 5 (Reference: Divisibility by 5 Rule)
72 is divisible by 6 (Reference: Divisibility by 6 Rule). 72 ÷ 6 = 12. Hence 5th lowest factor = 6 and 5thhighest factor = 12. So we can write as 1, 2, 3, 4, 6, . . . 12, 18, 24, 36, 72
vii.
72 is not divisible by 7 (Reference: Divisibility by 7 Rule)
viii.
72 is divisible by 8 (Reference: Divisibility by 8 Rule). 72 ÷ 8 = 9. Hence 6th lowest factor = 8 and 6thhighest factor = 9. Now our list is complete and the factors of 72 are 1, 2, 3, 4, 6, 8, 9 12, 18, 24, 36, 72
Exam ple2: F ind out the factors of 22 i.
Write down 1 and the number itself (22) as lowest and highest factors 1 . . . 22
ii.
22 is divisible by 2 (Reference: Divisibility by 2 Rule). 22 ÷ 2 = 11. Hence 2nd lowest factor = 2 and 2ndhighest factor = 11. So we can write as 1, 2 . . . 11, 22
iii.
22 is not divisible by 3 (Reference: Divisibility by 3 Rule).
iv.
22 is not divisible by 4 (Reference: Divisibility by 4 Rule).
v.
22 is not divisible by 5 (Reference: Divisibility by 5 Rule).
vi.
22 is not divisible by 6 (Reference: Divisibility by 6 Rule).
vii.
22 is not divisible by 7 (Reference: Divisibility by 7 Rule).
viii.
22 is not divisible by 8 (Reference: Divisibility by 8 Rule).
ix.
22 is not divisible by 9 (Reference: Divisibility by 9 Rule).
x.
22 is not divisible by 10 (Reference: Divisibility by 10 Rule). Now our list is complete and the factors of 22 are 1, 2, 11, 22
c.
Im portant Properties of F actors If a number is divisible by another number, then it is also divisible by all the factors of that number. Example : 108 is divisible by 36 because 106 ÷ 38 = 3 with remainder of 0. The factors of 36 are 1, 2, 3, 4, 6, 9 12, 18, 36 because each of these numbers divides 36 with a remainder of 0. Hence, 108 is also divisible by each of the numbers 1, 2, 3, 4, 6, 9, 12, 18, 36.
V.
What are P rime Numbers and Composite Numbers? a.
Prim e Num bers A prime number is a positve integer that is divisible byitself and 1 only. Prime numbers will have exactly two integer factors. Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, etc.
Please note the following facts Zero is not a prime number because zero is divisible by more than two factors. Zero can be divided by 1, 2, 3 etc. (0 ÷ 1 = 0, 0÷ 2 = 0 ...) One is not a prime number because it does not have two factors. It is divisible by only 1 b.
Com posite Num bers Composite numbers are numbers that have more than two factors. A composite number is divisible by at least one number other than 1 and itself. Examples: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, etc. Please note that zero and 1 are neither prime numbers nor composite numbers. Every whole number is either prime or composite, with two exceptions 0 and 1 which are neither prime nor composite
VI.
What are P rime Factorization and P rime factors ? a.
Prim e factor The factors which are prime numbers are called prime factors
b.
Prim e factorization Prime factorization of a number is the expression of the number as the product of its prime factors Exam ple 1: Prime factorization of 280 can be written as 280 = 2 × 2 × 2 × 5 × 7 = 23 × 5 × 7 and the prime factors of 280 are 2, 5 and 7 Exam ple 2: Prime factorization of 72 can be written as 72 = 2 × 2 × 2 × 3 × 3 = 23 × 32 and the prime factors of 72 are 2 and 3
c.
How to find out prim e factorization and prim e factors of a num ber Repeated Division Method : In order to find out the prime factorization of a number, repeatedly divide the number by the smallest prime number possible(2,3,5,7,11, ...) until the quotient is 1. Exam ple 1: Find out Prime factorization of 280 2 280
2 140
2 70
5 35
7
7
1
Hence, prime factorization of 280 can be written as 280 = 2 × 2 × 2 × 5 × 7 = 23 × 5 × 7 and the prime factors of 280 are 2, 5 and 7
Exam ple 2: Find out Prime factorization of 72 2 72
2 36
2 18
3 9
3 3
1
Hence, prime factorization of 72 can be written as 72 = 2 × 2 × 2 × 3 × 3 = 23 × 32 and the prime factors of 72 are 2 and 3 d.
Im portant Properties Every whole number greater than 1 can be uniquely expressed as the product of its prime factors. For example, 700 = 22 × 52 × 7
VII.
VIII.
Multiples Multiples of a whole number are the products of that number with 1, 2, 3, 4, and so on Example : Multiples of 3 are 3, 6, 9, 12, 15, ... If a number x divides another number y exactly with a remainder of 0, we can say that x is a factor of y and y is a multiple of x For instance, 4 divides 36 exactly with a remainder of 0. Hence 4 is a factor of 36 and 36 is a multiple of 4
What is L east Common Multiple (L CM) and how to find L CM a.
Least Com m on Multiple (LCM) Least Common Multiple (LCM) of two or more numbers is the smallest number that is a multiple of all the numbers Example: LCM of 3 and 4 = 12 because 12 is the smallest multiple which is common to 3 and 4 (In other words, 12 is the smallest number which is divisible by both 3 and 4) We can find out LCM using prime factorization method or division method
b.
How to find out LCM using prim e factorization m ethod Step1 : Express each number as a product of prime factors. Step2 : LCM = The product of highest powers of all prime factors Exam ple 1 : Find out LCM of 8 and 14 Step1 : Express each number as a product of prime factors. (Reference: Prime Factorization and how to find out Prime Factorization)
8 = 23 14 = 2 × 7 Step2 : LCM = The product of highest powers of all prime factors Here the prime factors are 2 and 7 The highest power of 2 here = 23 The highest power of 7 here = 7 Hence LCM = 23 × 7 = 56 Exam ple 2 : Find out LCM of 18, 24, 9, 36 and 90 Step1 : Express each number as a product of prime factors (Reference: Prime Factorization and how to find out Prime Factorization). 18 = 2 × 32 24 = 23 × 3 9 = 32 36 = 23 × 32 90 = 2 × 5 × 32 Step2 : LCM = The product of highest powers of all prime factors Here the prime factors are 2, 3 and 5 The highest power of 2 here = 23 The highest power of 3 here = 32 The highest power of 5 here = 5 Hence LCM = 23 × 32 × 5 = 360 c.
How to find out LCM using Div ision Method (shortcut) Step 1 : Write the given numbers in a horizontal line separated by commas. Step 2 : Divide the given numbers by the smallest prime number which can exactly divide at least two of the given numbers. Step 3 : Write the quotients and undivided numbers in a line below the first. Step 4 : Repeat the process until we reach a stage where no prime factor is common to any two numbers in the row. Step 5 : LCM = The product of all the divisors and the numbers in the last line. Exam ple 1 : Find out LCM of 8 and 14 2 8, 14
4, 7
Hence Least common multiple (L.C.M) of 8 and 14 = 2 × 4 × 7 = 56 Exam ple 2 : Find out LCM of 18, 24, 9, 36 and 90 2 18, 24, 9, 36, 90
2
9, 12, 9, 18, 45
3
9, 6, 9, 9, 45
3
3, 2, 3, 3, 15
1, 2, 1, 1, 5
Hence Least common multiple (L.C.M) of 18, 24, 9, 36 and 90 = 2 × 2 × 3 × 3 × 2 × 5 = 360
IX.
What is Highest Common Factor (HCF) or Greatest Common Measure (GCM) or Greatest Common Divisor (GCD) and How to find it out ?
a.
Highest Com m on F actor(H.C.F ) or Greatest Com m on Measure(G.C.M) or Greatest Com m on Divisor (G.C.D) Highest Common Factor(H.C.F) or Greatest Common Measure(G.C.M) or Greatest Common Divisor (G.C.D) of two or more numbers is the greatest number which divides each of them exactly. Example : HCF or GCM or GCD of 60 and 75 = 15 because 15 is the highest number which divides both 60 and 75 exactly. We can find out HCF using prime factorization method or division method
b.
How to find out HCF using prim e factorization m ethod Step1 : Express each number as a product of prime factors. (Reference: Prime Factorization and how to find out Prime Factorization) Step2 : HCF is the product of all common prime factors using the least power of each common prime factor. Exam ple 1 : Find out HCF of 60 and 75 (Reference:Prime Factorization and how to find out Prime Factorization) Step1 : Express each number as a product of prime factors. 60 = 22 × 3 × 5 75 = 3 × 52
Step2 : HCF is the product of all common prime factors using the least power of each common prime factor. Here, common prime factors are 3 and 5 The least power of 3 here = 3 The least power of 5 here = 5 Hence, HCF = 3 × 5 = 15 Exam ple 2 : Find out HCF of 36, 24 and 12 Step1 : Express each number as a product of prime factors. (Reference: Prime Factorization and how to find out Prime Factorization) 36 = 22 × 32 24 = 23 × 3 12 = 22 × 3 Step2 : HCF is the product of all common prime factors using the least power of each common prime factor. Here 2 and 3 are common prime factors. The least power of 2 here = 22 The least power of 3 here = 3 Hence, HCF = 22 × 3 = 12 Exam ple 3 : Find out HCF of 36, 27 and 80 Step1 : Express each number as a product of prime factors. (Reference: Prime Factorization and how to find out Prime Factorization) 36 = 22 × 32 27 = 33 80 = 24 × 5 Step2 : HCF = HCF is the product of all common prime factors using the least power of each common prime factor. Here you can see that there are no common prime factors. Hence, HCF = 1 c.
How to find out HCF using prim e factorization m ethod - By dividing the num bers (shortcut) Step 1 : Write the given numbers in a horizontal line separated by commas. Step 2 : Divide the given numbers by the smallest prime number which can exactly divide all of the given numbers. Step 3 : Write the quotients in a line below the first. Step 4 : Repeat the process until we reach a stage where no common prime factor exists for all of the numbers. Step 5 :We can see that the factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. Their product is the HCF Exam ple 1 : Find out HCF of 60 and 75 3 60, 75
5 20, 25
4, 5
We can see that the prime factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. no common prime factor is exists for the numbers came at the bottom. Hence HCF = 3 × 5 =15. Exam ple 2 : Find out HCF of 36, 24 and 12 2 36, 24, 12
2 18, 12, 6
3 9, 6, 3
3, 2, 1
We can see that the prime factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. no common prime factor is exists for the numbers came at the bottom. Hence HCF = 2 × 2 × 3 = 12. Exam ple 3 : Find out HCF of 36, 24 and 48 2 36, 24, 48
2 18, 12, 24
3 9, 6, 12
3, 2, 4
We can see that the prime factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. no common prime factor is exists for the numbers came at the bottom. Hence HCF = 2 × 2 × 3 = 12. d.
How to find out HCF using div ision m ethod(shortcut)
i.
To find out HCF of tw o giv en num bers using div ision m ethod Step 1: Divide the larger number by the smaller number Step 2: Divisor of step 1 is divided by its remainder
Step 3: Divisor of step 2 is divided by its remainder. Continue this process till we get 0 as remainder. Step 4: Divisior of the last step is the HCF.
ii.
To find out HCF of three giv en num bers using div ision m ethod Step 1: Find out HCF of any two numbers. Step 2: Find out the HCF of the third number and the HCF obtained in step 1 Step 3: HCF obtained in step 2 will be the HCF of the three numbers
iii.
To find out HCF of m ore than three num bers using div ision m ethod In a similar way as explained for three numbers, we can find out HCF of more than three numbers also
Exam ple 1 : Find out HCF of 60 and 75 60) 75 (1
60
15) 60 (4
60
0
Hence HCF of 60 and 75 = 15 Exam ple 2 : Find out HCF of 12 and 48 12) 48 (4
48
0
Hence HCF of 12 and 48 = 12 Exam ple 3 : Find out HCF of 3556 and 3224 3224) 3556 (1
3224
332) 3224 (9
2988
236) 332 (1
Hence HCF of 3556 and 3224 = 4 Exam ple 3 : Find out HCF of 9, 27, and 48 Taken any two numbers and find out their HCF first. Say, let's find out HCF of 9 and 27 initially. 9) 27 (3
27
0
Hence HCF of 9 and 27 = 9 HCF of 9 ,27, 48 = HCF of [(HCF of 9, 27) and 48] = HCF of [9 and 48] 9) 48 (5
45
3) 9 (3
9
0
Hence, HCF of 9 ,27, 48 = 3 Exam ple 4 : Find out HCF of 5 and 7 5) 7 (1
5
2) 5 (2
4
Hence HCF of 5 and 7 = 1 X.
How to calculate L CM and HCF for fractions Least Com m on Multiple (L.C.M.) for fractions
LCM for fractions =
LCM of Numerators HCF of Denominators
Exam ple 1: Find out LCM of
LCM =
LCM (1, 3, 3) 3 = 2 HCF (2, 8, 4)
Exam ple 2: Find out LCM of
LCM =
XI.
1 3 3 , , 2 8 4
2 3 , 5 10
LCM (2, 3) 6 = 5 HCF (5, 10)
Highest Common Multiple (H.C.F) for fractions
HCF for fractions =
HCF of Numerators LCM of Denominators
1) 2 (2
2
0
236
Exam ple 1: Find out HCF of
HCF =
HCF (3, 6, 9) 3 = 220 LCM (5, 11, 20)
Exam ple 2: Find out HCF of
HCF =
XII.
XIII.
3 6 9 , , 5 11 20
4 2 , 5 3
HCF (4, 2) 2 = 15 LCM (5, 3)
How to calculate L CM and HCF for Decimals Step 1 : Make the same number of decimal places in all the given numbers by suffixing zero(s) in required numbers as needed. Step 2 : Now find the LCM/HCF of these numbers without decimal. Step 3 : Put the decimal point in the result obtained in step 2 leaving as many digits on its right as there are in each of the numbers. Exam ple1 : Find the LCM and HCF of .63, 1.05, 2.1 Step 1 : Make the same number of decimal places in all the given numbers by suffixing zero(s) in required numbers as needed. i.e., the numbers can be writtten as .63, 1.05, 2.10 Step 2 : Now find the LCM/HCF of these numbers without decimal. Without decimal, the numbers can be written as 63, 105 and 210 . LCM (63, 105 and 210) = 630 HCF (63, 105 and 210) = 21 Step 3 : Put the decimal point in the result obtained in step 2 leaving as many digits on its right as there are in each of the numbers. i.e., here, we need to put decimal point in the result obtained in step 2 leaving two digits on its right. i.e., the LCM (.63, 1.05, 2.1) = 6.30 HCF (.63, 1.05, 2.1) = .21
How to compare fractions? a.
Type 1 : F ractions w ith sam e denom inators. Compare
3 1 and 5 5
These fractions have same denominator. So just compare the numerators. Bigger the numerator, bigger the number. 3 > 1. Hence
3 1 > 5 5
Exam ple 2: Compare
2 3 8 and and 7 7 7
These fractions have same denominator. So just compare the numerators. Bigger the numerator, bigger the number. 8 > 3 > 2. Hence
b.
8 3 2 > > 7 7 7
Type 2 : F ractions w ith sam e num erators. Exam ple 1: Compare
3 3 and 5 8
These fractions have same numerator. So just compare the denominators. Bigger the denominator, smaller the number. 8 > 5. Hence
3 3 < 8 5
Exam ple 2: Compare
7 7 7 and and 8 2 5
These fractions have same numerator. So just compare the denominators. Bigger the denominator, smaller the number. 8 > 5 > 2. Hence
c.
7 7 7 < < 8 5 2
Type 3 : F ractions w ith different num erators and denom inators. Exam ple 1: Compare
3 4 and 5 7
To compare such fractions, find out LCM of the denominators. Here, LCM(5, 7) = 35 Now , convert each of the given fractions into an equivalent fraction with 35 (LCM) as the denominator.
The denominator of
3 is 5. 5
5 needs to be multiplied with 7 to get 35. Hence,
4 is 7. 7
7 needs to be multiplied with 5 to get 35. Hence,
3 3×7 21 = = 5 5×7 35 The denominator of
4 4×5 20 = = 7 7×5 35 21 20 > 35 35 Hence,
3 4 > 5 7
Or Convert the fractions to decimals
3 = .6 5 4 = .5. . . (Need not find out the complete decimal value; just find out up to what is required for comparison. In this case the first digit itself is sufficient to do the 7
comparison) .6 > .5... Hence,
XIV.
3 4 > 5 7
Co-prime Numbers or R elatively P rime Numbers Two numbers are said to be co-prime (also spelled coprime) or relatively prime if they do not have a common factor other than 1. i.e., if their HCF is 1. Exam ple1: 3, 5 are co-prime numbers (Because HCF of 3 and 5 = 1) Exam ple2: 14, 15 are co-prime numbers (Because HCF of 14 and 15 = 1) A set of numbers is said to be pairwise co-prime (or pairwise relatively prime) if every two distinct numbers in the set are co-prime Exam ple1 : The numbers 10, 7, 33, 13 are pairwise co-prime, because HCF of any pair of the numbers in this is 1. HCF (10, 7) = HCF (10, 33) = HCF (10, 13) = HCF (7, 33) = HCF (7, 13) = HCF (33, 13) = 1. Exam ple2 : The numbers 10, 7, 33, 14 are not pairwise co-prime because HCF(10, 14) = 2 ≠ 1 and HCF(7, 14) = 7 ≠ 1. If a number is divisible by two co-prime numbers, then the number is divisible by their product also. Example 3, 5 are co-prime numbers (Because HCF of 3 and 5 = 1) 14325 is divisible by 3 and 5. 3 × 5 = 15 Hence 14325 is divisible by 15 also
If a number is divisible by more than two pairwise co-prime numbers, then the number is divisible by their product also. Example1 : The numbers 3, 4, 5 are pairwise co-prime because HCF of any pair of numbers in this is 1 1440 is divisible by 3, 4 and 5. 3 × 4 × 5 = 60. Hence 1440 is also divisible by 60 Example2 The numbers 3, 4, 9 are not pairwise co-prime because HCF (3, 9 ) = 3 ≠ 1 1440 is divisible by 3, 4 and 9. 3 X 4 X 9 = 108. However 1440 is not divisible by 108 as 3, 4, 9 are not pairwise co-prime
XV.
I mportant P oints to Note on L CM and HCF
Product of two numbers = Product of their HCF and LCM. Example LCM (8, 14) = 56
HCF (8, 14) = 2 LCM (8, 14) × HCF (8, 14) = 56 × 2 = 112 8 × 14 = 112 Hence LCM (8, 14) × HCF (8, 14) = 8 × 14
I mportant Concepts and Formulas - Sequence and Series 1.
A rithmetic P rogression (A .P .) (or A rithmetic Sequence) Arithmetic progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a constant, d to the preceding term. The constant d is called common difference.
An arithmetic progression is given by a, (a + d), (a + 2d), (a + 3d), ... where a = the first term , d = the common difference Examples for Arithmetic Progressions 1, 3, 5, 7, ... is an arithmetic progression (AP) with a = 1 and d = 2 7, 13, 19, 25, ... is an arithmetic progression (AP) with a = 7 and d= 6
a.
n th term of an arithm etic progression
t n = a + (n – 1)d where t n = nth term, a= the first term , d= common difference Ex am ple 1 : Find 10th term in the series 1, 3, 5, 7, ... a =1 d =3 – 1 =2 10th term, t 10 = a + (n-1)d = 1 + (10 – 1)2 = 1 + 18 = 19 Ex am ple 2 : Find 16th term in the series 7, 13, 19, 25, ... a =7 d = 13 – 7 = 6 16th term, t 16 = a + (n-1)d = 7 + (16 – 1)6 = 7 + 90 = 97
b.
Num ber of term s of an arithm etic progression
n=
(l − a) +1 d
where n = number of terms, a= the first term , l = last term, d= common difference Ex am ple : Find the number of terms in the series 8, 12, 16, . . .72 a =8 l = 72 d = 12 – 8 = 4
n=
c.
(l − a) (72 − 8) 64 +1 = +1 = + 1 = 16 + 1 = 17 d 4 4
Sum of first n term s in an arithm etic progression
Sn =
n n [ 2a + (n − 1)d ] = [ a + l ] 2 2
where a = the first term,
d= common difference,
l = t n = nth term = a + (n − 1)d
Ex am ple 1 : Find 4 + 7 + 10 + 13 + 16 + . . . up to 20 terms a =4 d =7 – 4 =3 Sum of first 20 terms, S20
=
n 20 [ 2a + (n − 1)d ] = [ (2 × 4) + (20 − 1)3 ] 2 2
= 10[8 + (19 × 3)] = 10[8 + 57] = 650
Ex am ple 2 : Find 6 + 9 + 12 + . . . + 30 a =6 l = 30 d =9 – 6 =3
n=
(l − a) (30 − 6) 24 +1 = +1 = +1 = 8+1 = 9 d 3 3
Sum, S = d.
n 9 9 [ a + l ] = [ 6 + 30 ] = × 36 = 9 × 18 = 162 2 2 2
Arithm etic Mean
If a, b, c are in AP, b is the Arithmetic Mean (A.M.) between a and c. In this case,
b=
1 (a + c) 2
The Arithmetic Mean (A.M.) between two numbers a and b =
1 (a + b) 2
If a, a1, a2 ... an, b are in AP we can say that a1, a2... an are the n Arithmetic Means between a and b.
e.
If a, b, c are in AP, 2b = a + c
To solve most of the problems related to A.P., the terms can be conveiently taken as 3 terms : (a – d), a, (a +d) 4 terms : (a – 3d), (a – d), (a + d), (a +3d) f.
5 terms : (a – 2d), (a – d), a, (a + d), (a +2d)
g.
T n = Sn - Sn-1
h.
If each term of an A.P. is increased, decreased , multiplied or divided by the same non-zero constant, the resulting sequence also will be in A.P.
i.
2.
In an A.P., sum of terms equidistant from beginning and end will be constant
Harmonic P rogression (H.P .) (or H armonic Sequence)
Non-zero numbers a 1 , a 2 , a 3 , . . . an are in H.P. if
1 1 1 1 , , , ... are in A.P. a1 a2 a3 an
Examples for Harmonic Progressions
1 1 1 , , , ⋯ is a harmonic progression (H.P.) 2 6 10
Three non-zero numbers a, b, c will be in HP, if
1 1 1 , , are in A.P. a b c
If a, (a+d), (a+2d), . . . are in A.P., nthterm of the A.P. = a + (n - 1)d Hence, if
1 1 1 1 , , , ⋯ are in H.P., nthterm of the H.P. = a a + d a + 2d a + (n − 1)d
If a, b, c are in HP, b is the Harmonic Mean(H.M.) between a and c In this case,
b=
2ac a +c
The Harmonic Mean (H.M.) between two numbers a and b =
2ab a +b
If a, a1, a2 ... an, b are in H.P. we can say that a1, a2 ... an are the n Harmonic Means between a and b.
If a, b, c are in HP,
3.
2 1 1 = + b a c
Geometric P rogression (G.P .) (or Geometric Sequence) A sequence of non-zero numbers is a Geometric Progression (G.P.) if the ratio of any term and its preceding term is always constant.
A Geometric Progression (G.P.) is given by a, ar, ar2, ar3, ... where a = the first term , r = the common ratio
Examples for Geometric Progressions 1, 3, 9, 27, ... is a geometric progression (G.P.) with a = 1 and r = 3 2, 4, 8, 16, ... is a geometric progression (G.P.) with a = 2 and r = 2
a.
n th term of a geom etric progression (G.P.)
t n = a r n−1 where t n = nth term, a= the first term , r = common ratio, n = number of terms Ex am ple 1 : Find the 10th term in the series 2, 4, 8, 16, ... a = 2,
r=
4 = 2, n = 10 2
10th term, t 10 = a r
n−1
= 2 × 210−1 = 2 × 29 = 2 × 512 = 1024
Ex am ple 2 : Find 5th term in the series 5, 15, 45, ... a = 5,
r=
15 = 3, n = 5 5
5th term, t 5 = a r
b.
n−1
= 5 × 35−1
= 5 × 34 = 5 × 81 = 405
Sum of first n term s in a geom etric progression (G.P.)
.
a(r n − 1) r−1
(if r > 1)
.
a(1 − r n ) 1−r
(if r < 1)
Sn =
where a= the first term , r = common ratio, n = number of terms
Ex am ple 1 : Find 4 + 12 + 36 + ... up to 6 terms
a = 4,
r=
12 = 3, n = 6 4 S6 =
Here r > 1. Hence,
Ex am ple 2 : Find
Here r < 1. Hence,
c.
1 1 + + 2 4
1+
1 ( ) 2 a = 1, r = 1
=
a(r n − 1) 4(36 − 1) 4(729 − 1) 4 × 728 = = = = 2 × 728 = 1456 r−1 3−1 2 2
1 , 2
S6 =
... up to 5 terms
n =5
n
a(1 − r ) = 1−r
1 1 [1 − ( ) 2 (1 −
5
] =
1 ) 2
1 ) 32 1 ( ) 2
(1 −
31 ) 32 = 1 ( ) 2 (
=
31 15 =1 16 16
Sum of an infinite geom etric progression (G.P.)
S∞ =
a 1−r
(if 0 < r < 1)
where a= the first term , r = common ratio
Ex am ple
a = 1,
: Find 1 + 1 ( ) 2 r= 1
1 1 1 + + + ...∞ 2 4 8
=
Here 0 < r < 1. Hence,
d.
1 2
S∞ =
a = 1−r
1 1 (1 − ) 2
=
1 1 ( ) 2
=2
Geom etric Mean
If three non-zero numbers a, b, c are in G.P., b is the Geometric Mean (G.M.) between a and c. In this case,
ac b = √ −−
The Geometric Mean (G.M.) between two numbers a and b = √
−− ab
(Note that if a and b are of opposite sign, their G.M. is not defined.)
e.
If a, b, c are in G.P., b2 = ac
f.
If a, b, c are in G.P.,
g.
In a G.P., product of terms equidistant from beginning and end will be constant.
a −b a = b−c b
To solve most of the problems related to G.P., the terms of the G.P. can be conveiently taken as
h.
4.
3 terms :
a , a, ar r
5 terms :
a a , , a, ar, ar2 2 r r
R elationship Betw een A rithmetic Mean, Harmonic Mean, and Geometric Mean of Tw o Numbers
If GM, AM and HM are the Geometic Mean, Arithmetic Mean and Harmonic Mean of two positive numbers respectively, then GM2 = AM × HM 5.
Some I nteresting P roperties to Note
Three numbers a, b and c are in AP if
b=
a +c 2
Three non-zero numbers a, b and c are in HP if
Three non-zero numbers a, b and c are in HP if
b=
2ac a +c
a −b a = b−c c
Let A, G and H be the A.M., G.M. and H.M. between two distinct positive numbers. Then i. A > G > H ii. A, G and H are in GP
If a series is both an A.P. and G.P., all terms of the series will be equal. In other words, it will be a constant sequence.
6.
P ow er Series : I mportant formulas
1.
1 + 1 + 1 + ⋯ n terms = ∑ 1 = n
2.
1+2+3+⋯+n = ∑n =
3.
12 + 22 + 32 + ⋯ + n 2 = ∑ n 2 =
4.
n 2 (n + 1) 1 + 2 + 3 +⋯ + n = ∑n = 4 3
3
3
3
n(n + 1) 2
3
n(n + 1)(2n + 1) 6 2
n(n + 1) ] =[ 2
2
I mportant Concepts and Formulas - Trigonometry I.
Trigonometric Basics
sin θ =
opposite side y = hypotenuse r
cos θ =
adjacent side x = hypotenuse r
tan θ =
opposite side y = adjacent side x
csc θ =
hypotenuse r = opposite side y
sec θ =
hypotenuse r = adjacent side x
cot θ =
adjacent side x = opposite side y
From Pythagorean theorem,
II.
x2 + y 2 = r 2
for the right angled triangle mentioned above
Basic Trigonometric Values
θ
θ
in degrees in radians
0°
0
sin θ
cos θ
0
1
tan θ 0
III.
30°
π 6
1 2
√3 2
45°
π 4
1 √2
1 √2
1
60°
π 3
√3 2
1 2
√3
90°
π 2
1
0
Not defined
180°
π
0
-1
0
270°
3π 2
-1
0
Not defined
360°
2π
0
1
0
Trigonometric Formulas
Degrees to Radians and vice versa
360° = 2π radian
Trigonom etry Quotient F orm ulas
tan θ =
sin θ cos θ
cot θ =
cos θ sin θ
1 √3
Trigonom etry - Reciprocal F orm ulas
csc θ =
1 sin θ
sec θ =
1 cos θ
cot θ =
1 tan θ
Trigonom etry - Pythagorean F orm ulas
sin 2 θ + cos2 θ = 1 sec2 θ − tan 2 θ = 1 csc2 θ − cot 2 θ = 1
Trigonom etry - Even-Odd F orm ulas
sin (−θ) = − sin θ cos (−θ) = cos θ tan (−θ) = − tan θ csc (−θ) = − csc θ sec (−θ) = sec θ cot (−θ) = − cot θ
Trigonom etry - Periodic F orm ulas if n is an integer,
sin (θ + 2πn) = sin θ cos (θ + 2πn) = cos θ tan (θ + 2πn) = tan θ csc (θ + 2πn) = csc θ sec (θ + 2πn) = sec θ cot (θ + 2πn) = cot θ (In these formulas, θ is mentioned in radians. If θ is in degrees, substitute 360° for
Trigonometry - Reduction Formulas F irst Quadrant
sin(
π − θ) = cos θ 2
π cos( − θ) = sin θ 2
Second Quadrant
sin(
π + θ) = cos θ 2
π cos( + θ) = − sin θ 2
tan(
π − θ) = cot θ 2
tan(
π + θ) = − cot θ 2
csc(
π − θ) = sec θ 2
csc(
π + θ) = sec θ 2
π sec( − θ) = csc θ 2 tan(
π − θ) = tan θ 2
Third Quadrant
π sec( + θ) = − csc θ 2 cot(
π + θ) = − tan θ 2
sin(π − θ) = sin θ cos(π − θ) = − cos θ tan(π − θ) = − tan θ csc(π − θ) = csc θ sec(π − θ) = − sec θ cot(π − θ) = − cot θ
F ourth Quadrant
sin(π + θ) = − sin θ
sin(2π − θ) = − sin θ
cos(π + θ) = − cos θ
cos(2π − θ) = cos θ
tan(π + θ) = tan θ
tan(2π − θ) = − tan θ
csc(π + θ) = − csc θ
csc(2π − θ) = − csc θ
sec(π + θ) = − sec θ
sec(2π − θ) = sec θ
2π )
cot(π + θ) = cot θ
cot(2π − θ) = − cot θ
(Note that in the above formulas, θ is mentioned in radians)
Trigonom etry - Sum -Difference F orm ulas
sin(θ1 + θ2 ) = sin θ1 cos θ2 + cos θ1 sin θ2 sin(θ1 − θ2 ) = sin θ1 cos θ2 − cos θ1 sin θ2 cos(θ1 + θ2 ) = cos θ1 cos θ2 − sin θ1 sin θ2 cos(θ1 − θ2 ) = cos θ1 cos θ2 + sin θ1 sin θ2 tan(θ1 + θ2 ) =
tan θ1 + tan θ2 1 − tan θ1 tan θ2
tan(θ1 − θ2 ) =
tan θ1 − tan θ2 1 + tan θ1 tan θ2
Trigonom etry - Double Angle F orm ulas
sin(2θ) = 2 sin θ cos θ cos(2θ) = cos2 θ − sin 2 θ = 1 − 2 sin 2 θ = 2 cos2 θ − 1 tan(2θ) =
2 tan θ 1 − tan 2 θ
sin θ = cos θ = tan θ =
eiθ − e−iθ 2i eiθ + e−iθ 2 eiθ − e−iθ (eiθ + e−iθ ) i
csc θ =
2i eiθ − e−iθ
sec θ =
2 eiθ + e−iθ
cot θ =
(eiθ + e−iθ ) i eiθ − e−iθ
Trigonom etry - Sum -to-Product F orm ulas
sin θ1 + sin θ2 = 2 sin (
θ1 + θ2 θ1 − θ2 ) cos ( ) 2 2
sin θ1 − sin θ2 = 2 cos (
θ1 + θ2 θ1 − θ2 ) sin ( ) 2 2
cos θ1 + cos θ2 = 2 cos (
θ1 + θ2 θ1 − θ2 ) cos ( ) 2 2
cos θ1 − cos θ2 = −2 sin (
θ1 + θ2 θ1 − θ2 ) sin ( ) 2 2
Trigonom etry - Product-to-Sum F orm ulas
sin θ1 sin θ2 =
1 [cos (θ1 − θ2 ) − cos(θ1 + θ2 )] 2
cos θ1 cos θ2 =
1 [cos (θ1 − θ2 ) + cos(θ1 + θ2 )] 2
sin θ1 cos θ2 =
1 [sin (θ1 + θ2 ) + sin(θ1 − θ2 )] 2
cos θ1 sin θ2 =
1 [sin (θ1 + θ2 ) − sin(θ1 − θ2 )] 2
1. (935421 x 625) = ? A. 542622125
B. 584632125
C. 544638125
D. 584638125
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
10 935421 × 625 = 935421 × 5 = 935421 × ( ) 2 4
4
=
935421 × 10000 = 584638125 16
2. Which of the following is a prime number ? A. 9
B. 8
C. 4
D. 2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : 2 is a prime number A prime number is a natural number greater than 1 which has no positive divisors other than 1 and itself. Hence the primer numbers are 2,3,5,7,11,13,17,...
3. What is the largest 4 digit number exactly divisible by 88? A. 9944
B. 9999
C. 9988
D. 9900
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Largest 4 digit number = 9999 9999 ÷ 88 = 113, remainder = 55 Hence largest 4 digit number exactly divisible by 88 = 9999 - 55
= 9944
4. {(481 + 426)2 - 4 x 481 x 426} = ? A. 3025
B. 4200
C. 3060
D. 3210
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
(a + b) 2 = a 2 + 2ab + b 2
(a - b) 2 = a 2 - 2ab + b 2
Here, the given statement is like (a + b)2 - 4ab where a= 481 and b = 426 (a + b)2 - 4ab = (a2 + 2ab + b2) - 4ab = a2 - 2ab + b2 = (a b)2 Hence {(481 + 426)2 - 4 x 481 x 426} = (481 - 426)2 = 552 = 3025
5. (64 - 12)2 + 4 x 64 x 12 = ? A. 5246
B. 4406
C. 5126
D. 5776
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
(a + b) 2 = a 2 + 2ab + b 2
(a - b) 2 = a 2 - 2ab + b 2
Here, the given statement is like (a - b)2 + 4ab where a= 64 and b = 12 (a - b)2 + 4ab = (a2 - 2ab + b2) + 4ab = a2 + 2ab + b2 = (a + b)2 Hence (64 - 12)2 + 4 x 64 x 12 = (64 + 12)2 = 762 = 5776
6. 121 x 54 = ? A. 68225
B. 75625
C. 72325
D. 71225
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
10 121 × 5 = 121 × ( ) 2 4
4
=
121 × 10000 = 7.5625 × 10000 = 75625 16
7. If (232 + 1) is completely divisible by a whole number, which of the following numbers is completely divisible by this number? A. (296 + 1)
B. (7 x 223 )
C. (216 - 1)
D. (216 + 1)
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let 232 = x. Then (232 + 1) = (x + 1) Assume that (x + 1) is completely divisible by a whole number,
N (296 + 1) = {(232)3 + 1} = (x3 + 1) = (x + 1)(x2 - x + 1) if (x + 1) is completely divisible by N, (x + 1)(x2 - x + 1) will also be divisible by N Hence (296 + 1) is completely divisible N
8. How many of the following numbers are divisible by 132 ? 264, 396, 462, 792, 968, 2178, 5184, 6336 A. 4
B. 3
C. 6
D. 8
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
If a number is divisible by two co-prime numbers, then the number is divisible by their product also. If a number is divisible by more than two pairwise co-prime numbers, then the number is divisible by their product also. Read more
If a number is divisible by another number, then it is also divisible by all the factors of that number. Read more
Here 3, 4 and 11 are pairwise co-prime numbers. 132 = 3 × 4 × 11 . Also 3,4 and 11 are factors of 132. Hence
if a number is divisible by 3, 4 and 11, the number will be divisible by their product 132 also. If a number is not divisible by 3 or 4 or 11, it is not divisible by 132 You must learn Divisibility Rules t o say whether a given number is divisible by another number without actually performing the division. throughdivisibility rules before proceeding
Please
go
further. 264 is divisible by 3, 4 and 11 => 264 is divisible by 132 396 is divisible by 3, 4 and 11 => 396 is divisible by 132 462 is divisible by 3 and 11, but not divisible by 4 => 462 is not divisible by 132 792 is divisible by 3, 4 and 11 => 792 is divisible by 132 968 is divisible by 4 and 11, but not divisible by 3 => 968 is not divisible by 132 2178 is divisible by 3 and 11, but not divisible by 4 => 2178 is not divisible by 132 5184 is divisible by 3 and 4, but not divisible by 11 => 5184 is not divisible by 132 6336 is divisible by 3, 4 and 11 => 6336 is divisible by 132 Hence, only 264, 396 ,792 and 6336 are divisible by 132. So the answer is 4
9. All prime numbers are odd numbers A. True
B. False
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 2 is even prime number
10. What is the unit digit in (6324)1797 × (615)316 × (341)476 ? A. 1
B. 2
C. 4
D. 0
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Unit digit in (6324)1797 = Unit digit in (4)1797 = Unit digit in [(42)898 × 4] = Unit digit in [16898 × 4] = Unit digit in (6 × 4) = 4 Unit digit in (615)316 = Unit digit in (5)316 = 5 Unit digit in (341)476 = Unit digit in (1)476 = 1 Hence, unit digit in (6324)1797 × (625)316 × (341)476 = Unit digit in [4 × 5 × 1] = 0
11. 5216 x 51 = ? A. 266016
B. 212016
C. 266436
D. 216314
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Normal way of multiplication may take time. Here are one alternative. 5216 x 51 = (5216 x 50) + 5216 = (5216 x 100/2) + 5216 = 521600/2 + 5216 = 260800 + 5216 = 266016
12. Which of the following number is divisible by 24 ? A. 31214
B. 61212
C. 512216
D. 3125832
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option D Explanation :
If a number is divisible by two co-prime numbers, then the number is divisible by their product also. If a number is divisible by more than two pairwise co-prime numbers, then the number is divisible by their product also. Read more
If a number is divisible by another number, then it is also divisible by all the factors of that number. Read more
24 = 3 × 8 where 3 and 8 are co-prime numbers. 3 and 8 are also factors of 24. Hence Hence if a number is divisible by 3, and 8, the number will be divisible by their product 24 also. If a number is not divisible by 3 or 8, it is not divisible by 24 You must learn Divisibility Rules t o say whether a given number is divisible by another number without actually performing the division. throughdivisibility rules before proceeding
Please
go
further. 31214 is not divisible by 3 and 8 => 31214 is not divisible 24 61212 is not divisible by 8 though it is divisible by 3 => 61212 is not divisible 24 512216 is not divisible by 3 though it is divisible by 8 => 512216 is not divisible 24
3125832 is divisible by 3 and 8 => 3125832 is divisible 24
719 × 719 + 347 × 347 − 719 × 347 719 × 719 × 719 + 347 × 347 × 347 1 25 A. B. 372 133 1 5 C. D. 1066 6 13.
=?
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
The given statement is in the form
a 2 + b2 − ab a 3 + b3
where a = 719 and b = 347
(Reference : Basic Algebraic Formulas)
a 2 + b2 − ab 3
a3 + b
=
( a2 + b2 − ab) 2
(a + b)(a 2 − ab + b )
=
1 1 1 1 = = = a +b 719 + 347 719 + 347 1066
14. If the number 481*673 is completely divisible by 9, what is the the smallest whole number in place of *? A. 3
B. 7
C. 5
D. 9
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let x be the smallest whole number in place of * Given that 481*673 to be completely divisible by 9, => (4 + 8 + 1 + x + 6 + 7 + 3) is divisible by 9 (Reference : Divisibility by 9) => (29 + x) is divisible by 9 x should be the smallest whole number, Hence, (29 + x) = 36 => x = 36 - 29 = 7
15. If n is a natural number, then (6n2 + 6n) is always divisible by: A. Both 6 and 12
B. 6 only
C. 12 only
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 6n2 + 6n = 6n(n + 1) Hence 6n2 + 6n is always divisible by 6 and 12 (∵ remember that n(n + 1) is always even)
16. 109 × 109 + 91 × 91 = ? A. 20162
B. 18322
C. 13032
D. 18662
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 2
2
2
(a + b) + (a − b) = 2(a 2 + b ) (Reference : Basic Algebraic Formulas)
1092 + 912 = (100 + 9) 2 + (100 − 9) 2 = 2(1002 + 92 ) = 2(10000 + 81) = 20162
17. When (6767 +67) is divided by 68, the remainder is A. 0
B. 22
C. 33
D. 66
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
(x n + 1) is divisible by (x + 1) only when n is odd => (6767 + 1) is divisible by (67 + 1)
=> (6767 + 1) is divisible by 68 => (6767 + 1) ÷ 68 gives a remainder of 0 => [(6767 + 1) + 66] ÷ 68 gives a remainder of 66 => (6767 + 67) ÷ 68 gives a remainder of 66
18.
(912 + 643) 2 + (912 − 643)2 (912 × 912 + 643 × 643)
A. 122
B. 2
C. 1
D. None of these
=?
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
(a + b) 2 + (a − b) 2 = 2(a 2 + b2 ) (912 + 643) 2 + (912 − 643)2 (912 × 912 + 643 × 643)
=
(912 + 643) 2 + (912 − 643)2 (9122 + 6432 )
19. What is the smallest prime number? A. 0
B. 1
C. 2
D. 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : smallest prime number is 2. 0 and 1 are neither prime numbers nor composite numbers.
20. (23341379 x 72) = ? A. 1680579288
B. 1223441288
C. 2142579288
D. 2142339288
Hide Answer | Notebook | Discuss Here is the answer and explanation
=
2(9122 + 6432 ) (9122 + 6432 )
=2
Answer : Option A Explanation : 23341379 x 72 = 23341379(70 + 2) = (23341379 x 70) + (23341379 x 2) = 1633896530 + 46682758 = 1680579288
21. If the number 5 * 2 is divisible by 6, then * = ? A. 2
B. 7
C. 3
D. 6
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
A number is divisible by 6 if it is divisible by both 2 and 3 Read More
Replacing * by x 5 x 2 is divisible by 2 (Reference : Divisibility by 2 rule) For 5 x 2 to be divisible by 3, 5 + x + 2 shall be divisible by 3(Reference : Divisibility by 3 rule) => 7 + x shall be divisible by 3 => x can be 2 or 5 or 8 From the given choices, answer = 2
1 2 3 ) + (1 − ) + (1 − ) +. . . up to n terms =? n n n n A. (n − 1) B. 2 1 1 C. (n − 1) D. (n + 1) 2 2 22.
(1 −
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
1 + 1 + 1 + ⋯ n terms = ∑ 1 = n 1+2+3+⋯+n = ∑n =
n(n + 1) 2
(Reference : Click Here)
(1 −
1 2 3 ) + (1 − ) + (1 − ) + ... up to n terms n n n
= (1 + 1 + 1 + ... up to n terms) − (
1 2 3 + + + ... up to n terms) n n n
= n−
1 (1 + 2 + 3 + ... up to n terms) n
= n−
1 n(n + 1) [ ] n 2
= n−
(n + 1) 2
=
(2n − n − 1) 2
=
n−1 2
23. What least number should be added to 1056, so that the sum is completely divisible by 23? A. 4
B. 3
C. 2
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : 1056 ÷ 23 = 45 with remainder = 21 Remainder = 21. 21 + 2 = 23. Hence 2 should be added to 1056 so that the sum will be
divisible by 23
24. 1398 x 1398 = ? A. 1624404
B. 1851404
C. 1951404
D. 1954404
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : 1398 x 1398 = (1398)2 = (1400 - 2)2 = 14002 - (2 × 1400 × 2) + 22 = 1960000 - 5600 + 4 = 1954404 Note : It will be great if you master speed maths techniques based on Vedic Mathematics, Trachtenberg System of Mathematics etc so that you can do these calculations even faster (Reference : Click Here)
25. On dividing a number by 56, we get 29 as remainder. On dividing the same number by 8, what will be the remainder ? A. 2
B. 3
C. 4
D. 5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : -----------------------------------------------------------------------------------Solution 1 -------------------------------------------------------------
-----------------------Number = 56x + 29 (∵ since the number gives 29 as remainder on dividing by 56) = (7 × 8 × x) + (3 × 8) + 5 Hence, if the number is divided by 8, we will get 5 as remainder. -----------------------------------------------------------------------------------Solution 2 -----------------------------------------------------------------------------------Number = 56x + 29 Let x = 1. Then the number = 56 × 1 + 29 = 85 85 ÷ 8 = 10, remainder = 5
26. ? + 3699 + 1985 - 2047 = 31111 A. 21274
B. 27474
C. 21224
D. 27224
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let x + 3699 + 1985 - 2047 = 31111 => x = 31111 - 3699 - 1985 + 2047 = 27474
27. the difference between a positive fraction and its reciprocal is 9/20 find the sum of that fraction and its reciprocal.
41 20 11 C. 20 A.
17 20 9 D. 20 B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : (Reference : Quadratic Equations and How to Solve Quadratic Equations)
Let the fraction = x
Then x −
1 9 = x 20
x2 − 1 9 = x 20
⇒
⇒ x2 − 1 =
9x 20
⇒ 20x 2 − 20 = 9x ⇒ 20x 2 − 9x − 20 = 0 −−−−−−− −b ± √ b2 − 4ac x= 2a
=
−−−−−−−−−−−−−−−−−− − 2 9 ± √ (−9) − 4 × 20 × (−20) 2 × 20
−−−−−− 9±√− 81 + 1600− 9 ± 41 50 32 = = = Or − 40 40 40 40
Given that the fraction is positive. Hence 50 5 = 40 4
x=
1 4 = x 5 x+
1 5 4 5×5+4×4 = + = x 4 5 20
=
41 20
28. How many 3 digit numbers are completely divisible 6 ? A. 146
B. 148
C. 150
D. 152
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option C Explanation : 100/
6 = 16, remainder = 4. Hence 2 more should be added to
100 to get the minimum 3 digit number divisible by 6. => Minimum 3 digit number divisible by 6 = 100 + 2 = 102 999/
6 = 166, remainder = 3. Hence 3 should be decreased from
999 to get the maximum 3 digit number divisible by 6. => Maximum 3 digit number divisible by 6 = 999 - 3 = 996 Hence, the 3 digit numbers divisible by 6 are 102, 108, 114,... 996 This is Arithmetic Progression with a = 102 ,d = 6, l=996
Number of terms =
(l − a) (996 − 102) 894 +1 = +1 = + 1 = 149 + 1 = 150 d 6 6
29. How many natural numbers are there between 43 and 200 which are exactly divisible by 6? A. 28
B. 26
C. 24
D. 22
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 43/
6 = 7, remainder = 1. Hence 5 more should be added to 43 to
get the minimum number divisible by 6 between 43 and 200. => Minimum number divisible by 6 between 43 and 200 = 43 + 5 = 48 200/
6 = 33, remainder = 2. Hence 2 should be decreased from
200 to get the maximum number divisible by 6 between 43 and 200.
=> Maximum number divisible by 6 between 43 and 200 = 200 2 = 198 Hence, natural numbers numbers divisible by 6 between 43 and 200 are 48, 54, 60,...198 This is Arithmetic Progression with a = 48, d = 6, l=198
Number of terms =
(l − a) (198 − 48) 150 +1 = +1 = + 1 = 25 + 1 = 26 d 6 6
30. What is the smallest 6 digit number exactly divisible by 111? A. 100010
B. 100011
C. 100012
D. 100013
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Smallest 6 digit number = 100000 100000/
111 = 900, remainder = 100. Hence 11 more should be
added to 100000 to get the smallest 6 digit number exactly divisible by 111 => smallest 6 digit number exactly divisible by 111 = 100000 + 11 = 100011
31. If x and y are positive integers such that (3x + 7y) is a multiple of 11, then which of the followings are divisible by 11? A. 9x + 4y C. 4x - 9y
B. x + y + 4 D. 4x + 6y
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : By hit and trial method, we get x=5 and y=1 such that 3x + 7y = 15 + 7 = 22 is a multiple of 11. Then
(4x + 6y) = (4 × 5 + 6 × 1) = 26 which is not divisible by 11 (x + y + 4) = (5 + 1 + 4) = 10 which is not divisible by 11 (9x + 4y) = (9 × 5 + 4 × 1) = 49 which is not divisible by 11 (4x - 9y) = (4 × 5 - 9 × 1) = 20 - 9 = 11 which is divisible by 11
32. if (64)2 - (36)2 = 10x, then x = ? A. 200
B. 220
C. 210
D. 280
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
a 2 − b2 = (a − b)(a + b) (64)2 - (36)2 = (64 - 36)(64 + 36) = 28 × 100 Given that (64)2 - (36)2 = 10x => 28 × 100 = 10x => x = 280
33.
852 × 852 × 852 − 212 × 212 × 212 852 × 852 + 852 × 212 + 212 × 212
A. 640
B. 620
C. 740
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
a 3 − b3 = (a − b)(a 2 + ab + b2 )
=?
Given Equation is in the form a 3 − b3
=
a 2 + ab + b2
a 3 − b3 a 2 + ab + b2
(a − b)(a 2 + ab + b2 )
where a = 852 and b = 212 = (a − b)
( a2 + ab + b2 )
Hence answer = (a - b) = (852 - 212) = 640
34.
2664 ÷ 12 ÷ 6 =?
A. 43
B. 41
C. 37
D. 33
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : 2664 ÷ 12 = 222 222 ÷ 6 = 37
OR
2664 ÷ 12 ÷ 6 = 2664 ×
35.
1 1 × = 37 12 6
2
(422 + 404) − (4 × 422 × 404) =?
A. None of these
B. 342
C. 324
D. 312
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
2
(a + b) − 4ab = (a 2 + 2ab + b2 ) − 4ab = (a2 − 2ab + b2 ) = (a − b)
2
2
Given Equation is in the form (a + b) − 4ab where a = 422 and b = 404 2
2
2
Hence answer = (a + b) − 4ab = (a − b) = (422 − 404) = 182 = 324
36. Which one of the following can't be the square of natural number ? A. 128242
B. 128881
C. 130321
D. 131044
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Square of a natural number cannot end with 2 . Hence 128242 cannot be the square of natural number
37. (32323 + 7344 + 41330) - (317 x 91) = ? A. 54210
B. 54250
C. 52150
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Its speed and accuracy which decides the winner. You can use various methods including Vedic mathematics to do complex calculations quickly. (32323 + 7344 + 41330) - (317 x 91) = 80997 - 28847 = 52150
38. (xn - an) is completely divisible by (x - a), if A. n is an even natural number
B. n is and odd natural number
C. n is any natural number
D. n is prime
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : (xn - an) is completely divisible by (x - a) for every natural number n
39. The number 97215*6 is completely divisible by 11. What is the smallest whole number in place of * ?
A. 4
B. 2
C. 1
D. 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : (Reference : Divisibility by 11 rule) From the above given "Divisibility by 11 rule", it is clear that if 97215*6 is divisible by 11, then (9 + 2 + 5 + 6) - (7 + 1 + *) is divisible by 11 => 22 - (8 + *) is divisible by 11 => (14 - *) is divisible by 11 The smallest whole number which can be substituted in the place of * to satisfy the above equation is 3 such that 14 - 3 = 11 and 11 is divisible by 11. Hence the answer is 3
40. (12 + 22 + 32 + ... + 102) = ? A. 395
B. 375
C. 55
D. 385
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : (Reference : Power Series : Important formulas)
12 + 22 + 32 + ⋯ + n 2 = ∑ n 2 =
n(n + 1)(2n + 1) 6
12 + 22 + 32 + ⋯ + 102 = =
n(n + 1)(2n + 1) 6
=
10(10 + 1)[(2 × 10) + 1] 6
10 × 11 × 21 10 × 11 × 7 = = 385 6 2
41. If the product 4864 × 9a2 is divisible by 12, then what is the value of a? A. 1
B. 2
C. 5
D. 6
Hide Answer | Discuss Here is the answer and explanation Answer : Option A Explanation : (Reference : Divisibility by 12)
A number is divisible by 12 if the number is divisible by both 3 and 4 A number is divisible by 3 if the sum of the digits is divisible by 3 A number is divisible by 4 if the number formed by the last two digits is divisible by 4.
4864 × 9a2 is divisible by 12 => 4864 × 9a2 is divisible by 3 and 4864 × 9a2 is divisible by 4 4864 is divisible by 4 (Because number formed by the last two digits = 64 which is divisible by 4) Hence 4864 × 9a2 will also be divisible by 4 4864 is not divisible by 3 (because 4 + 8 + 6 + 4 = 22 which is not divisible by 3) Hence 9a2 should be divisible by 3 such that 4864 × 9a2 is divisible by 3 => 9 + a + 2 is divisible by 3
=> 11 + a is divisible by 3 Hence a can be 1 or 4 or 7 such that 11 + a is divisible by 3 So, from the given choices, 1 is the answer
42. -88 × 39 + 312 = ? A. -3120
B. -3200
C. 3120
D. 3200
Hide Answer | Discuss Here is the answer and explanation Answer : Option A Explanation : Again a question on basic arithmetic. Many algebraic formulas and Vedic Maths can help you to solve such questions very fast. -88 × 39 + 312 = -2112 + 312 = -3432 + 312 = -3120
OR Since 88 can be written as 11 × 8, -88 × 39 + 312 = -11 × 8 × 39 + 312 = -11 × 312 + 312 (Reference : Multiplication by 11 using Speed Mathematics) = -3432 + 312 = -3120
OR -88 × 39 + 312 = -88 × (40-1) + 312 = -88 × 40 + 88 + 312 = -3520 + 88 + 312 = -3120
OR (using various Speed Mathematics Techniques)
43. 378 × ? = 252
2 3 1 C. 2 A.
B.
3 4
D. None of these
Hide Answer | Discuss Here is the answer and explanation Answer : Option A Explanation :
378 × ? = 252 252 126 14 2 ?= = = = 378 189 21 3
44. What least number should be subtracted from 13601 such that the remainder is divisible by 87 ? A. 27
B. 28
C. 29
D. 30
Hide Answer | Discuss Here is the answer and explanation Answer : Option C Explanation : 13601 ÷ 87 = 156, remainder = 29 Hence 29 is the least number which can be subtracted from 13601 such that the remainder is divisible by 87
45. Which one of the given numbers is completely divisible by 45? A. None of these
B. 165642
C. 202860
D. 112330
Hide Answer | Discuss Here is the answer and explanation Answer : Option C Explanation :
If a number is divisible by two co-prime numbers, then the number is divisible by their product also. If a number is divisible by more than two pairwise co-prime numbers, then the number is divisible by their product also. Read more
If a number is divisible by another number, then it is also divisible by all the factors of that number. Read more
We know that 45 = 9 × 5 where 9 and 5 are co-prime numbers. Also 9 and 5 are factors of 45. Hence if a number is divisible by 5 and 9, the number will be divisible by their product 45 also. If a number is not divisible by 5 or 9, it is not divisible by 45 You must learn Divisibility Rules t o say whether a given number is divisible by another number without actually performing the division. throughdivisibility rules before proceeding
Please
go
further. 112330 is divisible by 5 but not divisible by 9 => 112330 is not divisible by 45 202860 is divisible by 5 and 9 => 202860 is divisible by 45 165642 is not divisible by 5 and 9 => 165642 is not divisible by 45 Hence, 202860 is the answer
46. What is the remainder when 17200 is divided by 18 ? A. 3
B. 2
C. 1
D. 4
Hide Answer | Discuss Here is the answer and explanation Answer : Option C Explanation :
(x n − a n ) is completely divisible by (x + a) when n is even Read More... (17200 - 1200) is completely divisible by (17 + 1) as 200 is even. => (17200 - 1) is completely divisible by 18. Hence, when 17200 is divided by 18, we will get 1 as remainder.
47. 12 + 22 + 32 + ... + 82 = ? A. 204
B. 200
C. 182
D. 214
Hide Answer | Discuss Here is the answer and explanation Answer : Option A Explanation : (Reference : Power Series : Important formulas)
12 + 22 + 32 + ⋯ + n 2 = ∑ n 2 =
12 + 22 + 32 + ⋯ + 82 = =
n(n + 1)(2n + 1) 6
n(n + 1)(2n + 1) 6
=
8(8 + 1)[(2 × 8) + 1] 6
8 × 9 × 17 4 × 9 × 17 = = 4 × 3 × 17 = 204 6 3
48. 1 + 2 + 3 + ... + 12 = ? A. 66
B. 68
C. 76
D. 78
Hide Answer | Discuss Here is the answer and explanation Answer : Option D Explanation : (Reference : Power Series : Important formulas)
1+2+3+⋯+n = ∑n =
1 + 2 + 3 + ⋯ + 12 = =
n(n + 1) 2
n(n + 1) 2
=
12(12 + 1) 2
12 × 13 = 6 × 13 = 78 2
49. 13 + 23 + 33 + ... + 63 = ? A. 451
B. 441
C. 421
D. 401
Hide Answer | Discuss Here is the answer and explanation Answer : Option B Explanation : (Reference : Power Series : Important formulas)
n 2 (n + 1) 3 3 3 3 3 1 + 2 + 3 +⋯ + n = ∑n = 4
n(n + 1) ] 13 + 23 + 33 + ⋯ + 63 = [ 2 2
= [3 × 7] = 212 = 441
2
2
n(n + 1) ] =[ 2
6(7) ] =[ 2
2
2
50. Which one of the following is a prime number ? A. 307
B. 437
C. 247
D. 203
Hide Answer | Discuss Here is the answer and explanation Answer : Option A Explanation :
−−− √ 307 < 18 Prime numbers < 18 are 2, 3, 5, 7, 11, 13, 17 307 is not divisible by 2 307 is not divisible by 3 307 is not divisible by 5 307 is not divisible by 7 307 is not divisible by 11 307 is not divisible by 13 307 is not divisible by 17 Hence 307 is a prime number
−−− √ 437 < 21 Prime numbers < 21 are 2, 3, 5, 7, 11, 13, 17, 19 437 is not divisible by 2 437 is not divisible by 3 437 is not divisible by 5 437 is not divisible by 7 437 is not divisible by 11 437 is not divisible by 13 437 is not divisible by 17 But 437 is divisible by 19 => 437 is not a prime number
−−− √ 247 < 16 Prime numbers < 16 are 2, 3, 5, 7, 11, 13 247 is not divisible by 2 247 is not divisible by 3 247 is not divisible by 5 247 is not divisible by 7 247 is not divisible by 11 But 247 is divisible by 13 => 247 is not a prime number
−−− √ 203 < 15 Prime numbers < 15 are 2, 3, 5, 7, 11, 13 203 is not divisible by 2 203 is not divisible by 3 203 is not divisible by 5 But 203 is divisible by 7 => 203 is not a prime number
51. The difference of the squares of two consecutive even integers is divisible by which of the following numbers? A. 3
B. 6
C. 4
D. 7
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : --------------------------------------------------Solution 1 --------------------------------------------------Let the consecutive even integers are n and (n + 2) difference of the squares = (n + 2)2 - n2 = (n2 + 4n + 4) - n2 = 4n + 4 = 4(n + 1) which is divisible by 4 --------------------------------------------------Solution 2 (preferred way) --------------------------------------------------Just take any of the two consecutive even integers say 2 and 4 difference of the squares = 42 - 22 = 16 - 4 = 12 which is divisible by 4
52. Which one of the following numbers is completely divisible by 99? A. 115909
B. 115919
C. 115939
D. 115929
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
If a number is divisible by two co-prime numbers, then the number is divisible by their product also. If a number is divisible by more than two pairwise co-prime numbers, then the number is divisible by their product also. Read more
If a number is divisible by another number, then it is also divisible by all the factors of that number. Read more
We know that 99 = 9 × 11 where 9 and 11 are co-prime numbers. Also 9 and 11 are factors of 99. Hence if a number is divisible by 9 and 11, the number will be divisible by their product 99 also. If a number is not divisible by 9 or 11, it is not divisible by 99 You must learn Divisibility Rules t o say whether a given number is divisible by another number without actually performing the division. throughdivisibility rules before proceeding
Please
go
further. 115929 is divisible by both 9 and 11 => 115929 is divisible by 99 115939 is not divisible by 9 and 11 => 115939 is not divisible by 99 115919 is not divisible by 9 and 11 => 115919 is not divisible by 99 115909 is not divisible by 9 and 11 => 115909 is not divisible by
99 Hence, 115929 is the answer
612 × 612 × 612 + 321 × 321 × 321 612 × 612 − 612 × 321 + 321 × 321
53.
A. 933
B. 1000
C. 712
D. 843
=?
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
a 3 + b3 = (a + b)(a 2 − ab + b2 ) Read more...
Given Equation is in the form
(a 3 + b3 ) (a 2 − ab + b2 )
=
(a 3 + b3 ) (a 2 − ab + b2 )
(a + b) (a 2 − ab + b2 ) (a 2 − ab + b2 )
where a = 612 and b = 321
= (a + b)
Hence answer = (a + b) = 612 + 321 = 933
54. How many terms are there in the G.P. 4, 8, 16, 32, ... , 1024? A. 9
B. 8
C. 7
D. 6
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
n th term of a geom etric progression (G.P.)
t n = a r n−1 where t n = nth term, a= the first term , r = common ratio, n = number of terms Read more…
a=4 r=
8 =2 4
t n = 1024
t n = a r n−1 ⇒ 1024 = 4 × 2n−1 ⇒ 2n−1 = n−1
⇒2
1024 = 256 = 28 4 8
=2
⇒ n−1 = 8 ⇒ n = 8+1 = 9
55. 123 × 123 + 288 × 288 + 2 × 123 × 288 = ? A. 168151
B. 178121
C. 168921
D. 162481
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
2
(a + b) = a 2 + 2ab + b2 Read more...
Given Equation is in the form a 2 + b2 + 2ab where a = 123 and b = 288 a 2 + b2 + 2ab = (a + b) 2 2
2
Hence answer = (a + b) = (123 + 288) = 4112 = 168921
56. If a number is divided by 6 , 3 is the remainder . What is remainder if the the square of the number is divided by 6? A. 5
B. 4
C. 3
D. 2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the number be x Given that if the number is divided by 6 , 3 is the remainder => x ÷ 6 = k, remainder = 3 Hence x = 6k + 3 Square of the number = x2 = (6k + 3)2 =36k2 + 36k + 9
∵ (a + b)2 = a2 + 2ab + b2
=36k2 + 36k + 6 + 3 = 6(6k2 + 6k + 1) + 3 Hence, when the square of the number is divided by 6, we get 3 as remainder.
57. In a division sum, the remainder is 0 when a student mistook the divisor by 12 instead of 21 and obtained 35 as quotient. What is the correct quotient ? A. 25
B. 20
C. 15
D. 10
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let x be the number x ÷ 12 = 35 , remainder = 0 => x = 35 × 12
x 35 × 12 5 × 12 = = = 5 × 4 = 20 21 21 3
correct quotient =
58. 1531 × 132 + 1531 × 68 = ? A. 306000
B. 306100
C. 306200
D. 306400
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
a(b + c) = ab + ac
(Distributive Law)
Read more... 1531 × 132 + 1531 × 68 = 1531 (132 + 68) = 1531 × 200 = 306200
59. 100010 ÷ 1028 = ? A. 10
B. 100
C. 1000
D. 10000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
(a m ) n = amn = (a n ) m
a m . a n = a m+n
am = a m−n n a
Read more...
10
1000
28
10
=
(103 )
10
28
10
=
1030 28
10
= 102 = 100
60. 2 + 22 + 222 + 2.22 = ? A. 246
B. 248
C. 248.12
D. 248.22
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D
61. Which of the following numbers will completely divide (4915 - 1) ? A. 6
B. 7
C. 8
D. 9
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
(x n − a n ) is completely divisible by (x + a) when n is even Read More...
(4915 − 1) = [(72 )
15
− 1] = (730 − 1) = (730 − 130 )
which is completely divisible by (7 + 1) = 8
62. How many even prime numbers are there less than 50? A. 1
B. 15
C. 2
D. 16
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 2 is the only even prime number
63. How many prime numbers are there less than 50 ? A. 13
B. 14
C. 15
D. 16
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 are the prime numbers less than 50
64. What is the difference between local value and face value of 7 in the numerical 657903? A. 6993
B. 69993
C. 7000
D. 7
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : (Local value of 7) - (Face value of 7) = (7000 - 7) = 6993
65. 108 + 109 + 110 + ... + 202 = ? A. 14615
B. 14625
C. 14715
D. 14725
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Num ber of term s of an arithm etic progression
n=
(l − a) +1 d
where n = number of terms, a= the first term , l = last term, d= common difference Read more...
Sum of first n term s in an arithm etic progression
Sn =
n n [ 2a + (n − 1)d ] = [ a + l ] 2 2
where a = the first term, Read more…
d= common difference,
th
l = t n = n term = a + (n − 1)d
a = 108 l = 202 d = 109 − 108 = 1
n=
(l − a) (202 − 108) +1 = + 1 = 94 + 1 = 95 d 1
Sn =
n 95 95 × 310 [a +l]= [ 108 + 202 ] = = 95 × 155 = 14725 2 2 2
OR
(Reference : Power Series : Important formulas)
1+2+3+⋯+n = ∑n =
n(n + 1) 2
108 + 109 + 110+. . . +202 = [1 + 2 + 3 + ⋯ + 202] − [1 + 2 + 3 + ⋯ + 107] =[
202 × 203 107 × 108 ]− [ ] 2 2
= [101 × 203] − [107 × 54] = 20503 − 5778 = 14725
66. 23732 × 999 = ? A. 23708268
B. 22608258
C. 22608268
D. 23708258
Hide Answer | Notebook | Discuss
Here is the answer and explanation Answer : Option A Explanation : 23732 × 999 = 23732 × (1000 - 1) = (23732 × 1000) - (23732 × 1) = 23732000 - 23732 = 23708268 If you are interested to do fast calculations very fast, please go through speed maths topics
67. 123427201 - ? = 568794 A. 123527207
B. 223521407
C. 123527407
D. 122858407
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let 123427201 - x = 568794 x = 123427201 - 568794 = 122858407
68. 2210 × ? = 884
2 3 1 C. 5 A.
2 5 3 D. 4 B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Let 2210 × x = 884
x=
884 442 34 2 = = = 2210 1105 85 5
(Reference : Divisibility Rules)
69. If P and Q are odd numbers, then which of the following is even ? A. P + Q
B. PQ
C. P + Q + 1
D. PQ + 2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
The sum of two odd numbers is an even number
Read more...
Hence P + Q is an even number
OR
Just take any two odd numbers, say 1 and 3 P + Q = 1 + 3 = 4 = an even number P + Q + 1 = 1 + 3 + 1 = 5 = an odd number PQ = 1 × 3 = 3 = an odd number PQ + 2 = (1 × 3) + 2 = 5 = an odd number Hence P + Q is the answer
70. Which of the following numbers will completely divide (319 + 320 + 321 + 322) ? A. 25
B. 16
C. 11
D. 12
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option D Explanation : 319 + 320 + 321 + 322 = 319(1 + 3 + 32 + 33) = 319 × 40 = 318 × 3 × 4 × 10 = 318 × 12 × 10 which is divisible by 12
71. What is the largest 5 digit number exactly divisible by 94? A. 99922
B. 99924
C. 99926
D. 99928
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Largest 5 digit number = 99999 99999 ÷ 94 = 1063, remainder = 77 Hence largest 5 digit number exactly divisible by 94 = 99999 77 = 99922
72. What is the smallest 5 digit number exactly divisible by 94? A. 10052
B. 10054
C. 10056
D. 10058
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Smallest 5 digit number = 10000 10000 ÷ 94 = 106, remainder = 36 94 - 36 = 58.
ie, 58 should be added to 10000 to make it divisible by 94. => smallest 5 digit number exactly divisible by 94 = 10000 + 58 = 10058
73. 1234 - ? = 4234 - 3361 A. 351
B. 361
C. 371
D. 379
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let 1234 - x = 4234 - 3361 x = 1234 - 4234 + 3361 = 361
74. 320 ÷ 2 ÷ 3 = ? A. None of these
B. 53.33
C. 160
D. 106
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
320 ÷ 2 ÷ 3 = 160 ÷ 3 = 53.33 OR
320 ÷ 2 ÷ 3 = 320 ×
1 1 160 × = = 53.33 2 3 3
75. 476**0 is divisible by both 3 and 11. What are the non-zero digits in the hundred's and ten's places respectively? A. 8 and 5
B. 6 and 5
C. 8 and 2
D. 6 and 2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A
Explanation :
References 1. Divisibility by 3
2. Divisibility by 11
----------------------------------------------------------------------------Solution 1(Recom m ended) - hit and trial m ethod ----------------------------------------------------------------------------Just substitute the values in the missing places and apply divisibility rules. if a=6 and b=2, number = 476620 4 + 7 + 6 + 6 + 2 + 0 = 25 which is not divisible by 3 Hence 476620 is not divisible by 3 if a=8 and b=2, number = 476820 4 + 7 + 6 + 8 + 2 + 0 = 27 which is divisible by 3 Hence 476820 is divisible by 3 4 + 6 + 2 = 12 7 + 8 + 0 = 15 15 - 12 = 3 Hence 476820 is not divisible by 11 if a=6 and b=5, number = 476650 4 + 7 + 6 + 6 + 5 + 0 = 28 which is not divisible by 3 Hence 476650 is not divisible by 3 if a=8 and b=5, number = 476850 4 + 7 + 6 + 8 + 5 + 0 = 30 which is divisible by 3 Hence 476850 is divisible by 3 4 + 6 + 5 = 15 7 + 8 + 0 = 15
15 - 15 = 0 Hence 476850 is divisible by 11 So a=8 and b=5 is the answer ----------------------------------------------------------------------------Solution 2 ----------------------------------------------------------------------------Let the number be 476ab0 476ab0 is divisible by 3 => 4 + 7 + 6 + a + b + 0 is divisible by 3 => 17 + a + b is divisible by 3 ------------------------(Equation 1) 476ab0 is divisible by 11 [(4 + 6 + b) -(7 + a + 0)] is 0 or divisible by 11 => [3 + (b - a)] is 0 or divisible by 11 ------------------------(Equation 2) Substitute the values of a and b with the values given in the choices and select the values which satisfies both Equation 1 and Equation 2. if a=6 and b=2, 17 + a + b = 17 + 6 + 2 = 25 which is not divisible by 3 --- Does not meet equation 1 Hence this is not the answer if a=8 and b=2, 17 + a + b = 17 + 8 + 2 = 27 which is divisible by 3 --- meet equation 1 [3 + (b - a)] = [3 + (2 - 8)] = -3 which is neither 0 nor divisible by 11---Does not meet equation 2 Hence this is not the answer if a=6 and b=5, 17 + a + b = 17 + 6 + 5 = 28 which is not divisible by 3 --- Does not meet equation 1 Hence this is not the answer if a=8 and b=5, 17 + a + b = 17 + 8 + 5 = 30 which is divisible by 3 --- meet equation 1
[3 + (b - a)] = [3 + (5 - 8)] = 0 ---meet equation 2 Since these values satisfies both equation 1 and equation 2, this is the answer
76. On dividing 2272 as well as 875 by 3-digit number N, we get the same remainder. What is the sum of the digits of N? A. 11
B. 10
C. 9
D. 8
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let 2272 ÷ N = a, remainder = r => 2272 = Na + r ----------------------------(Equation 1) Let 875 ÷ N = b, remainder = r => 875 = Nb + r ----------------------------(Equation 1) (Equation 1) - (Equation 2) => 2272 - 875 = [Na + r] - [Nb + r] = NA - Nb = N(a - b) => 1397 = N(a - b) ----------------------------(Equation 3) It means 1397 is divisible by N But 1397 = 11 × 127 [Reference1 :how to find factors of a number?] [Reference2 :Prime Factorization?] You can see that 127 is the only 3 digit number which perfectly divides 1397 => N = 127 sum of the digits of N = 1 + 2 + 7 = 10
77. What is the sum of first ten prime numbers? A. 55
B. 101
C. 130
D. 129
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : (Reference : Prime Numbers) Required Sum = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 129
78. Which f the following numbers is exactly divisible by 11? A. 499774
B. 47554
C. 466654
D. 4646652
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Reference : Divisibility by 11 Rule Take 47554 4 + 5 + 4 = 13 7 + 5 = 12 13 - 12 = 1 1 is not divisible by 11 Hence 47554 is not divisible by 11 Take 466654 4 + 6 + 4 = 14 6 + 5 = 11 14 - 11 = 3 3 is not divisible by 11 Hence 466654 is not divisible by 11 Take 4646652
4 + 4 + 6 + 2 = 16 6 + 6 + 5 = 17 17 - 16 = 1 1 is not divisible by 11 Hence 4646652 is not divisible by 11 Take 499774 4 + 9 + 7 = 20 9 + 7 + 4 = 20 20 - 20 = 0 We got the difference as 0. Hence 499774 is divisible by 11
79. What is the sum all even natural numbers between 1 and 101? A. 5050
B. 2550
C. 5040
D. 2540
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Reference1 : Natural Numbers Reference2 : Arithmetic Progression (AP) and Related Formulas Required sum = 2 + 4 + 6+ . . . + 100 This is an arithmetic progression with a =2 d = (4 - 2) = 2
n=
(l − a) (100 − 2) 98 +1 = +1 = + 1 = 49 + 1 = 50 d 2 2
2 + 4 + 6 + ⋯ + 100 =
n 50 50 (a + l) = (2 + 100) = (102) = 50 × 51 = 2550 2 2 2
80. A boy multiplies 987 by a certain number and obtained 559981 as his answer. If in the answer both 9 are wrong , but the other digits are correct, then what will be the correct answer? A. 556581
B. 555681
C. 555181
D. 553681
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : ---------------------------------------------------------------------------Solution 1 ---------------------------------------------------------------------------The answer is divisible by 987. So we can use hit and trial method to find out the number divisible by 987 from the given choices 553681 ÷ 987 gives a remainder not equal to 0 555181 ÷ 987 gives a remainder not equal to 0 556581 ÷ 987 gives a remainder not equal to 0 But 555681 ÷ 987 gives 0 as remainder. Hence this is the answer ---------------------------------------------------------------------------Solution 2 (Recom m ended) ---------------------------------------------------------------------------The answer is divisible by 987.
If a number is divisible by two co-prime numbers, then the number is divisible by their product also. If a number is divisible by more than two pairwise co-prime numbers, then the number is divisible by their product also. Read more
If a number is divisible by another number, then it is also divisible by all the factors of that number. Read more
987 = 3 × 7 × 47 [Prime Factorization] Here 3, 7 and 47 are pairwise co-prime numbers. Also 3, 7 and 47 are factors of 987. Hence if a number is divisible by 3, 7 and 47, the number will be divisible by their product 987 also. If a number is not divisible by 3 or 7 or 47, it is not divisible by 987 You must learn Divisibility Rules t o say whether a given number is divisible by another number without actually performing the division. throughdivisibility rules before proceeding further. 556581 is divisible by 3 556581 is not divisible by 7 Hence 556581 is not divisible by 987 555181 is not divisible by 3
Please
go
Hence 555181 is not divisible by 987 553681 is not divisible by 3 Hence 553681 is not divisible by 987 555681 is divisible by 3 555681 is divisible by 7 555681 is divisible by 47 Hence 555681 is divisible by 987 also
81. Which one of the following cannot be the square of natural number ? A. 15186125824
B. 49873162329
C. 14936506225
D. 60625273287
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Square of a natural number cannot 60625273287 is the answer
end with 7.
82. 7128 + 1252 = 1202 + ? A. 6028
B. 1248
C. 2348
D. 7178
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : ? = 7128 + 1252 - 1202 = 7128 + 50 = 7178
83. 3 + 32 + 33 + ... + 38 = ? A. 9820
B. 9240
C. 9840
D. 9220
Hide Answer | Notebook | Discuss
Hence
Here is the answer and explanation Answer : Option C Explanation :
Sum of first n term s in a geom etric progression (G.P.)
.
a(r n − 1) r−1
(if r > 1)
.
a(1 − r n ) 1−r
(if r < 1)
Sn =
where a= the first term , r = common ratio, n = number of terms Read more... This is a Geometric Progression (GP) where a =3
r=
32 =3 3
n =8
a(r n − 1) 3(38 − 1) Sn = = r−1 3−1 3(38 − 1) 3(6561 − 1) = = 2 2 =
3 × 6560 = 3 × 3280 = 9840 2
84. 73411 × 9999 = ? A. 724836589
B. 724036589
C. 734036589
D. 734036129 | Discuss
Hide Answer | Notebook Here is the answer and explanation Answer : Option C Explanation : 73411 × 9999 = 73411(10000 - 1) = 734110000 - 73411 = 734036589
85. 32 + 33 + 34 + ... + 42 = ? A. 397
B. 407
C. 417
D. 427
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Num ber of term s of an arithm etic progression
n=
(l − a) +1 d
where n = number of terms, a= the first term , l = last term, d= common difference Read more...
Sum of first n term s in an arithm etic progression
Sn =
n n [ 2a + (n − 1)d ] = [ a + l ] 2 2
where a = the first term, Read more…
d= common difference,
l = t n = nth term = a + (n − 1)d
a = 32 l = 42 d = 33 − 32 = 1
n=
(l − a) (42 − 32) +1 = + 1 = 10 + 1 = 11 d 1
Sn =
n 11 11 × 74 [a +l]= [ 32 + 42 ] = = 11 × 37 = 407 2 2 2
OR
(Reference : Power Series : Important formulas)
1+2+3+⋯+n = ∑n =
n(n + 1) 2
32 + 33 + 34+. . . +42 = [1 + 2 + 3 + ⋯ + 42] − [1 + 2 + 3 + ⋯ + 31] =[
42 × 43 31 × 32 ]− [ ] 2 2
= [21 × 43] − [31 × 16] = 903 − 496 = 407
86. What is the digit in the unit place of the number represented by (795 - 358) A. 4
B. 3
C. 2
D. 1
Hide Answer | Notebook | Discuss
Here is the answer and explanation Answer : Option A Explanation : First let's find out the unit digit of 795 795 = [(74)23 × 73] Hence, Unit Digit of 795 = Unit Digit of (74)23 × Unit Digit of 73 ----(Equation 1) Unit Digit of [(74)23] = Unit Digit of [(7×7×7×7)23] = Unit Digit of [(9 × 9)23] of 49) = Unit Digit of [123]
(∵ 7×7=49 and 9 is the unit digit
(∵ 9×9=81 and 1 is the unit digit of 81)
= 1 ----(Equation 2) Unit Digit of 73 = Unit Digit of [7 × 7 × 7] = Unit Digit of [9 × 7] of 49)
(∵ 7 × 7 = 49 and 9 is the unit digit
= 3 (∵ 9 × 7 = 63 and 3 is the unit digit of 63) ---(Equation 3) Hence from Equation 1,Equation 2 and Equation 3, unit digit of 795 = 1 × 3 = 3 ----(Equation A) Similarly we can find out unit digit of 358 358 = [(34)14 × 32] Hence, Unit Digit of 358 = Unit Digit of (34)14 × Unit Digit of 32 ----(Equation 4) Unit Digit of [(34)14]
= Unit Digit of [(3×3×3×3)14] = Unit Digit of [(9 × 9)14] = Unit Digit of [114]
(∵ 3×3=9)
(∵ 9×9=81 and 1 is the unit digit of 81)
= 1 ----(Equation 5) Unit Digit of 32 = 9----(Equation 6) Hence from Equation 4,Equation 5 and Equation 6, Unit Digit of 358 = 1 × 9 = 9 ----(Equation B) We have already found out that Unit Digit of 795 = 3 (From equation A) and Unit Digit of 358 = 9 (From equation B) Hence, Unit Digit of (795 - 358) = Unit Digit of 795 - Unit Digit of 358 = unit digit of [larger number of last digit 3 - smaller number of last digit 9 ] =4
(∵ 795 > 358)
(∵ 4 is the unit digit when a smaller number of
last digit 9 is subtracted from a larger number of last digit 3. example : 113 - 19 = 94)
87. What is the unit digit in the number represented by [365 × 659 × 771] A. 1
B. 2
C. 3
D. 4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : First let's find out the unit digit of 365 365 = [(34)16 × 3]
Hence, Unit Digit of 365 = Unit Digit of (34)16 × 3 = Unit Digit of [(3×3×3×3)16] × 3 = Unit Digit of [(9 × 9)16] × 3 = Unit Digit of [116] × 3 of 81)
(∵ 3×3=9)
(∵ 9×9=81 and 1 is the unit digit
=1 × 3 = 3 ----(Equation A) Unit digit of 659 = 6 (∵ 6×6=36 and unit digit remains as 6 always) ----(Equation B) 771 = [(74)17 × 73] Hence, Unit Digit of 771 = Unit Digit of (74)17 × Unit Digit of 73 ----(Equation 1) Unit Digit of [(74)17] = Unit Digit of [(7×7×7×7)17] = Unit Digit of [(9 × 9)17] of 49) = Unit Digit of [117]
(∵ 7×7=49 and 9 is the unit digit
(∵ 9×9=81 and 1 is the unit digit of 81)
= 1 ----(Equation 2) Unit Digit of 73 = Unit Digit of [7 × 7 × 7] = Unit Digit of [9 × 7] of 49)
(∵ 7 × 7 = 49 and 9 is the unit digit
= 3 (∵ 9 × 7 = 63 and 3 is the unit digit of 63) ---(Equation 3) Hence from Equation 1,Equation 2 and Equation 3, unit digit of 771 = 1 × 3 = 3 ----(Equation C)
We have already found out that Unit Digit of 365 = 3 (From equation A) Unit Digit of 659 = 6 (From equation B) Unit Digit of 771 = 3 (From equation C) Hence, unit digit in the number represented by [365 × 659 × 771] = unit digit of [3 × 6 × 3] = unit digit of [8 × 3] 18) =4
(∵ 3 × 6 = 18 and 8 is the unit digit of
(∵ 8 × 3 = 24 and 4 is the unit digit of 24)
88. 112 × 112 + 88 × 88 = ? A. 26218
B. 20328
C. 20288
D. 24288
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
2
(a + b) = a 2 + 2ab + b2 2
(a − b) = a 2 − 2ab + b2 2
2
(a + b) + (a − b) = 2(a 2 + b2 ) Read more...
112 × 112 + 88 × 88 = 1122 + 882 = (100 + 12) 2 + (100 − 12) 2 = 2(1002 + 122 ) = 2(10000 + 144) = 2 × 10144 = 20288
89. 9312 x 9999 = ? A. 93110688
B. 93010688
C. 93110678
D. 83110688
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 9312 x 9999 = 9312(10000 - 1) = 93120000 - 9312 = 93110688
90. 112 × 112 - 88 × 88 = ? A. 4600
B. 4700
C. 4800
D. 4900
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
a 2 − b2 = (a − b)(a + b) Read more...
2
2
112 × 112 − 88 × 88 = 112 − 88
= (112 − 88)(112 + 88) = 24 × 200 = 4800
91. 1234 + 123 + 12 - ? = 1221 A. 148
B. 158
C. 168
D. 178
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : ? = 1234 + 123 + 12 - 1221 = 148
92. A three-digit number 4a3 is added to another three-digit number 984 to give a four digit number 13b7,
which is divisible by 11. What is the value of (a + b)? A. 9
B. 10
C. 11
D. 12
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : (Reference : Divisibility by 11 rule) 4a3 984 ---------13 b 7 => a + 8 = b -----------(Equation 1) 13b7 is divisible by 11 => (1 + b) - (3 + 7) is 0 or divisible by 11 => (b - 9) is 0 or divisible by 11 -----------(Equation 2) Assume that (b - 9) = 0 => b = 9 Substituting the value of b in Equation 1, a +8 =b a +8 =9 => a = 9 - 8 = 1 If a = 1 and b= 9, (a + b) = 1 + 9 = 10 10 is there in the given choices. Hence this is the answer.
93. 122 × 122 + 322 × 322 - 2 × 122 × 322 = ? A. 44000
B. 42000
C. 38000
D. 40000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
2
(a − b) = a 2 − 2ab + b2 Read more...
122 × 122 + 322 × 322 − 2 × 122 × 322 = 1222 − 2 × 122 × 322 + 3222 = (122 − 322) 2 = (−200) 2 = 40000
94. [22 + 42 + 62 + ... + 142) = ? A. 559
B. 363
C. 364
D. 560
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
12 + 22 + 32 + ⋯ + n 2 = ∑ n 2 = Read More ...
n(n + 1)(2n + 1) 6
22 + 42 + 62 + ⋯ + 142 2
2
2
= (2 × 1) + (2 × 2) + (2 × 3) + ⋯ + (2 × 7 ) 2
2
2
2
2
2
2
2
2
= (2 × 1 ) + (2 × 2 ) + (2 × 3 ) + ⋯ + (2 × 7 ) = 22 (12 + 22 + 32 + ⋯ + 72 ) = 22 [
=
n(n + 1)(2n + 1) 6
4 × 7 × 8 × 15 6
=
]= 4[
7(7 + 1)[(2 × 7) + 1] 6
4×7×8×5 2
]
= 2×7×8×5
= 56 × 10 = 560
95. The sum of the two numbers is 11 and their product is 24. What is the sum of the reciprocals of these numbers ?
7 12 11 C. 24
11 12 7 D. 8
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Let the numbers be x and y. Then x + y = 11 xy = 24 Hence, x+y 11 = xy 24 ⇒
1 1 11 + = y x 24
96. What is the difference between the place values of two sevens in the numeral 54709479 ?
A. 699930
B. 699990
C. 99990
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Required Difference = 700000 - 70 = 699930
97. Which of the following numbers completely divides (461 + 462 + 463 + 464) ? A. 11
B. 10
C. 9
D. 8
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 461 + 462 + 463 + 464 = 461[1 + 41 + 42 + 43] = 461[1 + 4 + 16 + 64] = 461 × 85 = 461 × 5 × 17 = 460 × 4 × 5 × 17 = 460 × 20 × 17 = 460 × 10 × 2 × 17 Hence this is completely divisible by 10
98. A number when divided by 75 leaves 34 as remainder. What will be the remainder if the same number is divided by 65? A. 3
B. 1
C. 6
D. 9
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option D Explanation : Let the number be x Let x ÷ 75 = p and remainder = 34 => x = 75p + 34 = (25p × 3) + 25 + 9 = 25(3p + 1) + 9 Hence, if the number is divided by 25, we will get 9 as remainder
99. The number 7490xy is divisible by 90. Find out (x + y). A. 4
B. 5
C. 6
D. 7
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
If a number is divisible by another number, then it is also divisible by all the factors of that number. Read more
A number is divisible by 10 if the last digit is 0. Read More A number is divisible by 9 if the sum of its digits is divisible by 9. Read More
90 = 10 × 9 7490xy is divisible by 90. Since 10 and 9 are factors of 90, 7490xy is divisible by 10 and 7490xy is divisible by 9
7490xy is divisible by 10. We know that A number is divisible by 10 if the last digit is 0. Hence y = 0 Thus we have the number 7490x0 which is divisible by 9 => sum of its digits of 7490x0 is divisible by 9 => 7 + 4 + 9 + 0 + x + 0 is divisble by 9 => 20 + x is divisible by 9 (Where x is a digit) => x = 7 Hence, (x + y) = (7 + 0) = 7
100. What is the smallest 3 digit prime number? A. 107
B. 100
C. 102
D. 101
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
1. Prime Numbers 2. Divisibility Rules
102 is divisible by 2 => 102 is not a prime number 100 is divisible by 2 => 100 is not a prime number
−−− √ 101 < 11 101 is not divisible by the prime numbers 2, 3, 5, 7. Hence 101 is a prime number Since 100 is not a prime number, 101 is the smallest 3 digit prime number
−−− √ 107 < 11 107 is also not divisible by the prime numbers 2, 3, 5, 7.
Hence 107 is also a prime number. But the smallest 3 digit prime number is 101
101. Which is the natural number nearest to 11720 and completely divisible by 58? A. 11716
B. 11712
C. 11718
D. 11714
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 11720 ÷ 58 = 202, remainder = 4 Hence the natural number nearest to 11720 and completely divisible by 58 = 11720 - 4 = 11716
102. 563124555 - ? = 232323 A. 562892232
B. 562892222
C. 562892212
D. 562892202
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : ? = 563124555 - 232323 = 562892232
103. What is the sum of first 200 natural numbers? A. 20120
B. 19901
C. 19900
D. 20100
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
1+2+3+⋯+n = ∑n = Read More...
n(n + 1) 2
Required Sum = 1 + 2 + 3+. . . +200 =
n(n + 1) 2
=
200 × 201 = 100 × 201 = 20100 2
=
200(200 + 1) 2
OR
Num ber of term s of an arithm etic progression
n=
(l − a) +1 d
where n = number of terms, a= the first term , l = last term, d= common difference Read more...
Sum of first n term s in an arithm etic progression
Sn =
n n [ 2a + (n − 1)d ] = [ a + l ] 2 2
where a = the first term, Read more…
d= common difference,
th
l = t n = n term = a + (n − 1)d
Required Sum, S n = 1 + 2 + 3+. . . +200 a=1 l = 200 d = 2−1 = 1
n = 200
Sn =
n 200 [a +l]= [ 1 + 200 ] = 100 × 201 = 20100 2 2
104. The number 1242*2 is completely divisible by 3. What is the smallest number in place of * ? A. 3
B. 2
C. 1
D. 0
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
A number is divisible by 3 if the sum of the digits is divisible by 3 Read more
1242*2 is divisible by 3 => 1 + 2 + 4 + 2 + * + 2 is divisible by 3 => 11 + * is divisible by 3 We need to find out the smallest value of * which satisfies the above equation. Hence, the smallest value of * is 1 such that 11 + 1 = 12 which is divisible by 3
105. Which of the given numbers is divisible by 3, 7, 9 and 11? A. None of these
B. 1890
C. 4230
D. 6237
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
1. Divisibility by 3 2. Divisibility by 7 2. Divisibility by 9 2. Divisibility by 11
1. Testing 4230 4 + 2 + 3 + 0 = 9. 9 is divisible by 3. Hence 4230 is also divisible by 3 4 + 2 + 3 + 0 = 9. 9 is divisible by 9. Hence 4230 is also divisible by 9 423 - (2 × 0) = 423 42 - (2 × 3) = 36 36 is not divisible by 7. Hence 4230 is not divisible by 7 Hence 4230 does not meet all divisibility conditions 2. Testing 1890 1 + 8 + 9 + 0 = 18. 18 is divisible by 3. Hence 1890 is also divisible by 3 1 + 8 + 9 + 0 = 18. 18 is divisible by 9. Hence 1890 is also divisible by 9 189 - (2 × 0) = 189 18 - (2 × 9) = 0
Hence 1890 is divisible by 7 1 + 9 = 10 8 +0 =8 10 - 8 = 2 2 is not divisible by 11. Hence 1890 is not divisible by 11 Hence 1890 does not meet all divisibility conditions 2. Testing 6237 6 + 2 + 3 + 7 = 18. 18 is divisible by 3. Hence 6237 is also divisible by 3 6 + 2 + 3 + 7 = 18. 18 is divisible by 9. Hence 6237 is also divisible by 9 623 - (2 × 7) = 609 60 - (2 × 9) = 42 42 is divisible by 7. Hence 6237 is also divisible by 7. 6 +3 =9 2 +7 =9 9 - 9 =0 Hence 6237 is divisible by 11 We got that 6237 is divisible by 3, 9, 7 and 11. Hence this is the answer.
106. 17 + 16 × 1.6 + 14 × 1.3 = ? A. 60.8
B. 68.8
C. 60.6
D. 59.6
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 17 + 16 × 1.6 + 14 × 1.3 = 17 + 25.6 + 18.2
= 60.8
143 × 72 = ? 107. 98 A. 1642
B. 1802
C. 2022
D. 1372
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
143 × 72 143 × 72 = 98 14 × 7 = 142 × 7 = 196 × 7 = 1372
108. What smallest number should be added to 8444 such that the sum is completely divisible by 7 ? A. 6
B. 5
C. 4
D. 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 8444 ÷ 7 = 1206, remainder = 2 7 - 2 =5 Hence, 5 should be added to 8444 such that the sum is completely divisible by 7.
109. 5332 × 992 = ? A. 5289344
B. 5289244
C. 5289214
D. 5289324
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
5332 × 992 = 5332(1000 - 8) = 5332000 - 42656 = 5289344 You can find many shortcuts to do calculations with in seconds. It is strongly recommend to go through Speed Mathematics which will save a lot of time when doing calculations.
110. 40% of 2/3 of a number is 32. What is the number? A. 160
B. 240
C. 80
D. 120
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Let x be the number. Then x×
2 40 × = 32 3 100
⇒ x = 32 ×
100 3 10 3 3 × = 32 × × = 8 × 10 × = 4 × 10 × 3 = 120 40 2 4 2 2
111. If a whole number n is divided by 4, we will get 3 as remainder. What will be the remainder if 2n is divided by 4 ? A. 4
B. 3
C. 2
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let n ÷ 4 = p , remainder = 3 => n = 4p + 3 2n = 2(4p + 3) = 8p + 6 = 8p + 4 + 2 = 4(2p + 1) + 2 Hence, if 2n is divided by 4, we will get 2 as remainder.
112. 241 × 999 = ? A. 240769
B. 230759
C. 230769
D. 240759
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : 241 × 999 = 241(1000 - 1) = 241000 - 241 = 240759 You can find many shortcuts to do calculations with in seconds. It is strongly recommend to go through Speed Mathematics which will save a lot of time when doing calculations.
113. The number 367505*8 is completely divisible by 8. What is the smallest whole number in place of *? A. 1
B. 2
C. 3
D. 4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
A number is divisible by 8 if the number formed by the last three digits is divisible by 8. Read more
367505*8 is divisible by 8 => 5*8 is divisible by 8 We need to find out the smallest value of * which satisfies the condition 5*8 is divisible by 8 518 is not divisible by 8 528 is divisible by 8. Hence the smallest value of * is 2 which satisfies the condition 5*8 is divisible by 8
114. 425 × ? = 170
1 2 3 C. 5 A.
2 5 2 D. 3
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 425 × ? = 170
⇒?=
170 34 2 = = 425 85 5
115. The unit digit in 7105 is A. 8
B. 7
C. 6
D. 5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 7105 = (74)26 × 7 Hence, Unit digit of 7105 = Unit digit of [(74)26 × 7] = Unit digit of [(7 × 7 × 7 × 7)26 × 7] = Unit digit of [(9 × 9)26 × 7] unit digit of 49) = Unit digit of [126 × 7] of 81) = Unit digit of [1 × 7] =7
(∵ 7 × 7 = 49 and 9 is the
(∵ 9 × 9 = 81 and 1 is the unit digit
(∵ Unit digit of 126 = 1)
116. Which of the following numbers is a prime number ? A. None of these
B. 377
C. 469
D. 176
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Reference : Divisibility Rules
−−− √ 469 < 22 Prime numbers < 22 are 2, 3, 5, 7, 11, 13, 17, 19 469 is not divisible by 2 469 is not divisible by 3 469 is not divisible by 5 But 469 is divisible by 7 Hence 469 is not a prime number
−−− √ 176 < 14 Prime numbers < 14 are 2, 3, 5, 7, 11, 13 176 is divisible by 2 Hence 176 is not a prime number
−−− √ 377 < 20 Prime numbers < 20 are 2, 3, 5, 7, 11, 13, 17, 19 377 is not divisible by 2 377 is not divisible by 3 377 is not divisible by 5 377 is not divisible by 7 377 is not divisible by 11 But 377 is divisible by 13
Hence 377 is not a prime number Hence answer is "None of these"
117. In a division sum, the divisor is 10 times the quotient and 5 times the remainder. If the remainder is 46, find out the dividend. A. 4426
B. 3426
C. 4336
D. 5336
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Given that Remainder = 46 Divisor is 5 times the remainder => Divisor = 5 × remainder = 5 × 46 = 230 Divisor is 10 times the quotient => Divisor = 10 × Quotient => 230 = 10 × Quotient
=> Quotient =
230 = 23 10
Dividend = (Divisor × Quotient) + Remainder = (230 × 23) + 46 = 5290 + 46 = 5336
118. The difference of two numbers is 1365. On dividing the larger number by the smaller, 6 is obtained as quotient and 15 as remainder. What is the smaller number ? A. 310
B. 330
C. 250
D. 270
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option D Explanation : Let the smaller number be x and the bigger number be (x + 1365) (x + 1365) ÷ x = 6, remainder = 15 => (x + 1365) = 6x + 15 => 5x = 1350 => x = 1350/5 = 270 Smaller number = x = 270
119. Difference between the squares of two consecutive odd integers is always divisible by A. 8
B. 7
C. 6
D. 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : ---------------------------------------------------------------Solution 1 ----------------------------------------------------------------Let two odd numbers be (2n - 1) and (2n + 1) Difference between the squares = (2n + 1)2 - (2n - 1)2 = [4n2 + 4n + 1] - [4n2 - 4n + 1] = 8n which is always divisible by 8 ---------------------------------------------------------------Solution 2 ----------------------------------------------------------------Take any two odd numbers, say 1 and 3
Difference between the squares = 32 - 12 = 9 - 1 = 8 From the given choices, now you can easily figure out the answer as 8
120. A number was divided successively in order by 4, 5 and 6. The remainders were 2, 3 and 4 respectively. What is the number? A. 224
B. 324
C. 304
D. 214
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let P be the number Given that P ÷ 4 = Q, remainder = 2 Q ÷ 5 = R, remainder = 3 R ÷ 6 = 1, remainder = 4 Hence R = 1 × 6 + 4 = 10 Q = 5R + 3 = (5 × 10) + 3 = 53 P = 4Q + 2 = (4 × 53) + 2 = 212 + 2 = 214
121. What is the sum of first 20 natural numbers? A. 220
B. 205
C. 190
D. 210
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
1+2+3+⋯+n = ∑n =
n(n + 1) 2
Read More...
Required Sum = 1 + 2 + 3+. . . +20 =
n(n + 1) 2
=
20 × 21 = 10 × 21 = 210 2
=
20(20 + 1) 2
OR
Num ber of term s of an arithm etic progression
n=
(l − a) +1 d
where n = number of terms, a= the first term , l = last term, d= common difference Read more...
Sum of first n term s in an arithm etic progression
Sn =
n n [ 2a + (n − 1)d ] = [ a + l ] 2 2
where a = the first term, Read more…
d= common difference,
l = t n = nth term = a + (n − 1)d
Required Sum, S n = 1 + 2 + 3+. . . +20 a=1 l = 20 d = 2−1 = 1
n = 20
Sn =
n 20 [a +l]= [ 1 + 20 ] = 10 × 21 = 210 2 2
122. ? - 23442 - 12411 = 2469 A. 38322
B. 37212
C. 37532
D. 38122
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : ? = 2469 + 12411 + 23442 = 38322
123. In dividing a number by 585, a student employed the method of short division. He divided the number successively by 5, 9 and 13 and got the remainders 4, 8, 12 respectively. What would have been the remainder if he had divided the number by 585? A. 144
B. 292
C. 24
D. 584
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let P be the number Given that
P ÷ 5 = Q, remainder = 4 Q ÷ 9 = R, remainder = 8 R ÷ 13 = 1, remainder = 12 Hence R = 1 × 13 + 12 = 25 Q = 9R + 8 = (9 × 25) + 8 = 225 + 8 = 233 P = 5Q + 4 = (5 × 233) + 4 = 1165 + 4 = 1169 P ÷ 585 = 1169 ÷ 585 = 1, remainder = 584
124. A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What will be the respective remainders if it is successively divided by 5 and 4? A. 1,2
B. 2, 3
C. 3, 2
D. 2,1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let P be the number Given that P ÷ 4 = Q, remainder = 1 Q ÷ 5 = 1, remainder = 4 Hence Q = 1 × 5 + 4 = 9 P = 4Q + 1 = (4 × 9) + 1 = 36 + 1 = 37 Now divide 37 successively by 5 and 4 respectively 37 ÷ 5 = 7, remainder = 2 7 ÷ 4 = 1, remainder = 3
125. What is the sum of even numbers between 9 and 53? A. 682
B. 672
C. 662
D. 702
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Num ber of term s of an arithm etic progression
n=
(l − a) +1 d
where n = number of terms, a= the first term , l = last term, d= common difference Read more...
Sum of first n term s in an arithm etic progression
Sn =
n n [ 2a + (n − 1)d ] = [ a + l ] 2 2
where a = the first term, Read more…
d= common difference,
l = t n = nth term = a + (n − 1)d
Required Sum = 10 + 12 + 14 + ⋯ + 52 This is an Arithmetic Progression with a = 10 l = 52 d = 12 − 10 = 2
n=
(l − a) (52 − 10) +1 = + 1 = 21 + 1 = 22 d 2
Sn =
n 22 [a +l]= [ 10 + 52 ] = 11 × 62 = 682 2 2
126. The number 13*48 is exactly divisible by 72. Find out the minimum value of * . A. 1
B. 2
C. 3
D. 4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
If a number is divisible by another number, then it is also divisible by all the factors of that number. Read more
A number is divisible by 8 if the number formed by the last three digits is divisible by 8. Read More A number is divisible by 9 if the sum of its digits is divisible by 9. Read More
72 = 8 × 9 Hence 8 and 9 are factors of 72 Hence, if the a number is divisible by 72, it must be divisible by 8 and 9 also 13*48 is divisible by 72. Since 9 and 8 are factors of 72, 13*48 is divisible by 9 and 13*48 is divisible by 8 13*48 is divisible by 9. => 1 + 3 + * + 4 + 8 is divisible by 9 => 16 + * is divisible by 9 -----------------------(Equation 1) 13*48 is divisible by 8 => *48 is divisible by 8 => * can be 2 or 4 or 6 or 8 -----------------------(Equation 2) We need to find out the minimum value of * which satisfies both equation 1 and 2 Take the value * = 2 from Equation 2 16 + * = 16 + 2 = 18 is divisible by 9. Hence * = 2 is the minimum value which satisfies both equation 1 and 2 and this is the answer
OR
Use hit and trial method
If a number is divisible by two co-prime numbers, then the number is divisible by their product also. Read more
Here 9 and 8 are pairwise co-prime numbers. 72 = 9 × 8. Hence, if a number is divisible by 9 and 8, the number will be divisible by their product 72 also. 1 is the minimum value given as the choices Substituting 1 in the place of *, we get 13148 13148 is not divisible by 9 => 13148 is not divisible by 72 Substituting 2 in the place of *, we get 13248 13248 is divisible by 9 13248 is divisible by 8 Hence 13248 is divisible by 72 Hence the minimum value of * is 2
127. Which of the following number is divisible by 3 but not by 9 ? A. 5271
B. 4122
C. 3141
D. 3222
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
A number is divisible by 3 if the sum of the digits is divisible by 3 Read More A number is divisible by 9 if the sum of its digits is divisible by 9. Read More
Take 3222 3 + 2 + 2 + 2 = 9 which is divisible by 3 and 9. Hence 3222 is divisible by 3 and 9 Take 3141 3 + 1 + 4 + 1 = 9 which is divisible by 3 and 9. Hence 3141 is divisible by 3 and 9 Take 4122 4 + 1 + 2 + 2 = 9 which is divisible by 3 and 9. Hence 4122 is divisible by 3 and 9 Take 5271 5 + 2 + 7 + 1 = 15 which is divisible by 3 , but not divisible by 9. Hence 5271 is divisible by 3 , but not divisible by 9
128. When a number is divided by 13, the remainder is 6. When the same number is divided by 7, then remainder is 1. What is the number ? A. 243 C. 312 Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Take 243 243 ÷ 7 = 34, Remainder = 5
B. 253 D. None of these
Hence this is not the answer Take 312 312 ÷ 7 = 44, Remainder = 4 Hence this is not the answer Take 253 253 ÷ 7 = 36, Remainder = 1 . 253 ÷ 13 = 19, Remainder = 6. This satisfies both the conditions given in the question. Hence this is the answer.
129. What is the sum of all two digit numbers divisible by 6? A. 805
B. 820
C. 790
D. 810
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Num ber of term s of an arithm etic progression
n=
(l − a) +1 d
where n = number of terms, a= the first term , l = last term, d= common difference Read more...
Sum of first n term s in an arithm etic progression
Sn =
n n [ 2a + (n − 1)d ] = [ a + l ] 2 2
where a = the first term,
d= common difference,
l = t n = nth term = a + (n − 1)d
Read more…
Required Sum = 12 + 18 + 24 + ⋯ + 96 This is an Arithmetic Progression with a = 12 l = 96 d=6
n=
(l − a) (96 − 12) +1 = + 1 = 14 + 1 = 15 d 6
Sn =
n 15 15 × 108 [a +l]= [ 12 + 96 ] = = 15 × 54 = 810 2 2 2
130. 2002 × 2002 = ? A. 4008004
B. 4006004
C. 4002004
D. 4004004
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 2002 × 2002 = 4008004
OR
2
(a + b) = a 2 + 2ab + b2 Read more... 2002 × 2002 = 20022 = (2000 + 2)2 = 2000 2 + (2 × 2000 × 2) + 22 = 4000000 + 8000 + 4 = 4008004
OR Using speed mathematics techniques given at Speed Mathematics which will save a lot of time when doing calculations.
131. Which natural number is completely divisible by 123 and nearest to 410081 A. 410082
B. 409959
C. 410078
D. 410071
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 410081 ÷ 123 = 3333, remainder = 122 Hence required number = 410081 + (123 - 122) = 410082
132. What is the difference between the place value and the face value of 6 in the numeral 296827? A. None of these
B. 5999
C. 994
D. 5994
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option D Explanation : Place value of 6 = 6000 Face value of 6 = 6 Difference = 6000 - 6 = 5994
133. The sum of a series, 27 + 36 + 45 + ... + 162 is 1512. What is the number of terms in the series? A. 14
B. 15
C. 16
D. 17
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Sum of first n term s in an arithm etic progression
Sn =
n n [ 2a + (n − 1)d ] = [ a + l ] 2 2
where a = the first term, Read more...
d= common difference,
l = t n = nth term = a + (n − 1)d
a = 27 d = 36 − 27 = 9 S n = 27 + 36 + 45+. . . +162 = 1512 ⇒
n [ 2a + (n − 1)d ] = 1512 2
⇒
n [ (2 × 27) + (n − 1)9 ] = 1512 2
⇒
n [ 54 + 9n − 9 ] = 1512 2
⇒ n(45 + 9n) = 3024 ⇒ 9n(5 + n) = 3024 ⇒ n(5 + n) = 336 -------(Equation 1) From here, you can either solve the quadratic equation using the techniques given here or just solve use hit and trial method ; both are given below. -------------------------------------------------------------------------------------Solution 1 - Solving the quadratic equation n(5 + n) = 336 to find out the value of n --------------------------------------------------------------------------------------
n(5 + n) = 336 n 2 + 5n − 336 = 0 −−−−−−− −b ± √ b2 − 4ac n= 2a
=
−−−−−−−−−−−−−−−−− −5 ± √ 52 − [4 × 1 × (−336)] (2 × 1)
−−−−−−−−− −−−− −5 ± √ (25 + 1344) −5 ± √ 1369 −5 ± 37 = = = 2 2 2 −5 + 37 = (∵ since n is the number of terms, only positive value is applicable here) 2 32 = = 16 2 = Hence, number of terms = 16 -------------------------------------------------------------------------------------Solution 2 - Solving the quadratic equation using hit and trial method -------------------------------------------------------------------------------------Take Equation 1 : n(5 + n) = 336
Take the values given in the choices and see which value satisfies the above equation If n = 14, n(5 + n) = 14 × 19 = 266 ? 336 If n = 15, n(5 + n) = 15 × 20 = 300 ? 336 If n = 16, n(5 + n) = 16 × 21 = 336. Hence n = 16 satisfies equation 1. Hence, number of terms = 16
1 2 3 −3+ +1− = ? 4 3 5 189 199 A. B. 60 60 129 169 C. D. 60 60 134.
5+
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
5+
1 2 3 −3+ +1− 4 3 5
= (5 − 3 + 1) + (
1 2 3 + − ) 4 3 5
= 3+
(1 × 15 + 2 × 20 − 3 × 12) 60
= 3+
19 60
=
180 + 19 60
=
199 60
135. 996ab is divisible by 80. What is (a + b) ? A. 3
B. 5
C. 6
D. 8
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
If a number is divisible by another number, then it is also divisible by all the factors of that number. Read more
A number is divisible by 8 if the number formed by the last three digits is divisible by 8. Read More A number is divisible by 9 if the sum of its digits is divisible by 9. Read More
80 = 10 × 8 996ab is divisible by 80. Since 10 and 8 are factors of 80, 996ab is divisible by 10 and 996ab is divisible by 8 996ab is divisible by 10. We know that A number is divisible by 10 if the last digit is 0. Hence b = 0 Thus we have the number 996a0 which is divisible by 8 => 6a0 is divisible by 8 => a = 0 or 4 or 8 Hence, (a + b) = (0 + 0) or (4 + 0) or (8 + 0) = 0 or 4 or 8
136. 123 - 333 + 321 - 111 = ?
A. 320
B. 100
C. 120
D. 0
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : 123 - 333 + 321 - 111 = 444 - 444 = 0
137. On multiplying a number by 3, the product is a number each of whose digits is 7. What is the smallest such number? A. 259129
B. 259219
C. 259279
D. 259259
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let's use hit and trial method here Smallest number from the given choices = 259129 259129 × 3 = 777387 Next highest number from the given choices = 259219 259219 × 3 = 777657 Next highest number from the given choices = 259259 259259 × 3 = 777777 Hence 259259 is the answer
138. On dividing a number by 357, 39 is obtained as remainder. On dividing the same number by 17, what will be the remainder ? A. 3
B. 4
C. 5
D. 6
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Let P be the number Let P ÷ 357 = a, remainder = 39 Then P = 357a + 39 = (17 × 21 × a) + 34 + 5 = (17 × 21 × a) + (17 × 2) + 5 = 17(21a + 2) + 5 Hence, if the same number is divided by 17, we will get 5 as remainder.
139.
444 666 × =? 3 4
A. 22622
B. 22642
C. 24622
D. 24642
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
444 666 × 3 4 =
444 × 666 3×4
= 111 × 222 = 24642
140. A student divides a number by 5 and get 3 as remainder. If his friend divides the square of the number by 5, what will be the remainder? A. 6
B. 5
C. 4
D. 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
2
(a + b) = a 2 + 2ab + b2 Read more... Let P be the number P ÷ 5 = a and remainder = 3 Then P = 5a + 3 P2 = (5a + 3)2 = (5a)2 + (2 × 5a × 3) + 32 = 25a2 + 30a + 9 = 25a2 + 30a + 5 + 4 = 5(5a2 + 6a + 1) + 4 Hence when P2 is divided by 5, we will get 4 as remainder
141. What is the unit digit in 771 ? A. 4
B. 3
C. 2
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 771 = (74)17 × 73 Hence, unit digit in 771 = Unit digit in [(74)17 × 73] = Unit digit in [(7 × 7 × 7 × 7 )17 × 73] = Unit digit in [(9 × 9)17 × 73] unit digit of 49)
(∵ 7 × 7 = 49 and 9 is the
= Unit digit in [117 × 73] digit of 81) = Unit digit in [1 × 73]
(∵ 9 × 9 = 81 and 1 is the unit
(∵ 117 = 1)
= Unit digit in [73] = Unit digit in [7 × 7 × 7] = Unit digit in [9 × 7] 49) =3
(∵ 7 × 7 = 49 and 9 is the unit digit of
(∵ 9 × 7 = 63 and 3 is the unit digit of 63)
142. What is the unit digit in the product (2344 × 6892 × 349 × 527 × 238) ? A. 1
B. 2
C. 3
D. 4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Unit digit of (2344 × 6892 × 349 × 527 × 238) = Unit digit of (4 × 2 × 9 × 7 × 8) = Unit digit of (8 × 9 × 7 × 8) = Unit digit of (2 × 7 × 8) digit of 72) = Unit digit of (4 × 8) of 14) =2
(∵ 4 × 2 = 8) (∵ 8 × 9 = 72 and 2 is the unit
(∵ 2 × 7 = 14 and 4 is the unit digit
(∵ 4 × 8 = 32 and 2 is the unit digit of 32)
143. Which is the common factor of (22121 + 19121) and (22231+ 19231) ? A. (22 - 19)
B. (22 + 19)
C. (22231 + 19231)
D. (22121 + 19121)
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B
Explanation :
(x n + a n ) is completely divisible by (x + a) when n is odd Read more... Hence, (22121 + 19121) is divisible by (22 + 19) because 121 is odd Similarly (22231 + 19231) is also divisible by (22 + 19) because 231 is odd Hence (22 + 19) is a common factor here
(232 + 323) 2 − (232 − 323)2 144. 232 × 323 A. 4
B. 8
C. 3424
D. 2344
=?
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
2
2
(a + b) = a 2 + 2ab + b 2
(a − b) = a 2 − 2ab + b2 Read more...
2
(a + b) − (a − b) Given expression is in the form ab
2
where a = 232 and b = 323
(a 2 + 2ab + b2 ) − (a 2 − 2ab + b2 ) = ab =
4ab ab
=4
145. A number when divided by 44, gives 432 as quotient and 0 as remainder. What will be the remainder
when dividing the same number by 31? A. 4
B. 0
C. 5
D. 8
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let P be the number P ÷ 44 = 432, remainder = 0 => P = 432 × 44 + 0 = 19008 P ÷ 31 = 19008 ÷ 31 = 613, remainder = 5
1. Which of the following statements is not correct? A. log(2×4×6) = log 2 + log 4 + log 6
B. log51 = 0
C. log(3+4) = log(3 × 4)
D. log55 = 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Logbb = 1. Hence log55 = 1 Logb1 = 0. Hence log51 = 0 log(a × b) = log a + log b similarly, log(a × b × c) = log a + log b + log c Hence log(2×4×6) = log 2 + log 4 + log 6 log(3+4) = log(3 × 4) is wrong LHS = log(3+4) = log 7 RHS = log(3 × 4) = log(12) log 7 ≠ log 12
2. Log5(0) = ? A. None of these
B. 5
C. 0
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : logb(0) is undefined
log √ 5 =? log 5 1 1 B. A. 2 √5 1 1 D. C. 8 4 3.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A
Explanation :
log √ 5 = log 5
1 log(5) 2 log 5
1 log 5 1 2 = = log 5 2
log √ 6 =? log √3 6 1 1 A. B. 3 2 2 3 D. C. 3 2 4.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
1 log(6) 2
1 log 6 log √ 6 3 2 = = = 3 1 1 2 log √ 6 log 6 3 log(6 ) 3
5.
If log
a b + log = log(a + b), then b a
A. a = b
B. a + b = 1
C. a - b = 1
D. a2 - b2 = 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
log
a b + log = log(a + b) b a
log (
a b × ) = log(a + b) b a
log(1) = log(a + b) a+b = 1
6. If log(64)= 1.806, log(16) = ? A. 1.204
B. 0.903
C. 1.806
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : log(64)= 1.806 => log(43) = 1.806 => 3log(4) = 1.806
⇒ log(4) =
1.806 3
log(16) = log(42 ) = 2 log(4) = 2 ×
1.806 = 2 × 0.062 = 1.204 3
7. If log 2 = 0.3010 and log 3 = 0.4771, What is the value of log51024? A. 4.31
B. 3.88
C. 3.91
D. 2.97
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
log (210 ) log 1024 log 5 1024 = = log 5 10 log ( ) 2 =
=
10 log(2) log 10 − log 2
10 × 0.3010 3.01 3010 = = = 4.31 1 − 0.3010 0.699 699
8. if log 2 = 0.30103 and log 3 = 0.4771, find the number of digits in (648)5. A. 15
B. 14
C. 13
D. 12
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option A Explanation : log(648)5 = 5 log(648) = 5 log(81 × 8) = 5 [ log(81) + log(8) ] = 5 [ log(34) + log(23) ] = 5 [ 4log(3) + 3log(2) ] = 5 [ 4 × 0.4771 + 3 × 0.30103 ] = 5(1.9084 + 0.90309) = 5 × 2.81149 ≈ 14.05 ie, log(648)5 ≈ 14.05 ie, its characteristic = 14 Hence, number of digits in (648)5 = 14+1 = 15
9. if log 2 = 0.30103, the number of digits in 2128 is A. 38
B. 39
C. 40
D. 41
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : log(2128) = 128log(2) = 128 × 0.30103 ≈ 38.4 ie, its characteristic = 38 Hence, number of digits in 2128 = 38+1 = 39
9 1 ) = − , find the value of x 32 8
10.
log x (
A. (
9 ) 32
8
32 ) C. ( 9
8
9 ) 32
2
32 ) D. ( 9
2
B. (
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C
Explanation :
log x (
9 1 )=− 32 8 9 32
⇒ x −1/8 = ⇒
1 x 1/8
9 32
=
32 9
⇒ x 1/8 =
8
32 x=( ) 9
9 1 log x ( ) = − , find the value of x 4 2 81 16 A. B. 16 9 16 9 C. D. 81 16 11.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
9 1 log x ( ) = − 4 2 9 4
⇒ x −1/2 = ⇒
1 x 1/2
=
4 9
⇒ x 1/2 = 4 x=( ) 9
9 4
2
=
16 81
12. if ax = by , then
log a x = log b y a x C. log = b y A.
B. None of these D.
log a y = log b x
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
a x = by log (a x ) = log (by ) x log a = y log b log a y = log b x
13. log2 512 = ? A. 10
B. 6
C. 9
D. 8
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : log2 512 = log2 (29) = 9
14. If logx y = 10 and log2 x = 1000, what is the value of y? A. 2100
B. 21000
C. 210000
D. 210
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : log2 x = 1000 => x = 21000 logx y = 10
=> y = x10 = (21000)10 = 2(1000 × 10) = 210000
15. if log102 = 0.3010, what is the value of log101600 ? A. None of these
B. 5.204
C. 1.204
D. 3.204
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : log101600 = log10(16 × 100) = log10(16) + log10(100) = log10(24) + log10(102) = 4 log10(2) + 2 = (4 × 0.3010) + 2 = 1.204 + 2 = 3.204
16.
1 1 1 + + =? log2 48 log 4 48 log 6 48
A. -1
B. 2
C. 0
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
1 1 1 + + log2 48 log 4 48 log 6 48 = log 48 2 + log48 4 + log 48 6 = log 48 (2 × 4 × 6) = log 48 (48) =1
17. If log10 5 + log10 (5x + 1) = log10 (x + 5) + 1, then x is equal to: A. 4
B. 3
C. 2
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : log10 5 + log10 (5x + 1) = log10 (x + 5) + 1 => log10 5 + log10 (5x + 1) = log10 (x + 5) + log10 10 => log10 [5(5x+1)] = log10 [10(x+5)] => 5(5x+1) = 10(x+5) => 5x+1 = 2(x+5) => 5x + 1 = 2x + 10 => 3x = 9 => x = 3
18.
If log 10 2 = a, what is the value of log10 (
A. -(a+2)
B. -(a+1)
C. (a+2)
D. (a+1)
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
1 ) 200
log 10 (
1 ) 200
= log 10 1 − log10 200 = 0 − log 10 (2 × 100) = −[log 10 2 + log10 100] = −(a + 2)
19. If log10 3 = 0.4771, then log3 10 is A.
1000 4771
B.
C. 1.4313
10000 4771
D. 0.4771
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
log 3 10 =
1 1 10000 = = log 10 3 0.4771 4771
20. If log5 (x2+x) - log5 (x+1) = 3, find the value of x A. 25
B. 125
C. 1/125
D. 1/25
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
log 5 (x 2 + x) − log5 (x + 1) = 3 ⇒ log 5 (
x2 + x )=3 x+1
⇒ log 5 [
x(x + 1) ]=3 x+1
⇒ log 5 x = 3 ⇒ x = 53 = 125
21.
Find the value of
1 log 10 125 − 2 log10 4 + log 10 32 3
A. 0
B. 1
C. 2
D. 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
1 log 10 125 − 2 log 10 4 + log 10 32 3 =log 10 (1251/3 ) − log 10 (42 ) + log 10 32 = log 10 5 − log10 16 + log 10 32 = log 10 (
5 × 32 ) 16
= log 10 (10) =1
22.
log (
a2 b2 c2 ) + log ( ) + log ( ) = ? bc ac ab
A. None of these
B. abc
C. 1
D. 0
Hide Answer | Notebook | Discuss
Here is the answer and explanation Answer : Option D Explanation :
a2 b2 c2 log ( ) + log ( ) + log ( ) bc ac ab a 2 b2 c2 = log ( × × ) = log(1) = 0 bc ac ab
23. if log2x = -6, x is equal to : A. 64
B.
1 32
C.
1 64
D. 32
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
log 2 x = −6 ⇒ x = (2)
24.
−6
=
1 1 = 6 64 2
If log 4 x + log2 x = 12, then x is equal to:
A. 1024
B. 256
C. 8
D. 16
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
log 4 x + log 2 x = 12 ⇒ ⇒
log x log x + = 12 log 4 log 2 log x log 22
+
log x = 12 log 2
⇒
log x log x + = 12 2 log 2 log 2
⇒
log x + 2 log x = 12 2 log 2
⇒
3 log x = 12 2 log 2 12 × 2 log 2 = 8 log 2 = log (28 ) = log(256) 3
⇒ log x = ⇒ x = 256
25. log(.001) (100) = ?
−2 3 −3 C. 2 A.
B.
3 2
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Let log (.001) (100)= p p
(.001) = 100 (
(
p
1 ) 1000 1
) 3
p
10
[(10) −3 ]
= 100
= 102 p
= 102
(10) −3p = 102 − 3p = 2 −2 3
p=
26. log5 200 × log200 125 equals : A. 5
B. 25
C. 3
D. 6
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
log 5 200 × log200 125 = =
log 200 log 125 × log 5 log 200
log 125 = log5 125 = log 5 (53 ) = 3 log 5
27. If log100[log3(log2 x)] = 1, x is equal to: A. None of these 100)
C. 2(3
B. 1 2 D. 3(2 )
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
log100[log3(log2 x)] = 1 log100[log3(log2 x)] = log100(100) log3(log2 x) = 100 log2 x = 3100 100)
x = 2(3
28. If log2[log3(log2 x)] = 1, x is equal to: A. 512
B. None of these
C. 256
D. 1024
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : log2[log3(log2 x)] = 1 log2[log3(log2 x)] = log2(2) log3(log2 x) = 2 log2 x = 32 = 9 x = 29 = 512
29. (log3 4) (log4 5) (log5 6) (log6 7) (log7 8) (log8 9) (log9 9) = ? A. 4
B. 0
C. 2
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
log 3 4 × log4 5 × log 5 6 × log 6 7 × log7 8 × log 8 9 × log 9 9 log 4 log 5 log 6 log 7 log 8 log 9 = × × × × × ×1 log 3 log 4 log 5 log 6 log 7 log 8 log 9 log 32 2 log 3 = = = =2 log 3 log 3 log 3
30. log(-2)(-2) = ? A. None of these
B. -1
C. 0
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : logbx is undefined for x ≤ 0
I mportan t Formu las - M ixtu res an d A lligation s 1.
Alligation It is the rule which enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a specified price.
2.
Mean Price Mean price is the cost price of a unit quantity of the mixture
3.
Suppose a container contains x of liquid from which y units are taken out and replaced by water. y After n operations, the quantity of pure liquid = [x(1 − ) x
4.
n
] units.
Rule of Alligation If two ingredients are mixed, then
(
Quantity of cheaper Quantity of dearer
)= (
C.P. of dearer - Mean Price ) Mean price - C.P. of cheaper
Cost Price(CP) of a unit quantity of cheaper (c)
Cost Price(CP) of a unit quantity of dearer (d)
Mean Price (m)
(d - m)
(m - c)
=> (Cheaper quantity) : (Dearer quantity ) = (d - m ) : (m - c)
1. A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container? A. 26 litres
B. 29.16 litres
C. 28 litres
D. 28.2 litres
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Assume that a container contains x of liquid from which y units are taken out and replaced by water. After n operations, the quantity of pure liquid
y = x(1 − ) x
n
4 ) Hence milk now contained by the container = 40(1 − 40 9 9 9 4×9×9×9 40 × × × = = 29.16 10 10 10 100
3
1 ) = 40(1 − 10
3
2. Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety of tea in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, what is the price of the third variety per kg ? A. Rs.182.50
B. Rs.170.5
C. Rs.175.50
D. Rs.180
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed in the ratio 1 : 1
So their average price =
(126 + 135) = 130.5 2
Hence let's consider that the mixture is formed by mixing two varieties of tea. one at Rs. 130.50 per kg and the other at Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. Now let's find out x.
By the rule of alligation, we can write as
Cost of 1 kg of 1st kind of tea 130.50
Cost of 1 kg of 2nd kind of tea x
Mean Price 153
(x - 153)
22.50
(x - 153) : 22.5 = 1 : 1 =>x - 153 = 22.50 => x = 153 + 22.5 = 175.5
3. A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5? A. 5litres, 7 litres
B. 7litres, 4 litres
C. 6litres, 6 litres
D. 4litres, 8 litres
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : ---------------------------------------Solution 1 ---------------------------------------Let x and (12-x) litres of milk be mixed from the first and second container respectively
Amount of milk in x litres of the the first container = .75x Amount of water in x litres of the the first container = .25x Amount of milk in (12-x) litres of the the second container = .5(12-x) Amount of water in (12-x) litres of the the second container = .5(12-x)
Ratio of water to milk = [.25x + .5(12-x)] : [.75x + .5(12-x)] = 3:5
⇒
⇒
(.25x + 6 − .5x) (.75x + 6 − .5x)
=
3 5
(6 − .25x) 3 = 5 (.25x + 6)
⇒ 30 − 1.25x = .75x + 18 ⇒ 2x = 12 ⇒x=6 Since x = 6, 12-x = 12-6 = 6 Hence 6 and 6 litres of milk should mixed from the first and second container respectively ---------------------------------------Solution 2 ----------------------------------------
Let cost of 1 litre milk be Rs.1 Milk in 1 litre mix in 1st can =
3 litre 4
Cost Price(CP) of 1 litre mix in 1st can = Rs.
Milk in 1 litre mix in 2nd can =
3 4
1 litre 2
Cost Price(CP) of 1 litre mix in 2nd can = Rs.
Milk in 1 litre of the final mix =
5 8
Cost Price(CP) of 1 litre final mix = Rs. => Mean price =
1 2
5 8
5 8
By the rule of alligation, we can write as
CP of 1 litre mix in 2nd can
CP of 1 litre mix in 1st can
1/2
3/4
Mean Price 5/8
3/4 - 5/8 = 1/8
5/8 - 1/2 = 1/8
=> mix in 2nd can :mix in 1st can = 1/8 : 1/8 = 1:1
ie, from each can,
1 × 12 = 6 litre should be taken 2
4. Two vessels A and B contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively. Find the ratio in which these mixture be mixed to obtain a new mixture in vessel C containing spirit and water in the ration 8:5? A. 3: 4
B. 4 : 3
C. 9 : 7
D. 7 : 9
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let Cost Price(CP) of 1 litre spirit be Rs.1 Quantity of spirit in 1 litre mixture from vessel A = 5/7 Cost Price(CP) of 1 litre mixture from vessel A = Rs. 5/7 Quantity of spirit in 1 litre mixture from vessel B = 7/13 Cost Price(CP) of 1 litre mixture from vessel B = Rs. 7/13 Quantity of spirit to be obtained in 1 litre mixture from vessel C = 8/13 Cost Price(CP) of 1 litre mixture from vessel C = Rs. 8/13 = Mean Price By the rule of alligation, we can write as
CP of 1 litre mixture from vessel A 5/7
CP of 1 litre mixture from vessel B 7/13
Mean Price 8/13
8/13 - 7/13 = 1/13
5/7 - 8/13 = 9/91
=> Mixture from Vessel A : Mixture from Vessel B = 1/13 : 9/91 = 7 : 9 = Required Ratio
5. The cost of Type 1 material is Rs. 15 per kg and Type 2 material is Rs.20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then what is the price per kg of the mixed variety of material? A. Rs. 19
B. Rs. 16
C. Rs. 18
D. Rs. 17
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C
Explanation : --------------------------------Solution 1 --------------------------------Cost Price(CP) of Type 1 material is Rs. 15 per kg Cost Price(CP) of Type 2 material is Rs. 20 per kg
Type 1 and Type 2 are mixed in the ratio of 2 : 3
Hence Cost Price(CP) of the resultant mixture = =
(15 × 2) + (20 × 3) (2 + 3)
(30 + 60) 90 = = 18 5 5
=> rice per kg of the mixed variety of material = Rs.18 --------------------------------Solution 2 --------------------------------Cost Price(CP) of Type 1 material is Rs. 15 per kg Cost Price(CP) of Type 2 material is Rs. 20 per kg Let Cost Price(CP) of resultant mixture be Rs.x per kg By the rule of alligation, we have
CP of Type 1 material 15
CP of Type 2 material 20
Mean Price x
(20-x) => Type 1 material : Type 2 material = (20-x) : (x-15) Given that Type 1 material : Type 2 material = 2 : 3 => (20-x) : (x-15) = 2 : 3
(x-15)
⇒
(20 − x) (x − 15)
=
2 3
⇒ 3 (20 − x) = 2 (x − 15) ⇒ 60 − 3x = 2x − 30 ⇒ 90 = 5x => x =
90 = 18 5
=>price per kg of the mixed variety of material = Rs.18
6. Find the ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs. 6.30 a kg. A. 4 : 3
B. 3 : 4
C. 2 : 3
D. 3 : 2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : CP of 1Kg 1st kind rice = Rs.7.20 CP of 1Kg 2nd kind rice = Rs.5.70 CP of 1Kg mixed rice = Rs.6.30 By the rule of alligation, we have
CP of 1Kg 1st kind rice 7.2
CP of 1Kg 2nd kind rice 5.7
Mean Price 6.3
6.3 - 5.7 = .6 Required Ratio = .6 : .9 = 6:9 = 2:3
7.2 - 6.3 = .9
7. 8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16 : 65. How much wine did the cask originally hold? A. 30 litres
B. 26 litres
C. 24 litres
D. 32 litres
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let initial quantity of wine = x litre
y After a total of 4 operations, quantity of wine = x(1 − ) x
n
8 = x(1 − ) x
4
Given that after a total of 4 operations, the ratio of the quantity of wine left in cask to that of water = 16 : 65
8 x(1 − ) x Hence we can write as x 4
⇒ (1 −
8 ) x
⇒ (1 −
8 2 )= x 3
⇒(
2 =( ) 3
4
=
16 81
4
x−8 2 )= x 3
⇒ 3x − 24 = 2x ⇒ x = 24
8. A jar full of whiskey contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohols and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is A.
4 3
B.
3 4
C.
3 2
D.
2 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Concentration of alcohol in 1st Jar = 40% Concentration of alcohol in 2nd Jar = 19% After the mixing, Concentration of alcohol in the mixture = 26% By the rule of alligation,
Concentration of alcohol in 1st Jar 40%
Concentration of alcohol in 2nd Jar 19%
Mean 26%
7
14
Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2
9. How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per Kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per Kg ? A. 60 Kg
B. 63 kg
C. 58 Kg
D. 56 Kg
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Selling Price(SP) of 1 Kg mixture= Rs. 9.24 Profit = 10%
Cost Price(CP) of 1 Kg mixture= =
100 (100 + P rofit%)
× SP =
100 × 9.24 (100 + 10)
100 92.4 × 9.24 = = Rs. 8.4 110 11
By the rule of alligation, we have
CP of 1 kg sugar of 1st kind CP of 1 kg sugar of 2nd kind Rs. 9 Rs. 7
Mean Price Rs.8.4
8.4 - 7 = 1.4
9 - 8.4 = .6
ie, to get a cost price of 8.4, the sugars of kind1 and kind2 should be mixed in the ratio 1.4 : .6 = 14 : 6 = 7 : 3 Let x Kg of kind1 sugar is mixed with 27 kg of kind2 sugar then x : 27 = 7 : 3
⇒
x 7 = 27 3
⇒ x = 27 ×
7 = 63 3
10. In what ratio should rice at Rs.9.30 per Kg be mixed with rice at Rs. 10.80 per Kg so that the mixture be worth Rs.10 per Kg ? A. 7 : 8
B. 8 : 7
C. 6 : 7
D. 7 ; 6
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : By the rule of alligation, we have
Cost of 1 kg rice of 1st kind Cost of 1 kg rice of 2nd kind 9.3 10.80
Mean Price 10
10.8-10 = .8
10 - 9.3 = .7
Required ratio = .8 : .7 = 8 : 7.
11. In what ratio must tea worth Rs. 60 per kg be mixed with tea worth Rs. 65 a Kg such that by selling the mixture at Rs. 68.20 a Kg ,there can be a gain 10%? A. 3 : 2
B. 2 : 3
C. 4 : 3
D. 3 : 4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Cost Price(CP) of 1 Kg mixture = Rs. 68.20 Profit = 10%
Cost Price(CP) of 1 Kg mixture= =
100 (100 + P rofit%)
× SP =
100 × 68.20 (100 + 10)
100 682 × 68.20 = = Rs. 62 110 11
By the rule of alligation:
CP of 1 kg tea of 1st kind 60
CP of 1 kg tea of 2nd kind 65
Mean Price 62
65 - 62 = 3 Hence required ratio = 3 : 2
62 - 60 = 2
12. A container contains a mixture of two liquids P and Q in the ratio 7 : 5. When 9 litres of mixture are drawn off and the container is filled with Q, the ratio of P and Q becomes 7 : 9. How many litres of liquid P was contained in the container initially? A. 23
B. 21
C. 19
D. 17
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Let's initial quantity of P in the container be 7x and initial quantity of Q in the container be 5x
Now 9 litres of mixture are drawn off from the container
Quantity of P in 9 litres of the mixtures drawn off = 9 ×
7 63 21 = = 12 12 4
Quantity of Q in 9 litres of the mixtures drawn off = 9 ×
5 45 15 = = 12 12 4
Hence Quantity of P remains in the mixtures after 9 litres is drawn off =7x −
21 4
Quantity of Q remains in the mixtures after 9 litres is drawn off =5x −
15 4
Since the container is filled with Q after 9 litres of mixture is drawn off, Quantity of Q in the mixtures = 5x −
15 21 + 9 = 5x + 4 4
Given that the ratio of P and Q becomes 7 : 9 ⇒ (7x −
21 21 ) : (5x + ) = 7 :9 4 4
21 ) 4 ⇒ 21 (5x + ) 4 (7x −
63x − (
7 9
9 × 21 7 × 21 ) = 35x + ( ) 4 4
28x = (
x= (
=
16 × 21 ) 4
16 × 21 ) 4 × 28
litres of P contained in the container initially = 7x = (
7 × 16 × 21 16 × 21 )= = 21 4 × 28 4×4
13. A vessel is filled with liquid, 3 parts of which are water and 5 parts of syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
1 3 1 C. 5 A.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
1 4 1 D. 6 B.
Let the quantity of the initial liquid in the vessel = 8 litre and quantity of water in the initial liquid = 3 litre and quantity of syrup in the initial liquid = 5 litre
Let x litre of the mixture is drawn off and replaced with water Quantity of water in the new mixture = 3 −
3x +x 8
Quantity of syrup in the new mixture = 5 −
5x 8
Given that in the new mixture, quantity of water = quantity of syrup ⇒ 3−
3x 5x +x = 5− 8 8
⇒
10x =2 8
⇒
5x =2 4
⇒x= ⇒
8 5
8 litre 5
Initially we assumed that the quantity of the initial liquid in the vessel = 8 litre for the ease of calculations. For that 8/5 litre of the mixture to be drawn off and replaced with water so that the mixture may be half water and half syrup Now, if the initial liquid in the vessel = 1 litre, quantity of the mixture to be drawn off and replaced with water so that the mixture may be half water and half syrup
=
8 1 1 × = 5 8 5
It means 1/5 of the mixture has to be drawn off and replaced with water so that the mixture may be half water and half syrup
14. In what ratio must water be mixed with milk costing Rs.12 per litre in order to get a mixture worth of Rs.8 per litre? A. 1 : 3
B. 2 : 2
C. 1 : 2
D. 3 : 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : By the rule of alligation, we have
Cost Price of 1 litre of water Cost Price of 1 litre of milk 0 12
Mean Price 8
12-8=4
8-0=8
Required Ratio = 4 : 8 = 1 : 2
15. In what ratio must tea at Rs.62 per Kg be mixed with tea at Rs. 72 per Kg so that the mixture must be worth Rs. 64.50 per Kg? A. 1 : 2
B. 2 : 1
C. 3 : 1
D. 1 : 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
By the rule of alligation, we have
Cost of 1 kg of 1st kind tea 62
Cost of 1 kg tea of 2nd kind tea 72
Mean Price 64.5
72-64.5=7.5
64.5-62=2.5
Required Ratio = 7.5 : 2.5 = 3 : 1
16. In what ratio must a grocer mix two varieties of pulses costing Rs.15 and Rs. 20 per kg respectively to obtain a mixture worth Rs.16.50 per Kg? A. 1 : 2
B. 2 : 1
C. 3 : 7
D. 7 : 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : By the rule of alligation,we have
CP of 1 kg of 1st variety pulse 15
CP of 1 kg of 2nd variety pulse 20
Mean Price 16.5
20-16.5 = 3.5
16.5-15=1.5
Required Ratio = 3.5 : 1.5 = 35 : 15 = 7 : 3
17. A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is A. 300
B. 400
C. 600
D. 500
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : By the rule of alligation,we have
Profit% by selling 1st part
Profit% by selling 2nd part
8
18
Net % profit 14
18-14=4
14-8=6
=> Quantity of part1 : Quantity of part2 = 4 : 6 = 2 : 3 Total quantity is given as 1000Kg
So Quantity of part2 (Quantity sold at 18% profit) = 1000 ×
3 = 600Kg 5
18. A dishonest milkman sells his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture? A. 25%
B. 20%
C. 22%
D. 24%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
-------------------------------------------------------------------------------Solution 1 - Using the concepts learned in Profit and Loss -------------------------------------------------------------------------------If a trader professes to sell his goods at cost price, but uses false weights, then Gain% = [
Error (True Value − Error)
× 100] %
Here Gain= 25% Here error = quantity of water he mixes in the milk = x Here the true value = true quantity of milk = T So the formula becomes 25 = ⇒1=
x × 100 (T − x)
x ×4 (T − x)
⇒ T − x = 4x ⇒ T = 5x percentage of water in the mixture =
x x 1 × 100 = × 100 = × 100 = 20% T 5x 5
-------------------------------------------------------------------------------Solution 2 - Using the concepts learned in Mixtures and Alligations -------------------------------------------------------------------------------Let CP of 1 litre milk = Rs.1 Given that SP of 1 litre mixture = CP of 1 Litre milk = Rs.1 Given than Gain = 25%
Hence CP of 1 litre mixture =
=
100 (100 + Gain%)
× SP
100 100 4 ×1 = = 125 5 (100 + 25)
By the rule of alligation, we have
CP of 1 litre milk 1
CP of 1 litre water 0
CP of 1 litre mixture 4/5
4/5 - 0 = 4/5
1- 4/5 = 1/5
=> Quantity of milk : Quantity of water = 4/5 : 1/5 = 4 : 1
Hence percentage of water in the mixture =
19.
1 × 100 = 20% 5
In what ratio must water be mixed with milk to gain 16
2 % on selling the mixture at cost price? 3
A. 6 : 1
B. 1 : 6
C. 1 : 4
D. 4 : 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Let CP of 1 litre milk = Rs.1 SP of 1 litre mixture = CP of 1 litre milk = Rs.1 Gain = 16
2 50 %= % 3 3
CP of 1 litre mixture =
=
100 50 (100 + ) 3
100 (100 + Gain%)
×1 =
100 350 ( ) 3
=
× SP
300 6 = 350 7
By the rule of alligation, we have
CP of 1 litre water 0
CP of 1 litre milk 1
CP of 1 litre mixture 6/7
1 - 6/7 = 1/7
6/7 - 0 = 6/7
Quantity of water : Quantity of milk = 1/7 : 6/7 = 1 : 6
20. In what ratio must rice at Rs.7.10 be mixed with rice at Rs.9.20 so that the mixture may be worth Rs.8 per Kg? A. 5 : 4
B. 2 : 1
C. 3 : 2
D. 4 : 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : By the rule of alligation, we have
CP of 1 kg Rice of 1st kind 7.1
CP of 1 kg Rice of 2nd kind 9.2
Mean Price 8
9.2 - 8 = 1.2
8 - 7.1 = .9
Required ratio = 1.2 : .9 = 12 : 9 = 4 : 3
21. How many Kg of rice at Rs.6.60 per Kg. be mixed with 56Kg of rice at Rs.9.60 per Kg to get a mixture worth Rs.8.20 per Kg A. 56 Kg
B. 52 Kg
C. 44 Kg
D. 49 Kg
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : By the rule of alligation, we have
Cost of 1 kg of 1st kind rice Cost of 1 kg of 2nd kind rice 6.6 9.6
Price of 1Kg of the mixture 8.2
9.6 - 8.2 = 1.4
8.2 - 6.6 = 1.6
Quantity of 1st kind rice : Quantity of 2nd kind rice = 1.4 : 1.6 =7 : 8 Quantity of 1st kind rice : 56 = 7 : 8
=> Quantity of 1st kind rice = 56 ×
7 = 49 8
22. How many litres of water must be added to 16 liters of milk and water contains 10% water to make it 20% water in it A. 4 litre
B. 2 litre
C. 1 litre
D. 3 litre
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : By the rule of alligation, we have
% Concentration of water in pure water : 100
% Concentration of water in the given mixture : 10
Mean % concentration 20
20 - 10 = 10
100 - 20 = 80
=> Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8 Here Quantity of the mixture = 16 litres => Quantity of water : 16 = 1 : 8
Quantity of water = 16 ×
1 = 2 litre 8
23. We have a 630 ml of mixture of milk and water in the ratio 7:2. How much water must be added to make the ratio 7:3? A. 70 ml
B. 60 ml
C. 80 ml
D. 50 ml
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : concentration of water in mixture1 = 2/9 (Since the ratio of milk and water = 7:2) ---item(1) concentration of water in pure water= 1 ---item(2) Now the above mentioned items are mixed to form a mixture2 where milk and water ratio = 7: 3 =>concentration of water in mixture2 = 3/10 By the rule of alligation, we have
concentration of water in mixture1 : 2/9
concentration of water in pure water : 1
Mean concentration 3/10
1 - 3/10 = 7/10
3/10 - 2/9 = 7/90
=> Quantity of mixture1 : Quantity of water = 7/10 : 7/90 = 1/10 : 1/90 = 1 : 1/9 Given that Quantity of mixture1 = 630 ml => 630 : Quantity of water = 1 : 1/9
⇒ Quantity of water = 630 ×
1 = 70 ml 9
24. 3 litre of water is added to 11 litre of a solution containing 42% of alcohol in the water. The percentage of alcohol in the new mixture is A. 25%
B. 20%
C. 30%
D. 33%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
---------------------------------------------------------Method-1 to solve the question ---------------------------------------------------------We have a 11 litre solution containing 42% of alcohol in the water => Quantity of alcohol in the solution=
11 × 42 100
Now 3 litre of water is added to the solution => Total Quantity of the new solution = 11 + 3 = 14
11 × 42 100 Percentage of alcohol in the new solution = × 100 14 =
11 × 3 = 33% 100
---------------------------------------------------------Method-2 to solve the question ---------------------------------------------------------%Concentration of alcohol in pure water = 0 %Concentration of alcohol in mixture = 42 quantity of water : Quantity of mixture = 3 : 11 Let the %Concentration of alcohol in the new mixture = x
By the rule of alligation, we have
%Concentration of alcohol : 0 %Concentration of alcohol in pure water in mixture : 42
Mean %Concentration
x
42 - x
x-0=x
But (42 - x) : x = 3 : 11
⇒
42 − x 3 = x 11
⇒
42 − x 3 = x 11
⇒ 42 × 11 − 11x = 3x ⇒ 14x = 42 × 11 ⇒ x = 3 × 11 = 33 => The percentage of alcohol in the new mixture is 33%
25. Rs.460 was divided among 41 boys and girls such that each boy Rs.12 and each girl got Rs.8. What is the number of boys? A. 33
B. 30
C. 36
D. 28
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : -----------------------------------------------------------------Solution 1 -----------------------------------------------------------------Assume that the number of boys = b and number of girls is g number of boys + number of girls = 41 => b + g = 41 ------------ (Equation 1) Given that each boy got Rs.12 and each girl got Rs.8 and Total amount = Rs.460
=> 12b + 8g = 460 -------- (Equation 2) Now we need solve Equation 1 and Equation 2 to get b and g
(Equation1) × 8 => 8b + 8g = 8 × 41 = 328 − − − − − − − −(Equation3) (Equation 2) - (Equation 3) = 4b = 460 - 328 = 132
=> b =
132 = 33 4
-----------------------------------------------------------------Solution 2 -----------------------------------------------------------------Given that Amount received by a boy = Rs.12 and Amount received by a girl =Rs.8 Total amount = 460 Given that number of boys + Number of girls = 41 Hence mean amount = 460/41 By the rule of alligation, we have
Amount received by a boy 12
Amount received by a girl 8
Mean amount 460/41
460/41 - 8 =132/41
12 - 460/41 = 32/41
Number of boys : Number of girls = 132/41 : 32/41 = 132 : 32 = 66 : 16 = 33 : 8 Given that number of boys + Number of girls = 41
Hence number of boys = 41 ×
33 = 33 41
26. A trader has 1600Kg of sugar. He sells a part at 8% profit and the rest at 12% profit. If he gains 11% on the whole , find the quantity sold at 12%. A. 1200 Kg
B. 1400 Kg
C. 1600 Kg
D. 800 Kg
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : By the rule of alligation, we have
% Profit by selling part1 8
% Profit by selling part2 12
Net % Profit 11
12 - 11 = 1
11 - 8 = 3
=>Quantity of part1 : Quantity of part2 = 1 : 3 Given that total quantity = 1600 Kg
Hence quantity of part2 (quantity sold at 12%) = 1600 ×
3 = 1200 4
27. In 40 litres of a mixture the ratio of milk to water is 7:1. In order to make the ratio of milk to water as 3:1, the quantity of water that should be added to the mixture will be
2 litre 3 2 C. 6 litre 3 A.
5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : By the rule of alligation, we have
1 litre 3
B.
4
D.
6 litre
Concentration of water in pure water : 1
Concentration of water in mixture : 1/8
Concentration of water in the final mixture 1/4
1/4 - 1/8 = 1/8
1 - 1/4 = 3/4
Quantity of water : Quantity of mixture = 1/8 : 3/4 = 1 : 6 Given that quantity of mixture = 40 litre =>Quantity of water : 40 = 1 : 6
⇒ Quantity of water = 40 ×
1 2 = 6 litre 6 3
28. Some amount out Rs.7000 was lent at 6% per annum and the remaining was lent at 4% per annum. If the total simple interest from both the fractions in 5 years was Rs.1600, the sum lent of 6% per annum was A. Rs. 2400
B. Rs. 2200
C. Rs. 2000
D. Rs. 1800
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Total simple interest received , I = Rs.1600 Principal , p = 7000 period, n = 5 years Rate of interest, r = ?
pnr 7000 × 5 × r => 1600 = 100 100 1600 × 100 160 32 => r = = = % 7000 × 5 35 7
Simple Interest, I =
By the rule of alligation, we have
Rate of interest % from part1 6
Rate of interest % from part2 4
Net rate of interest % 32/7
32/7 - 4 =4/7
6 - 32/7 = 10/7
=> Part1 : part2 = 4/7 : 10/7 = 4 : 10 = 2 : 5 Given that total amount = Rs.7000
The amount lent of 6% per annum (part1 amount) = 7000 ×
2 = Rs. 2000 7
29. In 1 kg mixture of iron and manganese 20% of manganese. How much iron should be added so that the proportion of manganese becomes 10% A. 1.5 Kg
B. 2 Kg
C. .5 Kg
D. 1 Kg
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : By the rule of alligation, we have
Percentage concentration of manganese in the mixture : 20
Percentage concentration of manganese in pure iron : 0
Percentage concentration of manganese in the final mixture 10
10 - 0 = 10
20 - 10 = 10
=> Quantity of the mixture : Quantity of iron = 10 : 10 = 1 : 1
Given that Quantity of the mixture = 1 Kg Hence Quantity of iron to be added = 1 Kg
30. John bought 20 kg of wheat at the rate of Rs.8.50 per kg and 35 kg at the rate of Rs.8.75 per kg. He mixed the two. Approximately at what price per kg should he sell the mixture to make 40% profit as the cost price? A. Rs.12
B. Rs.8
C. Rs.16
D. Rs.20
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
CP = 20 × 8.5 + 35 × 8.75 = 170 + 306.25 = 476.25 P rofit = 40% SP = =
(100 + P rofit%) 100
× CP =
(100 + 40) 140 × 476.25 = × 476.25 100 100
140 × 19.05 = 35 × 19.05 4
Total quantity = 20 + 35 = 55 Kg SP per Kg = ≈
35 × 19.05 7 × 19.05 = 55 11
7 × 19 133 ≈ ≈ 12 11 11
I mportan t Formu las - P artn ersh ips 1.
Partnership Two or more people can get together to do business by pooling resources. The deal is known as partnership. All the people who have invested money in the partnership are called partners.
2.
If investments of all the partners are for the same time period, the gain or loss is distributed among the partners = the ratio of their investments. Assume P and Q invest Rs. a and Rs. b respectively for a year in a partnership, then at the end of the year: (P's share of profit) : (Q's share of profit) = a : b.
3.
If investments are for different time periods, then equivalent capitals are calculated for a unit of time Assume A invests Rs. a for x months and B invests Rs. b for y months then, (A's share of profit) : (B's share of profit)= ax : by.
4.
A partner who manages the business is known as a working partner
5.
A partner who simply invests the money is a sleeping partner.
1. X and Y invest Rs.21000 and Rs.17500 respectively in a business. At the end of the year, they make a profit of Rs.26400. What is the share of X in the profit? A. Rs.14400
B. Rs.26400
C. Rs.12000
D. Rs.12500
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Ratio of the investment = 21000 : 17500 = 210 : 175 = 42 : 35 =6 : 5 Share of X in the profit = 26400 * (6/11) = 2400 * 6 = 14400
2. X starts a business with Rs.45000. Y joins in the business after 3 months with Rs.30000. What will be the ratio in which they should share the profit at the end of the year? A. 1:2
B. 2:1
C. 1:3
D. 3:1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Ratio in which they should share the profit = Ratio of the investments multiplied by the time period = 45000 * 12: 30000 * 9 = 45 * 12: 30 * 9 = 3*12: 2 * 9 = 2:1
3. Suresh started a business with Rs.20,000. Kiran joined him after 4 months with Rs.30,000. After 2 months, Suresh withdrew Rs.5,000 of his capital and 2 more months later, Kiran brought in Rs.20,000 more. What should be the ratio in which they should share their profits at the end of the year? A. 21:32
B. 32:21
C. 12:17
D. 17:12
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option A Explanation : Here capital is not the same.
Suresh invest 20000 for initial 6 months and 15000 for the next 6 months. Hence his term of ratio =( 20000*6 + 15000*6)
Kiran invest Rs.30000 for 4 months and Rs.50000 for next 4 months. Hence his term of ration = (30000*4 : 50000*4)
Suresh : Kiran = ( 20000*6 + 15000*6) : (30000*4 : 50000*4) = (20*6 + 15*6) : (30*4 + 50*4) = (20*3 + 15*3) : (30*2 : 50*2) = 105:160 = 21:32
4. Kamal started a business with Rs.25000 and after 4 months, Kiran joined him with Rs.60000. Kamal received Rs.58000 including 10% of profit as commission for managing the business. What amount did Kiran receive? A. 75000
B. 70000
C. 72000
D. 78000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Ratio of the profits = 25000*12 : 60000*8
= 25*12 : 60*8 = 5*3 : 12:2 = 5 : 4*2 = 5:8
Let the total profit = x. Then Kamal received 10x/100 = x/10 as managing the business
commission for
Remaining profit = x- x/10 = 9x/10 which is shared in the ration 5:8
Kamal's share = x/10 + (9x/10) * (5/13) = 58000 => x + 9x(5/13) = 580000 => x(1 + 45/13) = 580000 => x ( 58/13) = 580000 => x ( 1/13) = 10000 => x = 130000
Kiran's share = 130000 - 58000 = 72000
5. A and B started a partnership business investing Rs. 20,000 and Rs. 15,000 respectively. C joined them with Rs. 20,000 After six months. Calculate B's share in total profit of Rs. 25,000 earned at the end of 2 years from the starting of the business? A. 7500
B. 8500
C. 9000
D. 8000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : A : B : C = 20000*24 : 15000*24 : 20000*18 = 20*4 : 15*4 : 20*3 = 4*4 : 3*4 : 4*3 = 4:3:3
B's Share = 25000 * 3/10 = 7500
6. A starts a business with a capital of Rs. 85,000. B joins in the business with Rs.42500 after some time. For how much period does B join, if the profits at the end of the year are divided in the ratio of 3 : 1? A. 5 months
B. 6 months
C. 7 months
D. 8 months
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let B joins for x months. Then A:B = 85000*12 : x* 42500 = 3 : 1 => 850*12 : 425x= 3 : 1 => 850*12/ 425x = 3/1 = 3 => 850*4 /425x = 1 => 2*4/x = 1 => x = 8
7. A starts a business with Rs.40,000. After 2 months, B joined him with Rs.60,000. C joined them after some more time with Rs.1,20,000. At the end of the year, out of a total profit of Rs.3,75,000, C gets Rs.1,50,000 as his share. How many months after B joined the business, did C join? A. 4 months
B. 5 months
C. 6 months
D. 7 months
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Assume that C was there in the business for x months A:B:C = 40000*12 : 60000*10 : 120000*x = 40*12 : 60*10 : 120x = 40 : 5*10 : 10x
=8 : 10 : 2x =4 : 5 : x
C's share = 375000*x/(9+x) = 150000 => 375x/(9+x) = 150 => 15x = 6(9+x) => 5x = 18 + 2x => 3x = 18 => x = 18/3 = 6 It means C was there in the business for 6 months. Given that B joined the business after 2 months. Hence C joined after 4 months after B joined
8. A and B invest in a business in the ratio 3: 2. Assume that 5% of the total profit goes to charity. If A's share is Rs. 855, what is the total profit? A. 1400
B. 1500
C. 1600
D. 1200
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Assume that the total profit is x. Since 5% goes for charity, 95% of x will be divided between A and B in the ratio 3: 2 => A's profit = (95x/100) * (3/5) = 855 => 95x/100 = 855*5/3 = 285*5 = 1425 => x = 1425 * 100/95 = 285 * 100/19 = 1500 Hence the total profit = 1500
9. A, B and C invest in a partnership in the ratio: 7/2, 4/3, 6/5. After 4 months, A increases his share
50%. If the total profit at the end of one year be Rs. 21,600, then what is B's share in the profit? A. Rs. 2000
B. Rs. 3000
C. Rs. 4000
D. Rs. 5000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Ratio of the initial investment = 7/2 : 4/3 : 6/5 = 105 : 40 : 36
From this ratio, we can assume that actual initial investments of A, B and C are 105x, 40x and 36x respectively
A increases his share 50% after 4 months. Hence the ratio of their investments = (105x * 4) + (105x * 150/100 * 8) : 40x * 12 : 36x : 12 = 105 + (105 * 3/2 * 2) : 40*3 : 36 * 3 = 105 * 4 : 40 *3 : 36 * 3 = 35 * 4 : 40 : 36 = 35 : 10 : 9
B's share = total profit * (10/54) = 21,600 * 10/54 = 4000
10. A, B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Rs. 7400. What is the share of B in the profit. A. 2660
B. 1000
C. 2300
D. 4000
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option A Explanation : A is a working member and for that, he receive 5% of the profit = 5% of 7400 = 5*7400/100 = 370
Remaining amount = 7400-370 = 7030
Ratio of their investments = 6500*6 : 8400*5 : 10000*3 = 65*6 : 84*5 : 100*3 = 13*6 : 84 : 20*3 = 13*2 : 28 : 20 = 13 : 14 : 10
Share of B in the profit = 7030 * (14/37) = 190*14 = 2660
11. A, B, C subscribe Rs. 50,000 for a business. If A subscribes Rs. 4000 more than B and B Rs. 5000 more than C, out of a total profit of Rs. 35,000, what will be the amount A receives? A. 14200
B. 14700
C. 14800
D. 14500
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Total amount invested = 50000
Assume that investment of C = x. Then investment of B = 5000 + x , Investment of A = 4000+5000+x = 9000+x
x + 5000+x + 9000+x = 50000 => 3x + 14000 = 50000
=> 3x = 50000 – 14000 = 36000 => x = 36000/3 = 12000
=> Investment of C = x = 12000 , Investment of B = 5000 + x = 17000, Investment of A = 9000+x = 21000
Ratio of the investment of A : B : C = 21000 : 17000 : 12000 = 21 : 17 : 12 Share of A = Total Profit * 21/50 = 35000 * 21/50 = 700*21 = 14700
12. A, B, C rent a pasture. If A puts 10 oxen for 7 months, B puts 12 oxen for 5 months and C puts 15 oxen for 3 months for grazing and the rent of the pasture is Rs. 175, then how much amount should C pay as his share of rent? A. 45
B. 35
C. 55
D. 60
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : A : B: C = 10*7 : 12*5 : 15*3 = 2*7 : 12*1 : 3*3 = 14 : 12 : 9 amount that C should pay = 175 * (9/35) = 5*9 = 45
13. A and B entered into partnership with capitals in the ratio 4 : 5. After 3 months, A withdrew ¼ of his capital and B withdrew 1/5 of his capital. At the end of 10 months, the gain was Rs. 760. What is A's share in the profit? A. 310
B. 330
C. 370
D. 350
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B
Explanation : Ratio of the initial capital of A and B = 4 : 5 Hence we can take the initial capitals of A and B as 4x and 5x respectively
Ratio in which profit will be divided = (4x*3) + (3/4)*(4x)*7 : (5x*3)+ (4/5)*(5x)*7 = 12+21 : 15+28 = 33 : 43
A's share = 760 * 33/76 = 330
14. A starts a business with Rs. 3500. After 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2 : 3. B's contribution in the capital is A. 7000
B. 8000
C. 9000
D. 10000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Ratio in which profit is divided = 2:3 Assume that B's contribution to the capital = b => 3500*12 : b*7 = 2:3 => 3500*12/7b = 2/3 => b = (3500*12*3)/(2*7) = 500*6*3 = 9000
15. A, B and C shared the profit in a business in the ratio 5 : 7 : 8. They had partnered for 14 months, 8 months and 7 months respectively. What was the ratio of their investments? A. 10:12:14
B. 12:24:28
C. 20:22:12
D. 20:49:64
Hide Answer | Notebook | Discuss
Here is the answer and explanation Answer : Option D Explanation : Ratio in which A,B and C shared the profit = 5 : 7 : 8 Assume that the ratio of their investment = a:b:c Then 14a : 8b : 7c = 5 : 7 : 8
Good to go, we got the equation. Now it's a matter of time to find a, b and c
14a/8b = 5/7 => 98a = 40b => b = 98a/40 = 49a/20 ---(1)
14a/7c =5/8 => 2a/c =5/8 => 16a = 5c => c = 16a/5 ---(2)
a : b : c = a : 49a/20 : 16a/5 = 1 : 49/20 : 16/5 = 20 : 49 : 64
16. A and B started a partnership business investing capital in the ratio of 3 : 5. C joined in the partnership after six months with an amount equal to that of B. At the end of one year, the profit should be distributed among A, B and C in --- proportion. A. 10 : 5 : 4
B. 5 : 3 : 4
C. 3 : 4: 5
D. 6 : 10 : 5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Initial investment capital ratio of A and B = 3 : 5 Hence we can assume that initial capital of A and B are 3x and 5x respectively. Amount that C invest after 6 months = 5x (Since it is equal to B's investment)
Ratio in which profit should be distributed after 1 year = 3x*12 : 5x * 12 : 5x*6 => 3*12 : 5*12 : 5*6 = 6 : 10 : 5
17. A & B partner in a business , A contribute 1/4 of the capital for 15 months & B received 2/3 of the profit . For how long B's money was used A. 12 months
B. 10 months
C. 14 months
D. 16 months
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : B received 2/3 of the profit => A : B = 1 : 2
Let the total capital = x Then A's capital = x/4 B's capital = x – x/4 = 3x/4
Assume B's money was used for b months Then A:B = (x/4)*15 : (3x/4)*b = 1 : 2 => 15/4 : 3b/4 = 1 : 2 => 15 : 3b = 1 : 2
=> 5 : b = 1 : 2 => 5/b = 1/ 2 => b = 5*2 = 10
18. A , B , C started a partnership business by investing Rs 27000 , 72000 , 81000 respectively. At the end of the year , the profit were distributed among them. If C's share of profit is 36000, What is the total profit? A. 80000
B. 90000
C. 70000
D. 120000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : A:B:C = 27000:72000:81000 = 3:8:9
Let the total profit= p Then p * 9/20 = 36000 => p = 36000*20/9 = 80000
19. A & B started a partnership business. A's investment was thrice the investment of B and the period of his investment was two times the period of investments of B. If B received Rs 4000 as profit , what is their total profit? A. 28000
B. 30000
C. 32000
D. 34000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Suppose B's investment = x. Then A's investment= 3x
Suppose Bs period of investment = y, then A's period of investment = 2y
A : B = 3x * 2y : xy =6 : 1
Total profit * 1/7 = 4000 => Total profit = 4000*7 = 28000
20. P and Q invested in a business. The profit earned was divided in the ratio 2 : 3. If P invested Rs 40000, the amount invested by Q is A. 40000
B. 50000
C. 60000
D. 70000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the amount invested by Q = q 40000 : q = 2 : 3 => 40000/q = 2/3 => q = 40000 * (3/2) = 60000
21. A & B start a business jointly. A invests Rs 16000 for 8 month & B remains in a business for 4 months. Out of total profit, B claims 2/7 of the profit . How much money was contributed by B ? A. 11200
B. 12000
C. 12400
D. 12800
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : B's share = 2/7 of the profit => A:B = 5:2
(We can easily find out A:B as given above by doing quick mental calculations and it will be excellent if you can do as much as mental calculations possible. Just for explanation, see below steps to systematically find out A:B Let the total profit = x B's share = 2x/7 A's share = x – 2x/7 = 5x/7 A:B = 5x/7 : 2x/7 = 5:2)
Let the money contributed by B = b Then A:B = 16000*8 : b*4 = 5:2 => (16000*8)/4b = 5/2 => b = (16000*8*2)/ (4*5) = 800*8*2 = 12800
22. A and B starts a business investing Rs.85000 and Rs.15000 respectively. Find out the ratio in which the profit s should be shared. A. 10:3
B. 17:3
C. 3:10
D. 3:17
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Here A's and B's capitals are there for equal time. Hence A : B = 85000 : 15000 = 85 : 15 = 17 : 3
23. A , B , C start a business each investing Rs 20,000 . After 5 month A withdraws Rs 5000, B withdraws
Rs 4000 & C invests Rs 6000 more . At the end of the year, a total profit of Rs 69900 was recorded . Find the share of A A. 20600
B. 20700
C. 20500
D. 20400
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : A : B : C = (20000*5 +15000*7) : (20000*5 +16000*7) : (20000*5 +26000*7) = (20*5 + 15*7) : (20*5 + 16*7) : (20*5 + 26*7) = 205 : 212 : 282
A's share = 69900 * 205/(205+212+282) = 69900*205/699 = 20500
24. A invested Rs 76000 in a business. After few months, B joined him with Rs 57000. The total profit was divided between them in the ratio 2 : 1 at the end of the year. After how many months did B join? A. 2
B. 3
C. 4
D. 5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Assume that B joins for x months . Then A : B = 76000*12 : 57000*x = 2:1 => 76*12 : 57x = 2:1 => 76*12/57x = 2/1 = 2 => x = 76*12/57*2 = 76*4/19*2 = 8
Hence B was there in the business for 8 months, or joined after 12-8 = 4 months
25. Three partners A , B , C start a business . B's Capital is four times C's capital and twice A's capital is equal to thrice B's capital . If the total profit is Rs 16500 at the end of a year ,Find out B's share in it. A. 4000
B. 5000
C. 6000
D. 7000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Suppose C's capital = x then B's capital = 4x (Since B's Capital is four times C's capital) A's capital = 6x ( Since twice A's capital is equal to thrice B's capital)
A:B:C =6 x : 4x : x =6 : 4 : 1
B's share = 16500 * (4/11) = 1500*4 = 6000
26. In a business, A and C invested amounts in the ratio 2 : 1 , whereas the ratio between amounts invested by A and B was 3 : 2 . If Rs 157300 was their profit, how much amount did B receive? A. 48000
B. 48200
C. 48400
D. 48600
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Assume that investment of C = x Then, investment of A =2x Investment of B = 4x/3
A:B:C = 2x : 4x/3 : x = 2 : 4/3 : 1 =6 : 4 : 3
B's share = 157300 * 4/(6+4+3) = 157300*4/13 = 12100*4 = 48400
27. P , Q and R started a business by investing Rs 120000 , Rs 135000 & Rs 150000 respectively. Find the share of each, out of the annual profit of Rs 56700 A. 16800 , 18900 , 21000
B. 17850 , 18900 , 21000
C. 16800 , 18900 , 22000
D. 17850, 18500 , 22000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : P : Q : R = 120000 : 135000 : 150000 = 120:135:150 = 24:27:30 = 8:9:10
Share of P= 56700*8/27 = 2100*8 = 16800 Share of Q = 56700*9/27 = 2100*9 = 18900 Share of R = 56700*10/27 = 2100*10 = 21000
28. If 4 (P's Capital ) = 6 ( Q's Capital ) = 10 ( R's Capital ) , then out of the total profit of Rs 4650 , R will receive A. 600
B. 700
C. 800
D. 900
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let P's capital = p, Q's capital = q and R's capital = r Then 4p = 6q = 10r
=> 2p = 3q = 5r
=>q = 2p/3 r = 2p/5
P : Q : R = p : 2p/3 : 2p/5 = 15 : 10 : 6
R's share = 4650 * (6/31) = 150*6 = 900
29. P, Q, R enter into a partnership. P initially invests 25 lakh & adds another 10 lakhs after one year. Q initially invests 35 lakh & withdrawal 10 lakh after 2 years and R invests Rs 30 Lakhs . In what ratio should the profit be divided at the end of 3 years? A. 18:19:19
B. 18:18:19
C. 19:19:18
D. 18:19:19
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : P:Q:R = (25*1+35*2) : (35*2 : 25*1) : (30*3) = 95 : 95 : 90 = 19 : 19: 18
30. P , Q, R enter into a partnership & their share are in the ratio 1/2 : 1/3 : 1/4 , after two months , P withdraws half of the capitals & after 10 months , a profit of Rs 378 is divided among them . What is Q's share? A. 114
B. 120
C. 134
D. 144
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D
Explanation : The ratio of their initial investment = 1/2 : 1/3 : 1/4 = 6 : 4: 3
Let's take the initial investment of P, Q and R as 6x, 4x and 3x respectively
A:B:C = (6x * 2 + 3x * 10) : 4x*12 : 3x*12 = (12+30) : 4*12 : 3*12 =(4+10) : 4*4 : 12 = 14 : 16 : 12 =7 : 8 : 6
B's share = 378 * (8/21) = 18 * 8 = 144
I mportan t Formu las - P ercen tage 1.
Percentage Percent means for every 100 So, when percent is calculated for any value, it means that we calculate the value for every 100 of the reference value. percent is denoted by the symbol %. For example, x percent is denoted by x%
2.
x% =
x 100
Example :
3.
To express
Example :
4.
25% =
25 1 = 100 4
x x x as a percent,we have = ( × 100) % y y y 1 1 = ( × 100) % = 25% 4 4
If the price of a commodity increases by R%, the reduction in consumption so as not to increase the expenditure = [
5.
R (100 + R)
× 100] %
If the price of a commodity decreases by R%, the increase in consumption so as not to decrease the expenditure = [
R (100 − R)
× 100] %
6.
If the population of a town = P and it increases at the rate of R% per annum, then n R ) Population after n years = P (1 + 100
7.
If the population of a town = P and it increases at the rate of R% per annum, then P Population before n years = n R (1 + ) 100
8.
If the present value of a machine = P and it depreciates at the rate of R% per annum, Then n R ) Value of the machine after n years = P (1 − 100
9.
If the present value of a machine = P and it depreciates at the rate of R% per annum, Then P Value of the machine before n years = n R (1 − ) 100
1. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. What are the marks obtained by them? A. 42, 33
B. 42, 36
C. 44, 33
D. 44, 36
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the marks secured by them be x and (x + 9) Then sum of their marks = x + (x + 9) = 2x + 9 Given that (x + 9) was 56% of the sum of their marks
56 (2x + 9) 100 14 => (x + 9) = (2x + 9) 25
=> (x + 9) =
=> 25x + 225 = 28x + 126 => 3x = 99 => x = 33 Then (x + 9) = 33 + 9 = 42 Hence their marks are 33 and 42
2. If A = x% of y and B = y% of x, then which of the following is true? A. None of these C. Relationship determined.
B. A is smaller than B. between
A
and
B
cannot
E. A is greater than B. Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
A=
x xy y= .................(Equation 1) 100 100
B=
y yx x= .................(Equation 2) 100 100
be
D. If x is smaller than y, then A is greater than B.
From these equations, it is clear that A = B
3. If 20% of a = b, then b% of 20 is the same as: A. None of these
B. 10% of a
C. 4% of a
D. 20% of a
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : 20% of a = b
=> b =
20 a 100
20 a) 100 b b% of 20 = 20 = 100 100 4a = = 4% of a 100 (
20 =
20 × 20 × a 100 × 100
4. Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B. A. 2 : 1
B. 1 : 2
C. 1 : 1
D. 4 : 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
2 (6% of A + 8% of B) 3 5A 4B 2 6A 8B ) + = ( + 100 100 3 100 100 2 ⇒ 5A+ 4B = (6A+ 8B) 3 ⇒ 15A+ 12B = 12A+ 16B ⇒ 3A = 4B A 4 ⇒ = B 3 ⇒ A:B= 4 :3
5% of A + 4% of B =
5. Two employees X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week? A. Rs. 150
B. Rs. 300
C. Rs. 250
D. Rs. 200
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the amount paid to X per week = x and the amount paid to Y per week = y Then x + y = 550
But x = 120% of y =
120y 12y = 100 10
12y + y = 550 10 12 ⇒ y[ + 1] = 550 10 22y ⇒ = 550 10 ⇒ 22y = 5500 5500 500 ⇒ y= = = Rs.250 22 2
∴
6. Rahul went to a shop and bought things worth Rs. 25, out of which 30 Paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items? A. Rs. 15
B. Rs. 12.10
C. Rs. 19.70
D. Rs. 16.80
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Total cost of the items he purchased = Rs.25 Given that out of this Rs.25, 30 Paise is given as tax
=> Total tax incurred = 30 Paise = Rs.
30 100
Let the cost of the tax free items = x Given that tax rate = 6%
30 6 30 − x) = 100 100 100 ⇒ 6(25 − 0.3 − x) = 30 ⇒ (25 − 0.3 − x) = 5 ⇒ x = 25 − 0.3 − 5 = 19.7
∴ (25 −
7. The population of a town increased from 1,75,000 to 2,62,500 in a decade. What is the average percent increase of population per year? A. 4%
B. 6%
C. 5%
D. 50%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Increase in the population in 10 years = 2,62,500 - 1,75,000 = 87500
87500 8750 × 100 = = 50% 175000 175 50% Average % increase of population per year = = 5% 10 % ncrease in the population in 10 years =
8. Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get? A. 57%
B. 50%
C. 52%
D. 60%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Votes received by the winning candidate = 11628 Total votes = 1136 + 7636 + 11628 = 20400
Required percentage =
11628 11628 2907 969 × 100 = = = = 57% 20400 204 51 17
9. A fruit seller had some oranges. He sells 40% oranges and still has 420 oranges. How many oranges he had originally? A. 420
B. 700
C. 220
D. 400
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : He sells 40% of oranges and still there are 420 oranges remaining => 60% of oranges = 420
60 × Total Oranges = 420 100 Total Oranges ⇒ =7 100 ⇒ Total Oranges = 7 × 100 = 700 ⇒
10. A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
4 % 11 5 C. 45 % 11 A.
45
B.
45 %
D.
44
5 % 11
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Total runs scored = 110 Total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60 Total runs scored by running between the wickets = 110 - 60 = 50
Required % =
50 500 5 × 100 = = 45 % 110 11 11
11. What percentage of numbers from 1 to 70 have 1 or 9 in the unit's digit?
2 % 3
A.
20
C.
21%
B.
20%
D.
22
2 % 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Total numbers = 70 Total numbers in 1 to 70 which has 1 in the unit digit = 7 Total numbers in 1 to 70 which has 9 in the unit digit = 7 Total numbers in 1 to 70 which has 1 or 9 in the unit digit = 7 + 7 = 14
Required percentage =
14 140 × 100 = = 20% 70 7
12. In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, what was the number of valid votes that the other candidate got? A. 2800
B. 2700
C. 2100
D. 2500
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Total number of votes = 7500 Given that 20% of Percentage votes were invalid => Valid votes = 80%
Total valid votes = 7500 ×
80 100
1st candidate got 55% of the total valid votes. Hence the 2nd candidate should have got 45% of the total valid votes
=> Valid votes that 2nd candidate got = total valid votes × = 7500 ×
45 100
80 45 4 × = 75 × × 45 = 75 × 4 × 9 = 300 × 9 = 2700 100 100 5
13. In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State? A. 8200
B. 7500
C. 7000
D. 8000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : State A and State B had an equal number of candidates appeared. In state A, 6% candidates got selected from the total appeared candidates In state B, 7% candidates got selected from the total appeared candidates But in State B, 80 more candidates got selected than State A From these, it is clear that 1% of the total appeared candidates in State B = 80 => total appeared candidates in State B = 80 x 100 = 8000 => total appeared candidates in State A = total appeared candidates in State B = 8000
14. What percent of a day is 6 hours? A. 6.25%
B. 20%
C. 25%
D. 12.5%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Total hours in a day = 24
Required percentage =
6 × 100 = 25% 24
15. A student has to obtain 33% of the total marks to pass. He got 125 marks and failed by 40 marks. The maximum marks are
A. 600
B. 500
C. 400
D. 300
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Given that the student got 125 marks and still he failed by 40 marks => The minimum pass mark = 125 + 40 = 165 Given that minimum pass mark = 33% of the total mark
33 = 165 100 16500 => Total Mark = = 500 33
=> Total Mark ×
16. In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 of the number of students of 8 years of age which is 48. What is the total number of students in the school? A. 100
B. 102
C. 110
D. 90
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the total number of students = x Given that 20% of students are below 8 years of age then The number of students above or equal to 8 years of age = 80% of x -----(Equation 1) Given that number of students of 8 years of age = 48 ----(Equation 2) Given that number of students above 8 years of age = 2/3 of number of students of 8 years of age
=> number of students above 8 years of age =
2 × 48 = 32-----(Equation 3) 3
From Equation 1,Equation 2 and Equation 3,
80% of x = 48 + 32 = 80 ⇒
80x = 80 100
⇒
x =1 100
⇒ x = 100
17. In an examination, 5% of the applicants were found ineligible and 85% of the eligible candidates belonged to the general category. If 4275 eligible candidates belonged to other categories, then how many candidates applied for the examination? A. 28000
B. 30000
C. 32000
D. 33000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the number of candidates applied for the examination = x Given that 5% of the applicants were found ineligible. It means that 95% of the applicants were eligible (∴ 100% 5% = 95%)
Hence total eligible candidates =
95x 100
Given that 85% of the eligible candidates belonged to the general category It means 15% of the eligible candidates belonged to other categories(∴ 100% - 85% = 15%)
Hence Total eligible candidates belonged to other categories 15 95x 15 = total eligible candidates × = × 100 100 100 95x × 15 = 100 × 100 Given that Total eligible candidates belonged to other categories = 4275
⇒
95x × 15 = 4275 100 × 100
⇒
19x × 15 = 855 100 × 100
⇒
19x × 3 = 171 100 × 100
⇒
x×3 =9 100 × 100
⇒
x =3 100 × 100
⇒ x = 3 × 100 × 100 = 30000
18. A student multiplied a number by 3/5 instead of 5/3.What is the percentage error in the calculation? A. 64%
B. 32%
C. 34%
D. 42%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the number = 1 Then, ideally he should have multiplied 1 by 5/3. Hence the correct result was 1 x (5/3) = (5/3) By mistake, he multiplied 1 by 3/5. Hence the result with the error = 1 x (3/5) = (3/5)
Error =
5 3 25 − 9 16 − = = 3 5 15 15
16 ) 15 Error percentage error = × 100 = True Value 5 ( ) 3 16 × 3 × 100 16 × 100 = = = 16 × 4 = 64% 15 × 5 5×5 (
× 100
19. 270 students appeared for an examination, of which 252 passed. What is the pass percentage?
1 % 3 2 C. 92 % 3 A.
93
2 % 3
B.
93
D.
92%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Pass percentage =
252 2520 280 1 × 100 = = = 93 % 270 27 3 3
20. John's salary was decreased by 50% and subsequently increased by 50%. How much percent does he loss? A. 35%
B. 25%
C. 32%
D. 28%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let John's initial salary = Rs.100
Then, after decreasing 50%, john's salary = 100 × After the increase by 50%, john's salary = 50 ×
(100 − 50) 50 = 100 × = Rs.50 100 100
(100 + 50) 150 = 50 × = Rs.75 100 100
Actual loss = Rs.100 - Rs.75 = Rs.25
Loss percentage =
25 × 100 = 25% 100
21. How many litres of pure acid are there in 8 litres of a 20% solution? A. 2 litres
B. 1.4 litres
C. 1 litres
D. 1.6 litres
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Quantity of pure acid = 8 ×
20 = 1.6 100
22. The price of a car is Rs. 3,25,000. It was insured to 85% of its price. The car was damaged completely in an accident and the insurance company paid 90% of the insurance. What was the difference between the price of the car and the amount received ? A. Rs. 76,375
B. Rs. 34,000
C. Rs. 82,150
D. Rs. 70,000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Price of the car = Rs.3,25,000 Car insured to 85% of its price
=> Insured price = 325000 ×
85 100
Insurance company paid 90% of the insurance
⇒ Amount paid by Insurance company = Insured price × = 325000 ×
85 90 × = 325 × 85 × 9 = Rs.248625 100 100
90 100
Difference between the price of the car and the amount received = Rs.325000 - Rs.248625 = Rs.76375
23. If number x is 10% less than another number y and y is 10% more than 125, then find out the value of x. A. 123
B. 122
C. 122.25
D. 123.75
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : y is 10% more than 125
⇒ y = 125 ×
(100 + 10) 110 = 125 × 100 100
x is 10% less than another number y
⇒ x = y× =
(100 − 10) 90 110 90 = y× = 125 × × 100 100 100 100
125 × 110 × 90 125 × 11 × 9 1375 × 9 1375 × 9 = = = = 123.75 100 × 100 100 100 100
24. A housewife saved Rs. 2.50 in buying an item on sale. If she spent Rs. 25 for the item, approximately how much percent she saved in the transaction ? A. 9%
B. 10%
C. 7%
D. 6%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Actual Price = Rs.25 + Rs.2.50 = Rs.27.5 Saving = Rs.2.5
Percentage Saving = =
2.5 250 2500 × 100 = = 27.5 27.5 275
100 1 =9 % ≈ 9% 11 11
25. A pipe X is 30 meters and 45% longer than another pipe Y. find the length of the pipe Y. A. 20.12
B. 20.68
C. 20
D. 20.5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Length of pipe X = 30 meter Given that pipe X is 45% longer than Y Let the length of pipe Y = y
(100 + 45) 100 145 30 × 100 6 × 100 600 ⇒ 30 = y × => y = = = = 20.68 100 145 29 29
Then, Length of pipe X = y ×
26. On my sister's 15th birthday, she was 159 cm in height, having grown 6% since the year before. How tall was she the previous year? A. 150 cm
B. 140 cm
C. 142 cm
D. 154 cm
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Given that height on 15th birthday = 159 cm and growth = 6% Let the previous year height = x
Then height on 15th birthday = x ×
(100 + 6) 106 = x× 100 100
106 100 159 × 100 ⇒x= = 1.5 × 100 = 150 cm 106 ⇒ 159 = x ×
27. Q as a percentage of P is equal to P as a percentage of (P + Q). Find Q as a percentage of P. A. 62%
B. 50%
C. 75%
D. 66%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Q P = ......(Equation 1) P P +Q Since Q can be written as a certain percentage of P, we can assume that Q = KP KP P Hence equation 1 becomes = P P + KP 1 ⇒K= 1+K ⇒ K(K + 1) = 1 ......(Equation 2) Q KP Q as a percentage of P = × 100 = × 100 = 100K %. . . . . . (Equation3) P P From here we have two approaches. --------------------------------------------------------------------------------------Approach 1 - trial and error method --------------------------------------------------------------------------------------Here we use the values given in the choices to find out the answer Take 50% from the given choice 50 1 if 50% is the answer, 100K = 50 as per equation 3 ⇒ K = = 100 2 1 But if we substitute K as in equation 2, 2 1 1 1 3 3 ( + 1) = × = ≠ 1 K(K + 1) = 2 2 2 2 4 Given that
Now Take another choice, say 62% if 62% is the answer, 100K = 62 as per equation 3 ⇒ K =
62 100
62 in equation 2, 100 62 62 62 162 10044 ( K(K + 1) = + 1) = × = ≈1 100 100 100 100 10000 Hence 62% is the answer --------------------------------------------------------------------------------------Approach 2 - solving in the traditional way --------------------------------------------------------------------------------------Now let's solve the quadratic equation, equation 2 K(K + 1) = 1 K 2 +K−1 = 0 −−−−−−−−−−−−−− − 2 −−−−−− − 2 √ 1 −1 ± − [4 × 1 × (−1)] −b ± √ b − 4ac −1 ± √ 5 K= = = 2a 2×1 2 −1 ± 2.24 1.24 −3.24 = = or (avoiding -ve answer as here the answer should be positive) 2 2 2 = 0.62 From Equation 3, Q as a percentage of P = 100K % = 100 × .62 = 62% But if we substitute K as
28. If the price of petrol increases by 25% and Benson intends to spend only an additional 15% on petrol, by how much % will he reduce the quantity of petrol purchased? A. 8%
B. 7%
C. 10%
D. 6%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Assume that the initial price of 1 Litre petrol = Rs.100 ,Benson spends Rs.100 for petrol, such that Benson buys 1 litre of petrol
After the increase by 25%, price of 1 Litre petrol = 100 ×
(100 + 25) = Rs.125 100
Since Benson spends additional 15% on petrol, (100 + 15) amount spent by Benson = 100 × = Rs.115 100
Hence Quantity of petrol that he can purchase = Quantity of petrol reduced = (1 −
115 ) Litre 125 (1 −
Percentage Quantity of reduction = =
10 10 × 100 = × 4 = 2 × 4 = 8% 125 5
115 ) 125 1
115 Litre 125
× 100
29. Arun got 30% of the maximum marks in an examination and failed by 10 marks. However, Sujith who took the same examination got 40% of the total marks and got 15 marks more than the passing marks. What were the passing marks in the examination? A. 90
B. 250
C. 75
D. 85
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Let x is the maximum marks of the examination
Marks that Arun got = 30 % of x =
30x 100
Given that Arun failed by 10 marks 30x ⇒ Minimum Pass Mark = + 10......(Equation 1) 100
40x 100 Given that Sujith got 15 marks more than the passing marks 40x ⇒ = Minimum Pass Mark + 15 100 40x ⇒ Minimum Pass Mark = − 15......(Equation 2) 100 Marks that Sujith got = 40 % of x =
From equations 1 and 2, we have 30x 40x + 10 = − 15 100 100 10x ⇒ = 10 + 15 = 25 100 x ⇒ = 25 10 ⇒ x = 10 × 25 = 250 ⇒ Maximum marks of the examination = x = 250 Substituting the value of x in Equation 1, we have 30x 30 × 250 Minimum Pass Mark = + 10 = + 10 = 75 + 10 = 85 100 100
30. 30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years? A. 60%
B. 70%
C. 80%
D. 90%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let total number of men = 100
Then 80 men are less than or equal to 50 years old (Since 80% of the men are less than or equal to 50 years old) => 20 men are above 50 years old (Since we assumed total number of men as 100)
20% of the men above the age of 50 play football ⇒ Number of men above the age of 50 who play football = 20 ×
20 =4 100
Number of men who play football = 20 (Since 20% of all men play football)
4 × 100 = 20% 20 =>Percentage of men who play football less than or equal to the age 50 = 100% − 20% = 80% Percentage of men who play football above the age of 50 =
I mportan t Formu las - P ipes an d Cistern 1. Inlet of a tank or a cistern or reservoir Inlet is a pipe connected with a tank or cistern or reservoir. It is used to fill the tank. 2. Outlet of a tank or a cistern or reservoir Outlet is a pipe connected with a tank or cistern or reservoir. It is used to empty the tank.
1 x
3.
If a pipe can fill a tank in x hours, part filled in 1 hour =
4.
If a pipe can empty the tank in y hours, part emptied in 1 hour =
1 y
Let a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours. We can examine two cases here. Case 1: x < y In this case, the net part filled in 1 hour =
1 1 – x y
Case 2: y < x In this case, the net part emptied in 1 hour = 5.
1 1 – y x
1. Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:
9 hours 17 8 C. 2 hours 11 A.
13 hours 17 1 D. 4 hours 2
3
B.
1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Pipes A and B can fill a tank in 5 and 6 hours respectively =>Part filled by pipe A in hour = 1⁄5 and Part filled by pipe B in hour = 1⁄6 Pipe C can empty it in 12 hours => Part emptied by pipe C in 1 hour = 1⁄12
Net part filled by Pipes A,B and C together in 1 hour = i.e, the pipe can be filled in
60 9 =3 hours 17 17
1 1 1 17 + − = 5 6 12 60
2. Two pipes A and B can fill a cistern in 371⁄2 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if pipe B is turned off after: A. 5 min
B. 9 min
C. 10 min
D. 15 min
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Pipe A can fill a tank in 371⁄2 minutes = 75⁄2 minutes =>Part filled by pipe A in 1 minute = 2⁄75 Pipe B can fill a tank in 45 minutes => Part filled by pipe B in 1 minute = 1⁄45
=> Part filled by Pipe A and B in 1 minute = =
2 1 6+5 11 + = = 75 45 225 225
Assume that B is turned off after x minutes. i.e., for x minutes, both pipe A and B were open.
Part filled in x minutes by Pipe A and B = x ×
11 11x = 225 225
Now, the cistern must be filled in (30-x) minutes by pipe A alone
Part filled in (30-x) minutes by pipe A = (30 − x) ×
2(30 − x) 2 = 75 75
11x 2(30 − x) + =1 225 75 11x 2(30 − x) + =1 225 75 ⇒ 11x + 6(30 − x) = 225 ⇒ 11x + 180 − 6x = 225 ⇒ 5x = 45 x=9
3. A pump can fill a tank with water in 2 hours. Because of a leak, it took 22⁄3 hours to fill the tank. The leak can drain all the water of the tank in: A. 6 hours
B. 8 hours
C. 9 hours
D. 10 hours
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the leak can drain all the water of the tank in y hours Part of the tank filled by the pipe in 1 hr = 1/2 Part of the tank emptied by the leak in 1 hr = 1/y
1 1 3 − = 2 y 8 1 1 3 1 ⇒ = − = y 2 8 8 ⇒ y= 8 i.e., the leak can drain all the water of the tank in 8 hours
4. Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?
6 11 7 C. 11
5 11 8 D. 11
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Part of the tank filled by pipe A in 1 minute = 1⁄30 Part of the tank filled by pipe B in 1 minute = 1⁄20 Part of the tank filled by pipe C in 1 minute = 1⁄10 Here we have to find the proportion of the solution R. Pipe C discharges chemical solution R
Part of the tank filled by pipe C in 3 minutes = 3 ×
1 3 = 10 10
1 1 1 11 + + = 30 20 10 60 11 11 Part of the tank filled by pipe A,B and C together in 3 minute = 3 × = 60 20
Part of the tank filled by pipe A,B and C together in 1 minute =
( Required proportion = (
3 ) 10 11 ) 20
=
(3 × 20) (10 × 11)
=
6 11
5. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is: A. 30 hours
B. 15 hours
C. 10 hours
D. 6 hours
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the first pipe alone can fill the tank in x hours Then the second pipe can fill the tank in (x-5) hours and the third pipe can fill the tank in (x-5)-4 = (x-9) hours part filled by first pipe and second pipe together in 1 hr = part filled by third pipe in 1 hr
1 1 1 + = x x−5 x−9 (From here, better to find the value of x from the choices which will be easier. Or we can solve it as given below)
(x − 5)(x − 9) + x(x − 9) = x(x − 5) x 2 − 14x + 45 + x 2 − 9x = x 2 − 5x − 14x + 45 + x 2 − 9x = −5x x 2 − 18x + 45 = 0 (x − 15)(x − 3) = 0 x = 15 or 3 We can not take the value of x = 3 because, (x-9) becomes
negative which is not possible because the third pipe can fill the tank in (x-9) hours Hence, x = 15
6. Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately? A. 6 hours
B. 2 hours
C. 4 hours
D. 3 hours
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let pipe A alone can fill the tank in x hours Then pipe B can fill the tank in (x+6) hours
Part filled by pipe A in 1 hr = Part filled by pipe B in 1 hr =
1 x
1 x+6
Part filled by pipe A and pipe B in 1 hr =
1 1 + x x+6
It is given that pipes A and B together can fill the cistern in 4 hours. i.e., Part filled by pipes A and B in 1 hr = 1/4
1 1 1 + = x x+6 4 (From here, better to find the value of x from the choices which will be easier. Or we can solve it as given below)
4(x + 6) + 4x = x(x + 6) 4x + 24 + 4x = x 2 + 6x x 2 − 2x − 24 = 0 (x − 6)(x + 4) = 0 x = 6 or -4
Since x can not be negative, x = 6 i.e.,pipe A alone can fill the tank in 6 hours
7. Two pipes A and B can fill a tank in 12 and 24 minutes respectively. If both the pipes are used together, then how long will it take to fill the tank? A. 9 min
B. 8 min
C. 6 min
D. 4 min
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
1 12 1 Part filled by pipe B in 1 minute= 24
Part filled by pipe A in 1 minute=
Part filled by pipe A and pipe B in 1 minute=
1 1 1 + = 12 24 8
i.e., both the pipe together can fill the tank in 8 minutes
8. Two pipes A and B can fill a tank in 15 minutes and 40 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank? A. 10 min 10 sec
B. 25 min 20 sec
C. 14 min 40 sec
D. 20 min 10 sec
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
1 15 1 Part filled by pipe B in 1 minute= 40
Part filled by pipe A in 1 minute=
Part filled by pipe A and pipe B in 1 minute=
1 1 11 + = 15 40 120
pipe A and pipe B were open for 4 minutes.
Part filled by pipe A and pipe B in these 4 minutes = 4 × Remaining part to be filled = 1 −
11 19 = 30 30
11 11 = 120 30
19 ) 30 Time taken by pipe B to fill this remaining part = 1 ( ) 40 19 × 40 19 × 4 76 1 = = = = 25 minutes = 25 minutes 20 seconds 30 3 3 3 (
9. Two pipes can fill a tank in 25 and 30 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is: A. 250 gallons
B. 450 gallons
C. 120 gallons
D. 150 gallons
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Part filled by first pipe in 1 minute=
1 25
Part filled by second pipe in 1 minute=
1 30
Let the waste pipe can empty the full tank in x minutes
Then, part emptied by waste pipe in 1 minute=
1 x
All the three pipes can fill the tank in 15 minutes
i.e., part filled by all the three pipes in 1 minute= 1 1 + − 25 30 1 1 ⇒ = + x 25 ⇒ x = 150
⇒
1 15
1 1 = x 15 1 1 6 + 5 − 10 1 − = = 30 15 150 150
i.e, the waste pipe can empty the full tank in 150 minutes Given that waste pipe can empty 3 gallons per minute
ie, in 150 minutes, it can empty 150 × 3 = 450 gallons Hence, the volume of the tank = 450 gallons
10. A tank is filled in 10 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank? A. 70 hours
B. 30 hours
C. 35 hours
D. 50 hours
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the pipe A can fill the tank in x hours
Then pipe B can fill the tank in
x x hours and pipe C can fill the tank in hours 2 4
1 x 2 Part filled by pipe B in 1 hour = x 4 Part filled by pipe C in 1 hour = x
Part filled by pipe A in 1 hour =
Part filled by pipe A, pipe B and pipe C in 1 hour = i.e., pipe A, pipe B and pipe C can fill the tank in
1 2 4 7 + + = x x x x
x hours 7
Given that pipe A, pipe B and pipe C can fill the tank in 10 hours
x = 10 7 ⇒ x = 10 × 7 = 70 hours
=>
11. One pipe can fill a tank four times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in: A. 180 min
B. 144 min.
C. 126 min
D. 114 min
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A
Explanation :
Let the slower pipe alone can fill the tank in x minutes x Then the faster pipe can fill the tank in minutes 4 1 Part filled by the slower pipe in 1 minute = x 4 Part filled by the faster pipe in 1 minute = x 1 4 Part filled by both the pipes in 1 minute = + x x It is given that both the pipes together can fill the tank in 36 minutes 1 ⇒ Part filled by both the pipes in 1 minute = 36 1 4 1 + = x x 36 5 1 = x 36 x = 5 × 36 = 180 i.e.,the slower pipe alone fill the tank in 180 minutes
12. A tap can fill a tank in 4 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely? A. 3 hr
B. 1 hr 30 min
C. 2 hr 30 min
D. 2 hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : A tap can fill a tank in 4 hours => The tap can fill half the tank in 2 hours
Remaining part =
1 2
After half the tank is filled, three more similar taps are opened. Hence, total number of taps becomes 4.
Part filled by one tap in 1 hour =
1 4
1 =1 4 i.e., 4 taps can fill remaining half in 30 minutes Part filled by four taps in 1 hour = 4 ×
Total time taken = 2 hour + 30 minute = 2 hour 30 minutes
13. A tap can fill a tank in 4 hours. After half the tank is filled, two more similar taps are opened. What is the total time taken to fill the tank completely? A. 1 hr 20 min
B. 4 hr
C. 3 hr
D. 2 hr 40 min
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : A tap can fill a tank in 4 hours => The tap can fill half the tank in 2 hours
Remaining part =
1 2
After half the tank is filled, two more similar taps are opened. Hence, total number of pipes becomes 3.
Part filled by one tap in 1 hour =
1 4
1 3 = 4 4 1 ( ) 2 1 Time taken to fill the tank by 3 pipes = 2 3 ( ) 4 Part filled by three taps in 1 hour = 3 ×
=
4 2 == hour = 40 minutes 6 3
Total time taken = 2 hour + 40 minute = 2 hour 40 minutes
14. Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is: A. 10
B. 12
C. 14
D. 16
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : A, B and C can fill a tank in 6 hours.
⇒ Part filled by pipes A,B and C in 1 hr =
1 6
All these pipes are open for only 2 hours and then C is closed.
Part filled by pipes A,B and C in these 2 hours = Remaining part = 1 −
1 2 = 3 3
2 1 = 6 3
2 is filled by pipes A and B in 7 hours 3 2 ( ) 3 2 ⇒ Part filled by pipes A and B in 1 hr = − 7 21
This remaining part of
1 2 7−4 3 1 )= − = = 6 21 42 42 14 i.e., C alone can fill the tank in 14 hours
Part filled by pipe C in 1 hr = (
15. A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half? A. 15 min
B. 20 min
C. 27.5 min
D. 30 min
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
1 60 1 Part filled by pipe B in 1 minute = 40
Part filled by pipe A in 1 minute =
Part filled by both pipes A and pipe B in 1 minute =
1 1 2+3 5 1 + = = = 60 40 120 120 24
Suppose the tank is filled in x minutes Then, To fill the tanker from empty state, B is used for x/2 minutes and A and B is used for the rest x/2 minutes
x 1 x 1 × + × =1 2 40 2 24 ⇒
x 1 1 [ ]=1 + 2 40 24
⇒
x 8 × =1 2 120
⇒
x 1 × =1 2 15
x = 15 × 2 = 30 minutes
16. Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:
2 hours 3 1 C. 7 hours 2 A.
6
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
1 12 1 Part filled by pipe B in 1 hour = 15 1 Part filled by pipe C in 1 hour = 20
Part filled by pipe A in 1 hour =
In first hour, A and B is open In second hour, A and C is open then this pattern goes on till the tank fills
B. 6 hours D. 7 hours
1 1 9 3 + = = 12 15 60 20 1 1 8 2 Part filled by pipe A and pipe C in 1 hour = + = = 12 20 60 15 3 2 17 Part filled in 2 hour = + = 20 15 60 17 17 Part filled in 6 hour = ×3 = 60 20 17 3 )= Remaining part = (1 − 20 20
Part filled by pipe A and pipe B in 1 hour =
3 part needed to be filled. At this 7th hour, A and B is open 20 3 ( ) 20 3 Time taken by pipe A and B to fill this part = = 1 hour 20 3 ( ) 20
Now, 6 hours are over and only
Total time taken = 6 hour + 1 hour = 7 hour
17. A leak in the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 liters a minute. When the tank is full, the inlet is opened and due to the leak, the tank is empty in 24 hours. How many liters does the cistern hold? A. 4010 litre
B. 2220 litre
C. 1920 litre
D. 2020 litre
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Part emptied by the leak in 1 hour =
1 6
Net part emptied by the leak and the inlet pipe in 1 hour = Part filled by the inlet pipe in 1 hour =
1 1 1 − = 6 24 8
1 24
i.e., inlet pipe fills the tank in 8 hours = (8 × 60) minutes = 480 minutes Given that the inlet pipe fills water at the rate of 4 liters a minute Hence, water filled in 480 minutes = 480 × 4 = 1920 litre
i.e., The cistern can hold 1920 litre
18. A cistern can be filled by a tap in 3 hours while it can be emptied by another tap in 8 hours. If both the taps are opened simultaneously, then after how much time will the cistern get filled? A. 4.8 hr
B. 2.4 hr
C. 3.6 hr
D. 1.8 hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Part filled by the first tap in 1 hour =
1 3
Part emptied by the second tap 1 hour =
1 8
Net part filled by both these taps in 1 hour = i.e, the cistern gets filled in
1 1 5 − = 3 8 24
24 hours = 4.8 hours 5
19. Two taps A and B can fill a tank in 5 hours and 20 hours respectively. If both the taps are open then due to a leakage, it took 40 minutes more to fill the tank. If the tank is full, how long will it take for the leakage alone to empty the tank? A. 28 hr
B. 16 hr
C. 22 hr
D. 32 hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
1 5 1 Part filled by pipe B in 1 hour = 20
Part filled by pipe A in 1 hour =
Part filled by pipe A and B in 1 hour =
1 1 1 + = 5 20 4
i.e., A and B together can fill the tank in 4 hours Given that due to the leakage, it took 40 minutes more to fill the tank.
40 2 14 hour = 4 hour = hour 60 3 3 3 ⇒ Net part filled by pipe A and B and the leak in 1 hour = 14 1 3 1 ⇒ Part emptied by the leak in 1 hour = − = 4 14 28
i.e., due to the leakage, the tank got filled in 4
i.e., the leak can empty the tank in 28 hours
20. Bucket P has thrice the capacity as bucket Q. It takes 80 turns for bucket P to fill the empty drum. How many turns it will take for both the buckets P and Q, having each turn together to fill the empty drum? A. 30
B. 45
C. 60
D. 80
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let capacity of bucket P = x
Then capacity of bucket Q =
x 3
Given that it takes 80 turns for bucket P to fill the empty drum => capacity of the drum = 80x
Number of turns required if both P and Q are used =
80x x x+ 3
=
240x 240 = = 60 3x + x 4
21. A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 m3. The emptying of the tank is 10 m3 per minute higher than its filling capacity and the pump needs 8 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump? A. 20 m3 / min.
B. 40 m3 / min.
C. 50 m3 / min.
D. 60 m3 / min.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Let the filling capacity of the pump = x m3 / min. Then the emptying capacity of the pump = (x + 10) m3 / min.
2400 minutes x 2400 Time required for emptying the tank = minutes x + 10
Time required for filling the tank =
Pump needs 8 minutes lesser to empty the tank than it needs to fill it
2400 2400 − =8 x x + 10 300 300 ⇒ − =1 x x + 10 ⇒ 300(x + 10) − 300x = x(x + 10)
⇒
⇒ 3000 = x 2 + 10x ⇒ x 2 + 10x − 3000 = 0 (x + 60)(x − 50) = 0 x = 50 or -60
Since x can not be negative, x=50 i.e.,filling capacity of the pump = 50 m3 / min.
22. Two pipes A and B can separately fill a cistern in 40 minutes and 30 minutes respectively. There is a third pipe in the bottom of the cistern to empty it. If all the three pipes are simultaneously opened, then the cistern is full in 20 minutes. In how much time, the third pipe alone can empty the cistern? A. 120 min
B. 100 min
C. 140 min
D. 80 min
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
1 40 1 Part filled by pipe B in 1 minute = 30
Part filled by pipe A in 1 minute =
Net part filled by pipe A, pipe B and the third pipe in 1 hour =
1 20
⇒ Part emptied by the third pipe in 1 hour =
1 1 1 3+4−6 1 + − = = 40 30 20 120 120
i.e., third pipe alone can empty the cistern in 120 minutes
23. Two pipes A and B can fill a tank in 9 hours and 3 hours respectively. If they are opened on alternate hours and if pipe A is opened first, how many hours, the tank shall be full? A. 4 hr
B. 5 hr
C. 2 hr
D. 6 hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
1 9 1 Part filled by pipe B in 1 hour = 3
Part filled by pipe A in 1 hour =
Pipe A and B are opened alternatively.
1 1 1+3 4 + = = 9 3 9 9 4 8 Part filled in 4 hour = 2 × = 9 9 8 1 remaining part = 1 − = 9 9
Part filled in every 2 hour =
Now it is pipe A's turn.
1 ( ) 9 1 Time taken by pipe A to fill the remaining part = 9 1 ( ) 9
= 1 hour
Total time taken = 4 hour + 1 hour = 5 hour
24. 13 buckets of water fill a tank when the capacity of each bucket is 51 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 17 litres? A. 33
B. 29
C. 39
D. 42
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option C Explanation : --------------------------------------------solution 1 -------------------------------------------Capacity of the tank = (13 × 51) litre
Number of buckets required of capacity of each bucket is 17 litre =
13 × 51 = 13 × 3 = 39 17
--------------------------------------------solution 2 (using principles of chain rule) -------------------------------------------Let x be the number of buckets needed if the capacity of each bucket is 17 litres More capacity, less buckets (Indirect proportion) Capacity 51 : 17 :: x : 13 => 51 × 13 = 17 × x => 3 × 13 = x => x = 39
25. Pipe A can fill a tank in 8 hours, pipe B in 4 hours and pipe C in 24 hours. If all the pipes are open, in how many hours will the tank be filled? A. 2.4 hr
B. 3 hr
C. 4 hr
D. 4.2 hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Part filled by pipe A in 1 hour =
1 8
Part filled by pipe B in 1 hour =
1 4
Part filled by pipe C in 1 hour =
1 24
Part filled by pipe A, pipe B and pipe C in 1 hour =
1 1 1 10 + + = 8 4 24 24
i.e, pipe A, pipe B and pipe C together can fill the tank in
24 hours = 2.4 hours 10
26. Two pipes A and B can fill a tank is 8 minutes and 14 minutes respectively. If both the taps are opened simultaneously, and the tap A is closed after 3 minutes, then how much more time will it take to fill the tank by tap B? A. 6 min 15 sec
B. 5 min 45 sec
C. 5 min 15 sec
D. 6 min 30 sec
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
1 8 1 Part filled by pipe B in 1 minute = 14
Part filled by pipe A in 1 minute =
Part filled by pipe A and pipe B in 1 minute =
1 1 11 + = 8 14 56
Pipe A and pipe B were open for 3 minutes
11 33 = 56 56
Part filled by pipe A and pipe B in 3 minutes = 3 × Remaining part = 1 −
33 23 = 56 56
23 ) 56 Time taken by pipe B to fill this remaining part = 1 ( ) 14 23 × 14 23 3 = = minutes = 5 minutes = 5 minutes 45 seconds 56 4 4 (
27. A water tank is two-fifth full. Pipe A can fill a tank in 12 minutes and pipe B can empty it in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely? A. 2.8 min
B. 4.2 min
C. 4.8 min
D. 5.6 min
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Since pipe B is faster than pipe A, the tank will be emptied.
Part filled by pipe A in 1 minute =
1 12
Part emptied by pipe B in 1 minute =
1 6
Net part emptied by pipe A and pipe B in 1 minute = 2 ( ) 5 2 Time taken to empty of the tank = 5 1 ( ) 12
=
1 1 1 − = 6 12 12
2 × 12 = 4.8 min 5
28. Pipes A and B can fill a tank in 8 and 24 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in: A. 18 hr
B. 6 hr
C. 24 hr
D. 12 hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
1 8 1 Part filled by pipe B in 1 minute = 24
Part filled by pipe A in 1 minute =
Part emptied by pipe C in 1 minute =
1 12
Net part filled by pipe A, pipe B and pipe C in 1 minute = i.e, the tank will be filled in 12minutes
1 1 1 2 1 + − = = 8 24 12 24 12
29. One pipe can fill a tank 6 times as fast as another pipe. If together the two pipes can fill the tank in 22 minutes, then the slower pipe alone will be able to fill the tank in: A. 164 min
B. 154 min
C. 134 min
D. 144 min
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Let the slower pipe alone can fill the tank in x minutes x Then the faster pipe can fill the tank in minutes 6 1 Part filled by the slower pipe in 1 minute = x 6 Part filled by the faster pipe in 1 minute = x 1 6 Part filled by both the pipes in 1 minute = + x x It is given that both the pipes together can fill the tank in 22 minutes 1 ⇒ Part filled by both the pipes in 1 minute = 22 1 6 1 + = x x 22 7 1 = x 22 x = 22 × 7 = 154 i.e.,the slower pipe alone fill the tank in 154 minutes
30. Two pipes A and B can fill a tank in 2 and 6 minutes respectively. If both the pipes are used together, then how long will it take to fill the tank? A. 3 min
B. 2.5 min
C. 2 min
D. 1.5 min
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Part filled by the first pipe in 1 minute =
1 2
Part filled by the second pipe in 1 minute =
1 6
Net part filled by pipe A and pipe B in 1 minute =
1 1 2 + = 2 6 3
i.e, pipe A and B together can fill the tank in 3/2 minutes = 1.5 minutes
1. A bag contains 2 yellow, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
1 2 9 C. 11
10 21 7 D. 11
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Total number of balls = 2 + 3 + 2 = 7 Let S be the sample space. n(S) = Total number of ways of drawing 2 balls out of 7 = 7C2 Let E = Event of drawing 2 balls , none of them is blue. n(E) = Number of ways of drawing 2 balls , none of them is blue = Number of ways of drawing 2 balls from the total 5 (=7-2) balls = 5C2 (∵ There are two blue balls in the total 7 balls. Total number of non-blue balls = 7 - 2 = 5)
5×4 ( ) n(E) 5C 2 2×1 P(E) = = = 7C 2 n(S) 7×6 ( ) 2×1
=
5×4 10 = 7×6 21
2. A die is rolled twice. What is the probability of getting a sum equal to 9?
2 3 1 C. 3
2 9 1 D. 9
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces) Hence, total number of outcomes possible when a die is rolled twice, n(S) = 6 × 6 = 36 E = Getting a sum of 9 when the two dice fall = {(3, 6), {4, 5}, {5, 4}, (6, 3)} Hence, n(E) = 4
P(E) =
n(E) 4 1 = = 36 9 n(S)
3. Three coins are tossed. What is the probability of getting at most two tails?
7 8 1 C. 2 A.
1 8 1 D. 7 B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : -------------------------------------------------------------------------------------Solution 1 -------------------------------------------------------------------------------------Total number of outcomes possible when a coin is tossed = 2 (∵ Head or Tail) Hence, total number of outcomes possible when 3 coins are tossed, n(S) = 2 × 2 × 2 = 8
(∵ i.e., S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}) E = event of getting at most two Tails = {TTH, THT, HTT, THH, HTH, HHT, HHH} Hence, n(E) = 7
P(E) =
n(E) 7 = 8 n(S)
---------------------------------------------------------------------------------------Solution 2-------------------------------------------------------------------------
If n fair coins are tossed, n
Total number of outcomes in the sample space = 2
The probability of getting exactly r-number of heads when n coins are tossed =
n Cr 2n
The same formula can be used for Tails as well. Here n = 3 P(At most two tails) = P(0 Tail in 3 trials) + P(1 Tail in 3 trials) + P(2 Tail in 3 trials)
3C0
=
3
2
+
3C 1
+
3
2
3C 2 3
2
=
(3C 0 + 3C 1 + 3C2 )
=
23
(1 + 3 + 3) 23
7 8
=
---------------------------------------------------------------------------------------Solution 3 (Using Binom ial Probability distribution)-------------------------------
In a binomial experiment, The probability of achieving exactly r successes in n trials can be given by P (r successes in n trials) =
n ( ) pr q n−r r
where p = probability of success in one trial q = 1 - p = probability of failure in one trial
n ( ) r
= nCr=
n! (r!)(n − r)!
=
n(n − 1)(n − 2) ⋯ (n − r + 1) r!
(Read More ...)
Here, n = 3 p = probability of getting a Tail = 1/2 q = probability of getting a Head = 1/2 P(At most two tails) = P(0 Tail in 3 trials) + P(1 Tail in 3 trials) + P(2 Tail in 3 trials)
3 1 = ( )( ) 0 2
0
1 ( ) 2
3−0
3 1 = ( )( ) 0 2
3
3 1 + ( )( ) 1 2
3 1 + ( )( ) 1 2 3
1
1 ( ) 2
3 1 + ( )( ) 2 2
3
3−1
3 1 + ( )( ) 2 2
2
1 ( ) 2
3−2
3 3 3 1 = [( ) + ( ) + ( )] ( ) 0 1 2 2
3
= 7(
1 7 )= 8 23
4. When tossing two coins once, what is the probability of heads on both the coins?
1 4 3 C. 4
A.
B.
1 2
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : --------------------------------------------------------------------------------------
Solution 1 -------------------------------------------------------------------------------------Total number of outcomes possible when a coin is tossed = 2 (∵ Head or Tail) Hence, total number of outcomes possible when two coins are tossed, n(S) = 2 × 2 = 4 (∵ Here, S = {HH, HT, TH, TT}) E = event of getting heads on both the coins = {HH} Hence, n(E) = 1
P(E) =
n(E) 1 = 4 n(S)
---------------------------------------------------------------------------------------Solution 2-------------------------------------------------------------------------
If n fair coins are tossed, n
Total number of outcomes in the sample space = 2
The probability of getting exactly r-number of heads when n coins are tossed =
n Cr 2n
Here n = 2, r = 2
P(Exactly two Heads) =
2C2 22
=
1 4
---------------------------------------------------------------------------------------Solution 3 (Using Binom ial Probability distribution)-------------------------------
In a binomial experiment, The probability of achieving exactly r successes in n trials can be given by P (r successes in n trials) =
n ( ) pr q n−r r
where p = probability of success in one trial q = 1 - p = probability of failure in one trial
n ( ) r
= nCr=
n! (r!)(n − r)!
=
n(n − 1)(n − 2) ⋯ (n − r + 1) r!
(Read More ...)
Here, n = 2 p = probability of getting a Head = 1/2 q = probability of getting a Tail = 1/2
2 1 P(2 Heads in 2 Trials) = ( ) ( ) 2 2
2
1 ( ) 2
2−2
=
1 4
5. What is the probability of getting a number less than 4 when a die is rolled?
1 2 1 C. 3
A.
1 6 1 D. 4 B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces) i.e., n(S) = 6 E = Getting a number less than 4 = {1, 2, 3} Hence, n(E) = 3
n(E) 3 1 = = 6 2 n(S)
P(E) =
6. A bag contains 4 black, 5 yellow and 6 green balls. Three balls are drawn at random from the bag. What is the probability that all of them are yellow?
2 91 1 C. 8
1 81 2 D. 81
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Total number of balls = 4 + 5 + 6 = 15 Let S be the sample space. n(S) = Total number of ways of drawing 3 balls out of 15 = 15C3 Let E = Event of drawing 3 balls, all of them are yellow. n(E) = Number of ways of drawing 3 balls, all of them are yellow = Number of ways of drawing 3 balls from the total 5 = 5C3 (∵ there are 5 yellow balls in the total balls)
n(E) n(S)
P(E) =
=
5C 3 5C 2 = 15C 3 15C 3
[∵ nCr = nC(n - r). So 5C3 = 5C2. Applying
this for the ease of calculation]
5×4 ) 2×1 = 15 × 14 × 13 ( ) 3×2×1 (
=
5×4 15 × 14 × 13 ( ) 3
=
5×4 4 2 2 = = = 5 × 14 × 13 14 × 13 7 × 13 91
7. One card is randomly drawn from a pack of 52 cards. What is the probability that the card drawn is a face card(Jack, Queen or King)
1 13 3 C. 13
2 13 4 D. 13
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Total number of cards, n(S) = 52 Total number of face cards, n(E) = 12
P(E) =
n(E) 12 3 = = 52 13 n(S)
8. A dice is thrown. What is the probability that the number shown in the dice is divisible by 3?
1 6 1 C. 4 A.
1 3 1 D. 2 B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Total number of outcomes possible when a die is rolled, n(S) = 6 (∵ 1 or 2 or 3 or 4 or 5 or 6) E = Event that the number shown in the dice is divisible by 3 = {3, 6} Hence, n(E) = 2
P(E) =
n(E) 2 1 = = 6 3 n(S)
9. John draws a card from a pack of cards. What is the probability that the card drawn is a card of black suit?
1 2 1 C. 3
1 4 1 D. 13
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Total number of cards, n(S) = 52 Total number of black cards, n(E) = 26
P(E) =
n(E) 26 1 = = 52 2 n(S)
10. There are 15 boys and 10 girls in a class. If three students are selected at random, what is the probability that 1 girl and 2 boys are selected?
1 40 21 C. 46
1 2 7 D. 42
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let S be the sample space. n(S) = Total number of ways of selecting 3 students from 25 students = 25C3 Let E = Event of selecting 1 girl and 2 boys n(E) = Number of ways of selecting 1 girl and 2 boys 15 boys and 10 girls are there in a class. We need to select 2 boys from 15 boys and 1 girl from 10 girls Number of ways in which this can be done = 15C2 × 10C1 Hence n(E) = 15C2 × 10C1
P(E) =
n(E) n(S)
=
15C 2 × 10C1 25C 3
15 × 14 ) × 10 2×1 = 25 × 24 × 23 ( ) 3×2×1 (
=
=
15 × 14 × 10 25 × 24 × 23 ( ) 3
=
15 × 14 × 10 3 × 14 × 10 = 25 × 8 × 23 5 × 8 × 23
3 × 14 × 2 3 × 14 3×7 21 = = = 8 × 23 4 × 23 2 × 23 46
11. What is the probability of selecting a prime number from 1,2,3,... 10 ?
2 5 3 C. 5 A.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Total count of numbers, n(S) = 10
1 5 1 D. 7 B.
Prime numbers in the given range are 2,3,5 and 7 Hence, total count of prime numbers in the given range, n(E) = 4
P(E) =
n(E) 4 2 = = 10 5 n(S)
12. 3 balls are drawn randomly from a bag contains 3 black, 5 red and 4 blue balls. What is the probability that the balls drawn contain balls of different colors?
1 3 2 D. 11
3 11 1 C. 2
B.
A.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Total number of balls = 3 + 5 + 4 = 12 Let S be the sample space. n(S) = Total number of ways of drawing 3 balls out of 12 = 12C3 Let E = Event of drawing 3 different coloured balls To get 3 different coloured balls,we need to select one black ball from 3 black balls, one red ball from 5 red balls, one blue ball from 4 blue balls Number of ways in which this can be done = 3C1 × 5C1 × 4C1 i.e., n(E) = 3C1 × 5C1 × 4C1
P(E) =
=
n(E) n(S)
=
3C 1 × 5C 1 × 4C1 12C 3
3×5×4 12 × 11 × 10 ( ) 3×2×1
=
3×5×4 2 × 11 × 10
=
3×4 3 = 2 × 11 × 2 11
13. 5 coins are tossed together. What is the probability of getting exactly 2 heads?
5 16 7 D. 16
1 2 4 C. 11
B.
A.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : -------------------------------------------------------------------------------------Solution 1 -------------------------------------------------------------------------------------Total number of outcomes possible when a coin is tossed = 2 (∵ Head or Tail) Hence, total number of outcomes possible when 5 coins are tossed, n(S) = 25 E = Event of getting exactly 2 heads when 5 coins are tossed n(E) = Number of ways of getting exactly 2 heads when 5 coins are tossed = 5C2
P(E) =
n(E) 5C = 52 = n(S) 2
(
5×4 ) 2×1 2
5
=
5×2 2
5
=
5 2
4
=
5 16
---------------------------------------------------------------------------------------Solution 2-------------------------------------------------------------------------
If n fair coins are tossed, n
Total number of outcomes in the sample space = 2
The probability of getting exactly r-number of heads when
n Cr 2n
n coins are tossed =
Here n = 5, r = 2 Hence, Required probability =
n Cr 5C 2 = n = 2 25
(
5×4 ) 2×1
5×2
=
5
2
5
2
=
5 4
2
5 16
=
---------------------------------------------------------------------------------------Solution 3 (Using Binomial Probability distribution)------------------------------------
In a binomial experiment, The probability of achieving exactly r successes in n trials can be given by P (r successes in n trials) =
n ( ) pr q n−r r
where p = probability of success in one trial q = 1 - p = probability of failure in one trial
n ( ) r
= nCr=
n! (r!)(n − r)!
=
n(n − 1)(n − 2) ⋯ (n − r + 1) r!
(Read More ...)
Here, n = 5 p = probability of getting a head = 1/2 q = probability of getting a tail = 1/2
n P(2 Heads in 5 Trials) = ( ) pr q n−r r 5 1 = ( )( ) 2 2
2
1 ( ) 2
5−2
5 1 = ( )( ) 2 2
5
=
5×4 1 ( ) 2×1 2
5
=
10 5 = 16 25
14. What is the probability of drawing a "Queen" from a deck of 52 cards?
1 13 1 D. 3
1 2 1 C. 6
B.
A.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Total number of cards, n(S) = 52 Total number of "Queen" cards, n(E) = 4
P(E) =
n(E) 4 1 = = 52 13 n(S)
15. A card is randomly drawn from a deck of 52 cards. What is the probability getting an Ace or King or Queen?
3 13 1 C. 13 A.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A
2 13 1 D. 2 B.
Explanation : Total number of cards = 52 Total number of Ace cards = 4
P(Ace) =
4 1 = 52 13
Total number of King cards = 4
P(King) =
4 1 = 52 13
Total number of Queen cards = 4
P(Queen) =
4 1 = 52 13
Here, clearly the events of getting an Ace , King and Queen are mutually exclusive events. By Addition Theorem of Probability, we have P(Ace or King or Queen) = P (Ace) + P (King)+ P(Queen)
=
1 1 1 3 + + = 13 13 13 13
16. A card is randomly drawn from a deck of 52 cards. What is the probability getting a five of Spade or Club?
1 52 1 C. 26
1 13 1 D. 12
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : --------------------------------------------------------------------------------------Solution 1 --------------------------------------------------------------------------------------Total number of cards, n(S) = 52 E = event of getting a five of Spade or Club n(E) = 2 (∵ a five of Club, a five of Spade = 2 cards)
P(E) =
n(E) 2 1 = = 52 26 n(S)
---------------------------------------------------------------------------------------Total number of cards = 52 Total number of Spade Cards of Number 5 = 1 Total number of Club Cards of Number 5 = 1 P(Spade Cards of Number 5) =
P(Club Cards of Number 5) =
1 52
1 52
Here, clearly the events are mutually exclusive events. By Addition Theorem of Probability, we have P(Spade Cards of Number 5 or Club Cards of Number 5) = P(Spade Cards of Number 5) + P(Club Cards of Number 5)
=
1 1 1 + = 52 52 26
17. When two dice are rolled, what is the probability that the sum is either 7 or 11?
1 4 1 C. 9 A.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
2 5 2 D. 9 B.
Solution 2---------------------------------------------------------------------
Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces) Hence, total number of outcomes possible when two dice are rolled = 6 × 6 = 36 To get a sum of 7, the following are the favourable cases. (1, 6), (2, 5), {3, 4}, (4, 3), (5, 2), (6,1) => Number of ways in which we get a sum of 7 = 6
Number of ways in which we get a sum of 7 Total number of outcomes possible
P(a sum of 7) =
=
6 36
To get a sum of 11, the following are the favourable cases. (5, 6), (6, 5) => Number of ways in which we get a sum of 11 = 2
Number of ways in which we get a sum of 11 Total number of outcomes possible
P(a sum of 11) =
=
2 36
Here, clearly the events are mutually exclusive events. By Addition Theorem of Probability, we have P(a sum of 7 or a sum of 11) = P(a sum of 7) + P( a sum of 11)
=
6 2 8 2 + = = 36 36 36 9
18. A card is randomly drawn from a deck of 52 cards. What is the probability getting either a King or a Diamond?
4 13 1 C. 3
2 13 1 D. 2
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Total number of cards = 52 Total Number of King Cards = 4
P(King) =
4 52
Total Number of Diamond Cards = 13
P(Diamond) =
13 52
Total Number of Cards which are both King and Diamond = 1
P(King and Diamond) =
1 52
Here a card can be both a Diamond card and a King. Hence these are not mutually exclusive events. (Reference : mutually exclusive events) . By Addition Theorem of Probability, we have P(King or a Diamond) = P(King) + P(Diamond) – P(King and Diamond)
=
4 13 1 16 4 + – = = 52 52 52 52 13
19. John and Dani go for an interview for two vacancies. The probability for the selection of John is 1/3 and whereas the probability for the selection of Dani is 1/5. What is the probability that none of them are selected?
3 5 8 C. 15
7 12 1 D. 5
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let A = the event that John is selected and B = the event that Dani is selected. Given that P(A) = 1/3 and P(B) = 1/5
¯ is the event that A does not occur and We know that A the event that B does not occur
¯ is B
Probability that none of them are selected
¯ ∩B ¯) = P(A
(∵ Reference : Algebra of Events)
¯ ). P(B ¯) = P(A
(∵ Here A and B are Independent Events and refer theorem on independent events)
= [ 1 - P(A) ][ 1 - P(B)] = (1 −
=
1 1 ) (1 − ) 3 5
2 4 8 × = 3 5 15
20. John and Dani go for an interview for two vacancies. The probability for the selection of John is 1/3 and whereas the probability for the selection of Dani is 1/5. What is the probability that only one of them is selected?
3 5 2 C. 5
B. None of these
A.
D.
1 5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let A = the event that John is selected and B = the event that Dani is selected. Given that P(A) = 1/3 and P(B) = 1/5
¯ is the event that A does not occur and We know that A the event that B does not occur
¯ is B
Probability that only one of them is selected
¯ ) ∪ (B ∩ A ¯ )] = P [(A∩ B
(∵ Reference : Algebra of
Events)
¯ ) + P (B ∩ A ¯) = P (A∩ B
(∵ Reference : Mutually
Exclusive Events and Addition Theorem of Probability)
¯ ) + P(B)P(A ¯) = P(A)P(B
(∵ Here A and B are Independent Events and refer theorem on independent events)
= P(A) [1 − P(B)] + P(B) [1 − P(A)] =
1 1 1 1 1 4 1 2 4 2 2 (1 − ) + (1 − ) = × + × = + = 3 5 5 3 3 5 5 3 15 15 5
21. A letter is randomly taken from English alphabets. What is the probability that the letter selected is not a vowel?
5 25 5 C. 26
2 25 21 D. 26
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Total number of alphabets, n(S) = 26 Total number of characters which are not vowels, n(E) = 21
P(E) =
n(E) 21 = 26 n(S)
22. The probability A getting a job is 1/5 and that of B is 1/7 . What is the probability that only one of them gets a job?
11 35 2 C. 7 A.
12 35 1 D. 7 B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let A = Event that A gets a job and B = Event that B gets a job Given that P(A) = 1/5 and P(B) = 1/7 Probability that only one of them gets a job
¯ ) ∪ (B ∩ A ¯ )] = P [(A∩ B
(∵ Reference : Algebra of
Events)
¯ ) + P (B ∩ A ¯) = P (A∩ B
(∵ Reference : Mutually
Exclusive Events and Addition Theorem of Probability)
¯ ) + P(B)P(A ¯) = P(A)P(B
(∵ Here A and B are Independent Events and refer theorem on independent events)
= P(A) [1 − P(B)] + P(B) [1 − P(A)] =
1 1 1 1 1 6 1 4 6 4 10 2 (1 − ) + (1 − ) = × + × = + = = 5 7 7 5 5 7 7 5 35 35 35 7
23. A letter is chosen at random from the word 'ASSASSINATION'. What is the probability that it is a vowel?
4 13 7 C. 13
8 13 6 D. 13
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Total Number of letters in the word ASSASSINATION, n(S) = 13 Total number of Vowels in the word ASSASSINATION, n(E) = 6 (∵ 3 'A', 2 'I', 1 'O') Probability for getting a vowel, P(E) =
n(E) 6 = 13 n(S)
24. A letter is chosen at random from the word 'ASSASSINATION'. What is the probability that it is a consonant?
4 13 7 C. 13
8 13 6 D. 13
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Total Number of letters in the word ASSASSINATION, n(S) = 13 Total number of consonants in the word ASSASSINATION = 7
P(consonant) =
7 13
25. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
1 20 4 C. 20 A.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : -------------------------------------------------------------
9 20 1 D. 4 B.
-------------------------Solution 1 -------------------------------------------------------------------------------------Total number of tickets, n(S)= 20 To get a multiple of 3 , the favorable cases are 3,9,6,12,15,18. => Number of ways in which we get a multiple of 3 = 6
P(Multiple of 3) =
Number of ways in which we get a sum of multiple of 3 Total number of outcomes possible
=
6 20
To get a multiple of 5, the favorable cases are 5,10,15,20. => Number of ways in which we get a multiple of 5 = 4
P(Multiple of 5) =
Number of ways in which we get a multiple of 5 Total number of outcomes possible
=
4 20
There are some cases where we get multiple of 3 and 5. the favorable case for this is 15 => Number of ways in which we get a multiple of 3 and 5 = 1
P(Multiple of 3 and 5) =
Number of ways in which we get a multiple of 3 and 5 Total number of outcomes possible
=
1 20
Here a number can be both a a multiple of 3 and 5. Hence these are not mutually exclusive events. (Reference : mutually exclusive events) By Addition Theorem of Probability, we have P(multiple of 3 or 5) = P(multiple of 3) + P(multiple of 5)- P(multiple of 3 and 5)
=
6 4 1 9 + − = 20 20 20 20
---------------------------------------------------------------------------------------
Solution 2----------------------------------------------------------------------
To get a multiple of 3 or 5, the favorable cases are 3, 5, 6, 9, 10, 12,15,18, 20. =>Number of ways in which we get a multiple of 3 or 5 = 9
P(multiple of 3 or 5) =
Number of ways in which we get a multiple of 3 or 5 Total number of outcomes possible
=
9 20
26. One ball is picked up randomly from a bag containing 8 yellow, 7 blue and 6 black balls. What is the probability that it is neither yellow nor black?
1 3 1 C. 2
1 4 3 D. 4
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Total number of balls, n(S) = 8 + 7 + 6 = 21 n(E) = Number of ways in which a ball can be selected which is neither yellow nor black = 7 (∵ there are only 7 balls which are neither yellow nor black)
P(E) =
n(E) 7 1 = = 21 3 n(S)
27. Two cards are drawn together from a pack of 52 cards. The probability that one is a club and one is a diamond?
13 51 13 C. 102 A.
1 52 1 D. 26 B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : n(S) = Total number of ways of drawing 2 cards from 52 cards = 52C2 Let E = event of getting 1 club and 1 diamond. We know that there are 13 clubs and 13 diamonds in the total 52 cards.
Hence, n(E) = Number of ways of drawing one club from 13 and one diamond from 13 = 13C1 × 13C1
P(E) = =
n(E) 13C 1 × 13C 1 = 52C 2 n(S)
13 × 13 52 × 51 ( ) 2
=
13 × 13 13 13 = = 26 × 51 2 × 51 102
28. Two cards are drawn together at random from a pack of 52 cards. What is the probability of both the cards being Queens?
1 52 2 C. 221
1 221 1 D. 26
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : ---------------------------------------------------------------------------------------Solution 1 ---------------------------------------------------------------------------------------n(S) = Total number of ways of drawing 2 cards from 52 cards = 52C2 Let E = event of getting two Queens We know that there are total 4 Queens in the 52 cards Hence, n(E) = Number of ways of drawing 2 Queens out of 4=4C2
P(E) =
n(E) 4C 2 = 52C 2 n(S)
4×3 ) 2 = 52 × 51 ( ) 2 (
=
4×3 3 1 1 = = = 52 × 51 13 × 51 13 × 17 221
-----------------------------------------------------------------------------------------
Solution 2--------------------------------------------------------------------
Let A be the event of getting a Queen in the first draw Total number of Queens = 4 Total number of cards = 52
P(Queen in first draw) =
4 52
Assume that the first event is happened. i.e., a Queen is already drawn in the first draw and now B = event of getting a Queen in the second draw Since 1 Queen is drawn in the first draw, Total number of Queens remaining = 3 Since 1 Queen is drawn in the first draw, Total number of cards = 52 - 1 = 51
P(Queen in second draw) =
3 51
P(Queen in first draw and Queen in second draw) = P(Queen in first draw) × P(Queen in second draw)
=
4 3 1 1 1 × = × = 52 51 13 17 221
29. Two dice are rolled together. What is the probability of getting two numbers whose product is even?
17 36 3 C. 4 A.
Hide Answer | Notebook | Discuss Here is the answer and explanation
1 3 11 D. 25 B.
Answer : Option C Explanation : ---------------------------------------------------------------------------------------Solution 1 ---------------------------------------------------------------------------------------Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces) Hence, Total number of outcomes possible when two dice are rolled, n(S) = 6 × 6 = 36 Let E = the event of getting two numbers whose product is even = {(1,2), (1,4), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,2), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,2),(5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} Hence, n(E) = 27
P(E) =
n(E) 27 3 = = 36 4 n(S)
----------------------------------------------------------------------------------------- Solution 2 -----------------------------------------------------------------------------------------This problem can easily be solved if we know the following property of numbers
Multiplication Rules for Ev en and Odd Num bers 1. The product of even numbers is always even 2. The product of odd numbers is always odd 3. If there is at least one even number multiplied by any number of odd numbers, the product is always even More ...
From these properties, we know that the product will be an odd number only when both dice get odd numbers in each. In rest of the cases, product will be even Total number of outcomes possible when a die is rolled = 6 Total number of odd numbers = 3 (∵ 1 or 3 or 5)
P(Odd Number in first Die) =
3 1 = 6 2
Similarly, P(Odd Number in second Die) =
1 2
P(Odd product) = P(Odd number in first die and Odd Number in second die) = P(Odd number in first die).P(Odd number in second die) (∵ Here both these are Independent Events and refer theorem on independent events) =
1 1 1 × = 2 2 4
P(Even product) = 1 - P(Odd product) = 1 −
1 3 = 4 4
30. When two dice are tossed, what is the probability that the total score is a prime number?
1 4 2 C. 3 A.
1 3 5 D. 12 B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces) Hence, Total number of outcomes possible when two dice are rolled, n(S) = 6 × 6 = 36 Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ... etc
Let E = the event that the total is a prime number = {(1,1), (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (4,1), (4,3), (5,2), (5,6), (6,1), (6,5)} Hence, n(E) = 15
P(E) =
n(E) 15 5 = = 36 12 n(S)
31. There are 10 prizes and 25 blanks in a lottery. If John has taken a lottery, what is the probability for him to get a prize?
4 7 2 C. 7
3 7 1 D. 7
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Total number of outcomes possible, n(S) = 10 + 25 = 35 Total number of prizes, n(E) = 10
P(E) =
n(E) 10 2 = = 35 7 n(S)
32. Six dice are tossed together. What is the probability of getting the same face in all the dice? A. C.
17
B.
65 1
D.
66
1
65 7 64
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Total number of outcomes possible when a die is tossed = 6 (∵ any one face out of the 6 faces) Hence, total number of outcomes possible when 6 dice are thrown, n(S) = 66 n(E) = Number of ways of getting same face in all the dice =6C1 = 6
P(E) =
n(E) 6 1 = 6 = 5 n(S) 6 6
33. Six dice are tossed together. What is the probability of getting different faces in all of the dice? A.
1
65
B.
1
66
C.
6!
D.
65
6! 66
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Total number of outcomes possible when a die is tossed = 6 (∵ any one face out of the 6 faces) Hence, total number of outcomes possible when 6 dice are thrown, n(S) = 66 n(E) = Number of ways of getting different faces in all the dice = The number of arrangements of 6 numbers 1,2,3,4,5,6 by taking all at a time = 6!
P(E) =
n(E) n(S)
=
6! 66
34. Six persons enter a lift on the ground floor of a nine floor apartment. Assuming that each of them independently and with equal probability can leave the lift at any floor beginning with the first, what is the probability that all the six persons are leaving the lift at different floors? A.
C.
8!
86 8C 6 8
6
B.
D.
8!
87 8P 6 86
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Apart from the ground floor, there are 8 floors Let's find out the total num ber of w ays in w hich all the six persons can leave the lift at eight different floors The 1st person can leave the lift in any of the 8 floors (8 ways) The 2nd person can leave the lift in any of the remaining 7 floors (7 ways) The 3nd person can leave the lift in any of the remaining 6 floors (6 ways)
... The 6th person can leave the lift in any of the remaining 3 floors (3 ways) Total number of ways = 8 × 7 × 6 × 5 × 4 × 3 = 8P6 (In fact, from the definition of permutations itself, we will be able to directly say that the number of ways in which all the six persons can leave the lift at 8 different floors = 8P6) n(E)= Total Num ber of w ays in w hich all the six persons can leav e the lift at eight different floors = 8 P 6 Now w e w ill find out the total num ber of w ay s in w hich each of the six persons can leav e the lift at any of the eight floors The 1st person can leave the lift in any of the 8 floors (8 ways) The 2nd person can leave the lift in any of the 8 floors (8 ways) The 3nd person can leave the lift in any of the 8 floors (8 ways) ... The 6th person can leave the lift in any of the 8 floors (8 ways) Total number of ways = 8 × 8 × 8 × 8 × 8 × 8 = 86 i.e., The total num ber of w ays in w hich each the six persons can leave the lift at any of the eight floors = n(S) = 8 6
P(E) =
n(E) 8P = 66 n(S) 8
35. Raja has 4 tickets of a lottery for which 12 tickets were sold and 3 prices are to be given. What is the probability that Raja will win at least one price?
41 55 12 C. 55 A.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
43 55 14 D. 55 B.
Let E be the event that Raja will not won any price n(E) = number of ways in which Raja will not won any price = number of ways of taking 4 tickets from 12, none of them win a price = number of ways of taking 4 tickets from 9 (=12-3) = 9C4 n(S)= number of ways of taking 4 tickets from 12 = 12C4
Probability that Raja will not win any price =
=
n(E) n(S)
9C 4 12C 4
9×8×7×6 ) 4×3×2×1 = 12 × 11 × 10 × 9 ( ) 4×3×2×1 (
=
9×8×7×6 12 × 11 × 10 × 9
=
8×7 4×7 2×7 14 = = = 2 × 11 × 10 11 × 10 11 × 5 55
Probability that Raja will at least one price = 1 - Probability that Raja will not win any price
= 1−
14 41 = 55 55
36. A basket contains 15 apples and 10 oranges out of which 4 apples and 2 oranges are defective. If a person takes two fruits at random, what is the probability that either both are apples or both are good
221 300 1 C. 2
B. None of these
A.
D.
312 401
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Number of ways in which two apples can be taken from 15 apples = 15C2 Number of ways in which two fruits can be taken from 25 (=15+10) fruits = 25C2
P(both are apples)=
15C 2 25C 2
Number of ways in which two fruits taken can be taken from 19 good fruits = 19C2 (∵ Total good fruits = (15+10) – (4+2) = 19) Number of ways in which two fruits can be taken from 25(=15+10) fruits = 25C2
P(Both are good)=
19C 2 25C 2
P(Both are Apples and both are good) = ? There are 11 (=15-4) good apples. Number of ways in which two apples can be taken from these 11 good apples = 11C2 Number of ways in which two fruits can be taken from 25 (=15+10) fruits = 25C2
11C 2 25C 2
P(Both are Apples and both are good) =
Here the fruits can be apples and good. Hence these are not mutually exclusive events. (Reference : mutually exclusive events). By addition theorem on probability, we have
P ( Both are Apples or Both are Good) = P(Both are Apples ) + P(Both are Good) – P(Both are Apples and both are good)
15C 2 19C 2 11C2 15C 2 + 19C 2 − 11C 2 + − = 25C 2 25C 2 25C2 25C 2 15 × 14 19 × 18 11 × 10 )+ ( )− ( ) 2×1 2×1 2×1 = 25 × 24 ( ) 2×1 (
=
(105 + 171 − 55) 25 × 12
=
=
(15 × 7) + (19 × 9) − (11 × 5) 25 × 12
221 300
37. A bag contains 4 blue, 5 white and 6 green balls. Two balls are drawn at random. What is the probability that both the balls are blue? A.
2 35
B.
1 17
C.
1 15
D.
2 21
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option A Explanation : Let E be the event that both the balls are blue n(E) = Number of ways in which 2 balls can be drawn from 4 blue balls = 4C2 n(S) = Number of ways in which 2 balls can be drawn from 15 balls (∵ 4 + 5 + 6 = 15) = 15C2
P(E) =
n(E) 4C 2 = 15C 2 n(S)
4×3 ) 2×1 = 15 × 14 ( ) 2×1 (
=
4×3 4 2 2 = = = 15 × 14 5 × 14 5×7 35
38. A bag contains 4 blue, 5 white and 6 green balls. Two balls are drawn at random. What is the probability that one ball is white?
10 21 3 C. 4 A.
1 2 2 D. 35 B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 1 white ball can be drawn from 5 white balls in 5C1 = 5 ways 1 ball from the remaining 10 balls can be drawn in 10C1 = 10 ways Hence, one white ball and one other colour ball can be drawn in 5 × 10 = 50 ways i.e.,n(E) = 50 n(S) = Number of ways in which 2 balls can be drawn from 15 balls (∵ 4 + 5 + 6 = 15) = 15C2
P(E) =
n(E) 50 = 15C 2 n(S)
50 15 × 14 ( ) 2×1
=
=
50 × 2 10 × 2 10 10 = = = 15 × 14 3 × 14 3×7 21
39. Three boys P,Q and R are to speak at a function along with 5 others. If all of them speak in random order, what is the probability that P speaks before Q and Q speaks before R?
2 9 1 C. 6 A.
1 3 1 D. 9 B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : n(S) = Total num ber of w ays in w hich 8 boy s can speak =8 P 8 = 8! Now , let's find out the total num ber of w ay s in w hich 8 boys can speak w here P speaks before Q and Q speaks before R. Let the 8 boys be A, B, C, D, E, P, Q, R. There are 8 positions to be filled. Number of options for A = 8 (Any of the 8 positions.) Number of options for B = 7 (Any of the 7 positions.) Number of options for C = 6 (Any of the 6 positions.) Number of options for D = 5 (Any of the 5 positions.) Number of options for E = 4 (Any of the 4 positions.) Since P must speak before Q, and Q must speak before R, the 3 remaining positions must be occupied as P-Q-R. i.e, there is only 1 way of doing this Total number of ways = (8 × 7 × 6 × 5 × 4 × 1)= 8P5 i.e., n(E) = Total num ber of w ays in w hich 8 boy s can speak w here P speaks before Q and Q speaks before R = 8 P 5
P(E) =
n(E) 8P = 5 = n(S) 8!
(
8! ) 3! 8!
=
1 1 = 3! 6
40. There is a point in a circle. What is the probability that this point is closer to its circumference than to the centre?
1 3 1 C. 2
1 4 3 D. 4
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Consider a circle with radius r as shown above. Let P be any point in this circle. If P is in the inner circle of radius r/2 (i.e, in segment A), it is closer to the centre of the circle (of radius r) If P is outside the inner circle of radius r/2 (i.e, in segment B), it is closer to the circumference of the circle (of radius r) Area of segment B = Area of Circle of radius r – Area of the circle of radius r/2
r = πr 2 − π( ) 2
2
= πr 2 −
πr 2 3πr 2 = 4 4
Area of the circle of radius r = πr 2
Required Probability =
Area of segment B Total Area of the circle of radius r
( =
3πr 2 ) 4 πr 2
=
3 4
I mportan t Formu las - P robability or Ch an ce 1.
Probability or Chance
Probability or chance is a common term used in day-to-day life. For example, we generally say, 'it may rain today'. This statement has a certain uncertainty. Probability is quantitative measure of the chance of occurrence of a particular event.
2.
Experim ent
An experiment is an operation which can produce well-defined outcomes.
3.
Random Ex perim ent
If all the possible outcomes of an experiment are known but the exact output cannot be predicted in advance, that experiment is called a random experiment. Ex am ples i.
Tossing of a fair coin When we toss a coin, the outcome will be either Head (H) or Tail (T)
ii.
Throw ing an unbiased die Die is a small cube used in games. It has six faces and each of the six faces shows a different number of dots from 1 to 6. Plural of die is dice. When a die is thrown or rolled, the outcome is the number that appears on its upper face and it is a random integer from one to six, each value being equally likely.
iii.
Draw ing a card from a pack of shuffled cards A pack or deck of playing cards has 52 cards which are divided into four categories as given below Spades (♠)
Clubs (♣)
Hearts (♥)
Diamonds (♦)
Each of the above mentioned categories has 13 cards, 9 cards numbered from 2 to 10, an Ace, a King, a Queen and a jack Hearts and Diamonds are red faced cards whereas Spades and Clubs are black faced cards. Kings, Queens and Jacks are called face cards iv.
4.
Taking a ball random ly from a bag containing balls of different colours
Sam ple Space
Sample Space is the set of all possible outcomes of an experiment. It is denoted by S. Ex am ples
5.
i.
When a coin is tossed, S = {H, T} where H = Head and T = Tail
ii.
When a dice is thrown, S = {1, 2 , 3, 4, 5, 6}
iii.
When two coins are tossed, S = {HH, HT, TH, TT} where H = Head and T = Tail
Event
Any subset of a Sample Space is an event. Events are generally denoted by capital letters A, B , C, D etc. Ex am ples
6.
i.
When a coin is tossed, outcome of getting head or tail is an event
ii.
When a die is rolled, outcome of getting 1 or 2 or 3 or 4 or 5 or 6 is an event
Equally Likely Events
Events are said to be equally likely if there is no preference for a particular event over the other. Ex am ples
7.
i.
When a coin is tossed, Head (H) or Tail is equally likely to occur.
ii.
When a dice is thrown, all the six faces (1, 2, 3, 4, 5, 6) are equally likely to occur.
Mutually Ex clusive Events
Two or more than two events are said to be mutually exclusive if the occurrence of one of the events excludes the occurrence of the other This can be better illustrated with the following examples i.
When a coin is tossed, we get either Head or Tail. Head and Tail cannot come simultaneously. Hence occurrence of Head and Tail are mutually exclusive events.
ii.
When a die is rolled, we get 1 or 2 or 3 or 4 or 5 or 6. All these faces cannot come simultaneously. Hence occurrences of particular faces when rolling a die are mutually exclusive events. Note : If A and B are mutually exclusive events, A set.
iii.
∩ B = ϕ where ϕ represents empty
Consider a die is thrown and A be the event of getting 2 or 4 or 6 and B be the event of getting 4 or 5 or 6. Then A = {2, 4, 6} and B = {4, 5, 6} Here A
∩ B ≠ ϕ . Hence A and B are not mutually exclusive events.
8.
Independent Events
Events can be said to be independent if the occurrence or non-occurrence of one event does not influence the occurrence or non-occurrence of the other. Ex am ple : When a coin is tossed twice, the event of getting Tail(T) in the first toss and the event of getting Tail(T) in the second toss are independent events. This is because the occurrence of getting Tail(T) in any toss does not influence the occurrence of getting Tail(T) in the other toss.
9.
Sim ple Ev ents
In the case of simple events, we take the probability of occurrence of single events. Ex am ples
10.
i.
Probability of getting a Head (H) when a coin is tossed
ii.
Probability of getting 1 when a die is thrown
Com pound Ev ents
In the case of compound events, we take the probability of joint occurrence of two or more events. Ex am ples i.
11.
When two coins are tossed, probability of getting a Head (H) in the first toss and getting a Tail (T) in the second toss.
Exhaustiv e Ev ents
Exhaustive Event is the total number of all possible outcomes of an experiment. Ex am ples i.
When a coin is tossed, we get either Head or Tail. Hence there are 2 exhaustive events.
ii.
When two coins are tossed, the possible outcomes are (H, H), (H, T), (T, H), (T, T). Hence there are 4 (=22) exhaustive events.
iii.
12.
When a dice is thrown, we get 1 or 2 or 3 or 4 or 5 or 6. Hence there are 6 exhaustive events.
Algebra of Ev ents
Let A and B are two events with sample space S. Then i.
A ∪ B is the event that either A or B or Both occur. (i.e., at least one of A or B occurs)
ii.
A
iii.
¯ is the event that A does not occur A
iv.
¯ ∩B ¯ is the event that none of A and B occurs A
∩ B is the event that both A and B occur
Ex am ple : Consider a die is thrown , A be the event of getting 2 or 4 or 6 and B be the event of getting 4 or 5 or 6. Then A = {2, 4, 6} and B = {4, 5, 6} A
∪ B = {2, 4, 5, 6}
A
∩ B = {4, 6}
¯ = {1, 3, 5} A ¯ = {1, 2, 3} B ¯ ∩B ¯ = {1,3} A
13.
Probability of en Event
Let E be an event and S be the sample space. Then probability of the event E can be defined as P(E) =
n(E) n(S)
where P(E) = Probability of the event E, n(E) = number of ways in which the event can occur and n(S) = Total number of outcomes possible
Exam ples
i.
A coin is tossed once. What is the probability of getting Head? Total number of outcomes possible when a coin is tossed = n(S) = 2 (∵ Head or Tail) E = event of getting Head = {H}. Hence n(E) = 1
P(E) = ii.
n(E) n(S)
=
1 2
Two dice are rolled. What is the probability that the sum on the top face of both the dice will be greater than 9? Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces) Hence, total number of outcomes possible two dice are rolled, n(S) = 6 × 6 = 36 E = Getting a sum greater than 9 when the two dice are rolled = {(4, 6), {5, 5}, {5, 6}, {6, 4}, {6, 5}, (6, 6)} Hence, n(E) = 6
P(E) =
14.
n(E) 6 1 = = 36 6 n(S)
Im portant form ulas
P(S) = 1 0
≤ P (E) ≤ 1
P(ϕ ) = 0
15.
(∵ Probability of occurrence of an impossible event = 0)
Addition Theorem
Let A and B be two events associated with a random experiment. Then P(A U B) = P(A) + P(B) – P(A
∩ B)
If A and B are mutually exclusive events, then P(A U B) = P(A) + P(B) because for mutually exclusive events, P(A ∩ B) = 0
If A and B are two independents events, then 16.
P(A
∩ B) = P(A).P(B)
Ex am ple : Two dice are rolled. What is the probability of getting an odd number in one die and getting an even number in the other die? Total number of outcomes possible when a die is rolled, n(S) = 6 (∵ any one face out of the 6 faces) Let A be the event of getting the odd number in one die = {1,3,5}. => n(A)= 3
P(A) =
n(A) 3 1 = = 6 2 n(S)
Let B be the event of getting an even number in the other die = {2,4, 6}. => n(B)= 3
P(B) =
n(B) 3 1 = = 6 2 n(S)
Required Probability, P(A
Let A be any event and occur). Then
¯ ) = 1 - P(A) P(A
17.
Odds on an ev ent
∩ B) = P(A).P(B) =
1 1 1 × = 2 2 4
¯ be its complementary event (i.e., A ¯ is the event that A does not A
Let E be an event associat ed with a random experiment. Let and y outcomes are not favourable to E, then
x outcomes are favourable to E
x x : y , i.e., and y y Odds against E are y : x , i.e., x x P(E) = x+y Odds in favour of E are
P(E¯ ) =
y x+y
Ex am ple : What are the odds in favour of and against getting a 1 when a die is rolled? Let E be an event of getting 1 when a die is rolled Outcomes which are favourable to E,
x=1
Outcomes which are not favourable to E, Odds in favour of getting 1 = Odds against getting 1 =
18.
Conditional Probability
x 1 = y 5
x y 5 = = y x 1
y= 5
Let A and B be two events associated with a random experiment. Then, probability of the occurrence of A given that B has already occurred is called conditional probability and denoted by P(A/B) Example : A bag contains 5 black and 4 blue balls. Two balls are drawn from the bag one by one without replacement. What is the probability of drawing a blue ball in the second draw if a black ball is already drawn in the first draw? Let A be the event of drawing black ball in the first draw and B be the event of drawing a blue ball in the second draw. Then, P(B/A) = Probability of drawing a blue ball in the second draw given that a black ball is already drawn in the first draw. Total Balls = 5 + 4 = 9 Since a black ball is drawn already, total number of balls left after the first draw = 8 total number of blue balls after the first draw = 4
P(B/A) =
19.
4 1 = 8 2
Binom ial Probability distribution A binomial experiment is a probability experiment which satisfies the following requirements. 1. Each trial can have only two outcomes. These outcomes can be considered as either success or failure. 2. There must be a fixed number of trials. 3. The outcomes of each trial must be independent of each other. 4. The probability of a success must remain the same for each trial.
In a binomial experiment, The probability of achieving exactly r successes in n trials can be given by P (r successes in n trials) =
n ( ) pr q n−r r
where p = probability of success in one trial q = 1 - p = probability of failure in one trial
n ( ) r
20.
= nCr=
n! (r!)(n − r)!
More Shortcut F orm ulas
=
n(n − 1)(n − 2) ⋯ (n − r + 1) r!
If n fair coins are tossed, n
Total number of outcomes in the sample space = 2
The probability of getting exactly r-number of heads when n coins are tossed =
nCr 2n
I mportan t Formu las - P roblems on Train s 5 m/s 18
1.
x km/hr = x ×
2.
y m/s = y ×
3.
Speed = distance/time
18 km/hr 5
s = d/t
4.
velocity = displacement/time v = d/t
5.
Time taken by a train x meters long to pass a pole or standing man or a post = Time taken by the train to travel x meters.
6.
Time taken by a train x meters long to pass an object of length y meters = Time taken by the train to travel (x + y) metres.
7.
Suppose two trains or two objects are moving in the same direction at v1 m/s and v2 m/s where v1 > v2, then their relative speed = (v1 – v2) m/s
8.
Suppose two trains or two objects are moving in opposite directions at v1 m/s and v2 m/s , then their relative speed = (v1+ v2) m/s
9.
Assume two trains of length x metres and y metres are moving in opposite directions at v1 m/s and v2 m/s, Then The time taken by the trains to cross each other = (x+y) / (v1+v2) seconds
10.
Assume two trains of length x metres and y metres are moving in the same direction at at v1 m/s and v2 m/s where v1 > v2, Then The time taken by the faster train to cross the slower train = (x+y) / (v1-v2) seconds
11.
Assume that two trains (objects) start from two points P and Q towards each other at the same time and after crossing they take p and q seconds to reach Q and P respectively. Then,
A's speed: B's speed = √
q :√ p
1. A train is running at a speed of 40 km/hr and it crosses a post in 18 seconds. What is the length of the train? A. 190 metres
B. 160 metres
C. 200 metres
D. 120 metres
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Speed of the train, v = 40 km/hr = 40000/3600 m/s = 400/36 m/s Time taken to cross, t = 18 s Distance Covered, d = vt = (400/36)× 18 = 200 m Distance covered is equal to the length of the train = 200 m
2. A train ,130 meters long travels at a speed of 45 km/hr crosses a bridge in 30 seconds. The length of the bridge is A. 270 m
B. 245 m
C. 235 m
D. 220 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Assume the length of the bridge = x meter Total distance covered = 130+x meter total time taken = 30s speed = Total distance covered /total time taken = (130+x)/30 m/s => 45 × (10/36) = (130+x)/30 => 45 × 10 × 30 /36 = 130+x => 45 × 10 × 10 / 12 = 130+x => 15 × 10 × 10 / 4 = 130+x => 15 × 25 = 130+x = 375 => x = 375-130 =245
3. A train has a length of 150 meters . it is passing a man who is moving at 2 km/hr in the same direction of the train, in 3 seconds. Find out the speed of the train. A. 182 km/hr
B. 180 km/hr
C. 152 km/hr
D. 169 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Length of the train, l = 150m Speed of the man , Vm= 2 km/hr Relative speed, Vr = total distance/time = (150/3) m/s = (150/3) × (18/5) = 180 km/hr Relative Speed = Speed of train, Vt - Speed of man (As both are moving in the same direction) => 180 = Vt - 2 => Vt = 180 + 2 = 182 km/hr
4. A train having a length of 240 m passes a post in 24 seconds. How long will it take to pass a platform having a length of 650 m? A. 120 sec
B. 99 s
C. 89 s
D. 80 s
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : v = 240/24 (where v is the speed of the train) = 10 m/s t = (240+650)/10 = 89 seconds
5. A train 360 m long runs with a speed of 45 km/hr. What time will it take to pass a platform of 140 m long? A. 38 sec
B. 35 s
C. 44 sec
D. 40 s
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Speed = 45 km/hr = 45×(10/36) m/s = 150/12 = 50/4 = 25/2 m/s Total distance = length of the train + length of the platform = 360 + 140 = 500 meter Time taken to cross the platform = 500/(25/2) = 500×2/25 = 40 seconds
6. Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively . If they cross each other in 23 seconds, what is the ratio of their speeds? A. Insufficient data
B. 3 : 1
C. 1 : 3
D. 3 : 2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the speed of the trains be x and y respectively length of train1 = 27x length of train2 = 17y
Relative speed= x+ y Time taken to cross each other = 23 s => (27x + 17 y)/(x+y) = 23 => (27x + 17 y)/ = 23(x+y) => 4x = 6y
=> x/y = 6/4 = 3/2
7. A jogger is running at 9 kmph alongside a railway track in 240 meters ahead of the engine of a 120 meters long train . The train is running at 45 kmph in the same direction. how much time does it take for the train to pass the jogger? A. 46
B. 36
C. 18
D. 22
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Distance to be covered = 240+ 120 = 360 m Relative speed = 36 km/hr = 36×10/36 = 10 m/s Time = distance/speed = 360/10 = 36 seconds
8. Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. If the faster train passes the slower train in 36 seconds,what is the length of each train? A. 88
B. 70
C. 62
D. 50
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Assume the length of each train = x Total distance covered for overtaking the slower train = x+x = 2x Relative speed = 46-36 = 10km/hr = (10×10)/36 = 100/36 m/s Time = 36 seconds 2x/ (100/36) = 36 => (2x × 36 )/100 = 36 => x = 50 meter
9. Two trains having length of 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions (on parallel tracks). The time which they take to cross each other, is A. 10.8 s
B. 12 s
C. 9.8 s
D. 8 s
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Distance = 140+160 = 300 m Relative speed = 60+40 = 100 km/hr = (100×10)/36 m/s Time = distance/speed = 300 / (100×10)/36 = 300×36 / 1000 = 3×36/10 = 10.8 s
10. Two trains are moving in opposite directions with speed of 60 km/hr and 90 km/hr respectively. Their lengths are 1.10 km and 0.9 km respectively. the slower train cross the faster train in --- seconds A. 56
B. 48
C. 47
D. 26
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Relative speed = 60+90 = 150 km/hr (Since both trains are moving in opposite directions) Total distance = 1.1+.9 = 2km Time = 2/150 hr = 1//75 hr = 3600/75 seconds = 1200/25 = 240/5 = 48 seconds
11. A train passes a platform in 36 seconds. The same train passes a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, The length of the platform is A. None of these
B. 280 meter
C. 240 meter
D. 200 meter
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option C Explanation : Speed of the train = 54 km/hr = (54×10)/36 m/s = 15 m/s Length of the train = speed × time taken to cross the man = 15×20 = 300 m Let the length of the platform = L Time taken to cross the platform = (300+L)/15 => (300+L)/15 = 36 => 300+L = 15×36 = 540 => L = 540-300 = 240 meter
12. A train moves past a post and a platform 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train? A. 79.2 km/hr
B. 69 km/hr
C. 74 km/hr
D. 61 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let x is the length of the train and v is the speed Time taken to move the post = 8 s => x/v = 8 => x = 8v --- (1)
Time taken to cross the platform 264 m long = 20 s (x+264)/v = 20 => x + 264 = 20v ---(2)
Substituting equation 1 in equation 2, we get 8v +264 = 20v
=> v = 264/12 = 22 m/s = 22×36/10 km/hr = 79.2 km/hr
13. Two trains having equal lengths, take 10 seconds and 15 seconds respectively to cross a post. If the length of each train is 120 meters, in what time (in seconds) will they cross each other when traveling in opposite direction? A. 10
B. 25
C. 12
D. 20
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : speed of train1 = 120/10 = 12 m/s speed of train2 = 120/15 = 8 m/s if they travel in opposite direction, relative speed = 12+8 = 20 m/s distance covered = 120+120 = 240 m time = distance/speed = 240/20 = 12 s
14. Two trains, one from P to Q and the other from Q to P, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is A. 2 : 3
B. 2 :1
C. 4 : 3
D. 3 : 2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Ratio of their speeds = Speed of first train : Speed of second train
−− √ 16 = √9 = 4:3
15. A train having a length of 1/4 mile , is traveling at a speed of 75 mph. It enters a tunnel 3 ½ miles long. How long does it take the train to pass through the tunnel from the moment the front enters to the moment the rear emerges? A. 3 min
B. 4.2 min
C. 3.4 min
D. 5.5 min
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Total distance = 3 ½ + ¼ = 7/2 + ¼ = 15/4 miles Speed = 75 mph Time = distance/speed = (15/4) / 75 hr = 1/20 hr = 60/20 minutes = 3 minutes
16. A train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the train? A. 270 m
B. 210 m
C. 340 m
D. 130 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Speed= 72 kmph = 72×10/36 = 20 m/s Distance covered = 250+ x where x is the length of the train Time = 26 s (250+x)/26 = 20 250+x = 26×20 = 520 m x = 520-250 = 270 m
17. A train overtakes two persons who are walking in the same direction to that of the train at 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. What is the length of the train? A. 62 m
B. 54 m
C. 50 m
D. 55 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let x is the length of the train in meter and v is its speed in kmph x/9 = ( v-2)(10/36) ---(1) x/10 =( v-4) (10/36) --- (2)
Dividing equation 1 with equation 2 10/9 = (v-2)/(v-4) => 10v - 40 = 9v - 18 => v = 22 Substituting in equation 1, x/9 = 200/36 => x = 9×200/36 = 50 m
18. A train is traveling at 48 kmph . It crosses another train having half of its length , traveling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. What is the length of the platform? A. 500 m
B. 360 m
C. 480 m
D. 400 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Speed of train1 = 48 kmph Let the length of train1 = 2x meter Speed of train2 = 42 kmph Length of train 2 = x meter (because it is half of train1's length) Distance = 2x + x = 3x Relative speed= 48+42 = 90 kmph = 90×10/36 m/s = 25 m/s
Time = 12 s Distance/time = speed => 3x/12 = 25 => x = 25×12/3 = 100 meter Length of the first train = 2x = 200 meter
Time taken to cross the platform= 45 s Speed of train1 = 48 kmph = 480/36 = 40/3 m/s Distance = 200 + y where y is the length of the platform => 200 + y = 45×40/3 = 600 => y = 400 meter
19. A train having a length of 270 meter is running at the speed of 120 kmph . It crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train? A. 320 m
B. 190 m
C. 210 m
D. 230 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Relative speed = 120+80 = 200 kmph = 200×10/36 m/s = 500/9 m/s time = 9s Total distance covered = 270 + x where x is the length of other train (270+x)/9 = 500/9 => 270+x = 500 => x = 500-270 = 230 meter
20. Two trains, each 100 m long are moving in opposite directions. They cross each other in 8 seconds. If one is moving twice as fast the other, the speed of the faster train is A. 75 km/hr
B. 60 km/hr
C. 35 km/hr
D. 70 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Total distance covered = 100+100 = 200 m Time = 8 s let speed of slower train is v . Then the speed of the faster train is 2v (Since one is moving twice as fast the other) Relative speed = v + 2v = 3v 3v = 200/8 m/s = 25 m/s => v = 25/3 m/s Speed of the faster train = 2v = 50/3 m/s = (50/3)×(36/10) km/hr = 5×36/3 = 5×12 = 60 km/hr
21. Two stations P and Q are 110 km apart on a straight track. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed of 25 kmph. At what time will they meet? A. 10.30 a.m
B. 10 a.m.
C. 9.10 a.m.
D. 11 a.m.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Assume both trains meet after x hours after 7 am Distance covered by train starting from P in x hours = 20x km Distance covered by train starting from Q in (x-1) hours = 25(x-1) Total distance = 110 => 20x + 25(x-1) = 110 => 45x = 135
=> x= 3 Means, they meet after 3 hours after 7 am, ie, they meet at 10 am
22. A train overtakes two persons walking along a railway track. The first person walks at 4.5 km/hr and the other walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train? A. 81 km/hr
B. 88 km/hr
C. 62 km/hr
D. 46 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let x is the length of the train in meter and y is its speed in kmph
x/8.4 = (y-4.5)(10/36) ---(1) x/8.5 = (y-5.4)(10/36) ---(2) Dividing 1 by 2 8.5/8.4 = (y-4.5)/ (y-5.4) => 8.4y - 8.4 × 4.5 = 8.5y - 8.5×5.4 .1y = 8.5×5.4 - 8.4×4.5 => .1y = 45.9-37.8 = 8.1 => y = 81 km/hr
23. A train , having a length of 110 meter is running at a speed of 60 kmph. In what time, it will pass a man who is running at 6 kmph in the direction opposite to that of the train A. 10 sec
B. 8 sec
C. 6 sec
D. 4 sec
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Distance = 110 m Relative speed = 60+6 = 66 kmph (Since both the train and the man are in moving in opposite direction) = 66×10/36 mps = 110/6 mps Time = distance/speed = 110/(110/6) = 6 s
24. A 300 metre long train crosses a platform in 39 seconds while it crosses a post in 18 seconds. What is the length of the platform? A. 150 m
B. 350 m
C. 420 m
D. 600 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Length of the train = distance covered in crossing the post = speed × time = speed × 18 Speed of the train = 300/18 m/s = 50/3 m/s Time taken to cross the platform = 39 s (300+x)/(50/3) = 39 s where x is the length of the platform 300+x = (39 × 50) / 3 = 650 meter x = 650-300 = 350 meter
25. A train crosses a post in 15 seconds and a platform 100 m long in 25 seconds. Its length is A. 150 m
B. 300 m
C. 400 m
D. 180 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Assume x is the length of the train and v is the speed x/v = 15 => v = x/15
(x+100)/v = 25 => v = (x+100)/25 Ie, x/15 = (x+100)/25 => 5x = 3x+ 300 => x = 300/2 = 150
26. A train , 800 meter long is running with a speed of 78 km/hr. It crosses a tunnel in 1 minute. What is the length of the tunnel (in meters)? A. 440 m
B. 500 m
C. 260 m
D. 430 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Distance = 800+x meter where x is the length of the tunnel Time = 1 minute = 60 seconds Speed = 78 km/hr = 78×10/36 m/s = 130/6 = 65/3 m/s Distance/time = speed (800+x)/60 = 65/3 => 800+x = 20×65 = 1300 => x = 1300 - 800 = 500 meter
27. Two train each 500 m long, are running in opposite directions on parallel tracks. If their speeds are 45 km/hr and 30 km/hr respectively, the time taken by the slower train to pass the driver of the faster one is A. 50 sec
B. 58 sec
C. 24 sec
D. 22 sec
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Relative speed = 45+30 = 75 km/hr = 750/36 m/s = 125/6 m/s
We are calculating the time taken by the slower train to pass the driver of the faster one .Hence the distance = length of the smaller train = 500 m Time = distance/speed = 500/(125/6) = 24 s
28. Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. If the fast train completely passes a man sitting in the slower train in 5 seconds, the length of the fast train is : A.
19 m
C.
13
2 m 9
7 m 9
B.
27
D.
33 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Relative speed = 40-20 = 20 km/hr = 200/36 m/s = 100/18 m/s Time = 5 s Distance = speed × time = (100/18) × 5 = 500/18 m = 250/9 = length of the fast train
27
7 m = 9
29. Two trains are running in opposite directions in the same speed. The length of each train is 120 meter. If they cross each other in 12 seconds, the speed of each train (in km/hr) is A. 42
B. 36
C. 28
D. 20
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Distance covered = 120+120 = 240 m Time = 12 s Let the speed of each train = v. Then relative speed = v+v = 2v
2v = distance/time = 240/12 = 20 m/s Speed of each train = v = 20/2 = 10 m/s = 10×36/10 km/hr = 36 km/hr
30. A train 108 m long is moving at a speed of 50 km/hr . It crosses a train 112 m long coming from opposite direction in 6 seconds. What is the speed of the second train? A. 82 kmph
B. 76 kmph
C. 44 kmph
D. 58 kmph
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Total distance = 108+112 = 220 m Time = 6s Relative speed = distance/time = 220/6 m/s = 110/3 m/s = (110/3) × (18/5) km/hr = 132 km/hr => 50 + speed of second train = 132 km/hr => Speed of second train = 132-50 = 82 km/hr
31. How many seconds will a 500 meter long train moving with a speed of 63 km/hr, take to cross a man walking with a speed of 3 km/hr in the direction of the train ? A. 42
B. 50
C. 30
D. 28
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Distance = 500m Speed = 63 -3 km/hr = 60 km/hr = 600/36 m/s = 50/3 m/s Time taken = distance/speed = 500/(50/3) = 30 s
I mportan t Formu las - P rofit an d Loss 1.
Cost price (CP) is the price at which an article is purchased.
2.
Selling price (SP) is the price at which an article is sold.
3.
If SP > CP, it is a profit or gain
4.
If CP > SP, it is a loss.
5.
Gain or Profit = SP – CP
6.
Loss = CP - SP
7.
Loss or gain is always reckoned on CP
8.
Profit Percentage (Profit %) =
9.
Loss Percentage (Loss %) =
10.
(CP - SP) Loss × 100 = × 100 CP CP
In the case of a gain or profit,
SP =
11.
(SP - CP) Profit × 100 = × 100 CP CP
(100 + Gain%) 100
× CP
CP =
100 (100 + Gain %)
SP =
(100 − Loss%) × CP 100
CP =
100 × SP (100 − Loss%)
× SP
In the case of a loss,
12.
If an article is sold at a gain of 20%, then SP = 120% of CP
13.
If an article is sold at a loss of 20% then SP = 80% of CP
14.
If an article is sold at a loss of 20% then SP = 80% of CP
15.
If a person sells two items at the same price; one at a gain of x % and another at a loss of x %, then the seller always incurs a loss expressed as
Loss % = (
16.
Common Loss and Gain % 10
)
2
=(
x ) 10
2
If a trader professes to sell his goods at cost price, but uses false weights, then
Gain% = [
Error ( True Value - Error )
× 100] %
1. John buys an old scooter for Rs. 4700 and spends Rs. 800 on its repairs. If he sells the scooter for Rs. 5800, what is his gain percent? A. C.
12% 4 4 % 7
10% 5 D. 5 % 11 B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Cost Price (CP) = 4700 + 800 = Rs. 5500. Selling Price (SP) = Rs. 5800. Gain = (SP) - (CP) = 5800 - 5500= Rs. 300.
Gain% = (
Gain 300 300 60 5 ) × 100 = ( ) × 100 = = =5 CP 5500 55 11 11
2. The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, find out the value of x A. 15
B. 25
C. 18
D. 16
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the Cost Price (CP) of one article = 1 => CP of x articles = x ------------------------------(Equation 1)
CP of 20 articles = 20
Given that cost price of 20 articles is the same as the selling price of x articles => Selling price (SP) of x articles = 20--------------(Equation 2)
Given that Profit = 25%
SP − CP 25 1 )= = ------------( Equation 3) CP 100 4
⇒ (
Substituting equations 1 and 2 in equation 3,
⇒
(20 − x) 1 = x 4
⇒ 80 − 4x = x ⇒ 5x = 80 ⇒x=
80 = 16 5
3. If selling price is doubled, the profit triples. What is the profit percent? A.
100
C.
66
2 3
Hide Answer | Notebook | Discuss
B.
105
D.
120
1 3
Here is the answer and explanation Answer : Option A Explanation : Let the CP = x , SP = y profit = SP –CP = y-x
If SP is doubled, SP = 2y Now Profit = SP –CP = 2y - x
Given that If selling price is doubled, the profit triples
⇒ 2y– x = 3(y − x) ⇒ 2y– x = 3y − 3x ⇒ y = 2x ⇒ P rofit% =
P rofit × 100 CP
=
y− x × 100 x
=
2x– x × 100 x
=
x × 100 x
= 100%
4. In a shop, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, find out approximately what percentage of the selling price is the profit? A. 250%
B. 100%
C. 70%
D. 30%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
\begin{align} &\text{Let the CP = 100} \\ \\ &Profit = \dfrac{320}{100} \times 100 = 320 \\\\ &SP = CP + Profit = 100 + 320 = 420 \\ \\ \\ &\text{If the cost increases by 25%, New CP}
5. A vendor bought bananas at 6 for a rupee. How many for a rupee must he sell to gain 20%? A. 3
B. 4
C. 5
D. 6
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
CP of 6 bananas = 1 Gain = 20% SP of 6 bananas = 1 ×
120 12 = 100 10
⇒ The number of bananas he should sell for Rs.1 for a gain of 20% =
10 10 ∗6= =5 12 2
6. The percentage profit earned by selling an item for Rs. 1920 is equal to the percentage loss incurred by selling the same item for Rs. 1280. At what price should the item be sold to make 25% profit? A. Insufficient Data
B. Rs. 3000
C. Rs. 2000
D. Rs. 2200
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Let CP = x Percentage profit earned by selling an item for Rs. 1920 =
SP − CP CP
=
1920 − x × 100 x
× 100
Percentage loss incurred by selling the same item for Rs. 1280 =
CP − SP CP
=
x − 1280 × 100 x
× 100
Given that Percentage profit earned by selling an item for Rs. 1920 = Percentage loss incurred by selling the same item for Rs. 1280 ⇒
1920 − x x − 1280 × 100 = × 100 x x
⇒
1920 − x x − 1280 = x x
⇒ 1920– x = x– 1280 ⇒ 2x = 1920 + 1280 = 3200 3200 2
⇒x= = 1600
Required Selling Price = CP × = 1600 ×
125 100
125 5 = 1600 × 100 4
= 400 × 5 = 2000
7. An exporter expects a gain of 22.5% on his cost price. If in a week, his sale was of Rs. 392, what was his profit? A. Insufficient Data
B. Rs. 80
C. Rs. 90
D. Rs. 72
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
SP = 392 Gain = 22.5% CP = =
100 100 100 × SP = × 392 = × 392 100 + 22.5 122.5 100 + Gain%
1000 40 40 × 392 = × 392 = × 56 = 40 × 8 = 320 1225 49 7
Profit = SP − CP = 392 − 320 = 72
8. A man buys a scooter for Rs. 1400 and sells it at a loss of 15%. What is the selling price of the scooter? A. Rs. 1240
B. Rs. 1190
C. Rs. 1090
D. Rs. 1130
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
SP = 1400 ×
85 = 14 × 85 = 1190 100
9. Murali purchased 20 dozens of toys at the rate of Rs. 375 per dozen. He sold each one of them at the rate of Rs. 33. Find out his profit percentage. A. 3.5
B. 5.6
C. 4.1
D. 3.4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
CP of 1 toy =
375 12
SP of 1 toy = 33 Profit = SP − CP = 33 −
375 12
P rofit Profit% = × 100 = CP
= (33 ×
375 ) 12 375 12
(33 −
× 100
12 4 − 1) × 100 = (33 × − 1) × 100 375 125
7 7 28 3 × 100 = × 4 = = 5 % = 5.6% 125 5 5 5
=
10. Some items were bought at 6 items for Rs. 5 and sold at 5 items for Rs. 6. What is the gain percentage?
1 3
A.
44%
B.
33
C.
31
2 3
D.
30%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
CP of 1 item =
5 6
SP of 1 item =
6 5
6 5 − SP − CP 36 5 6 Gain% = × 100 = × 100 = ( − 1) × 100 CP 5 25 6 11 × 100 = 11 × 4 = 44 25
11. On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. What is the cost price of a ball? A. Rs. 43
B. Rs. 60
C. Rs. 55
D. Rs. 34
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B
Explanation : Loss = (CP of 17 balls) - (SP of 17 balls) = (CP of 17 balls) 720
Given that Loss = (CP of 5 balls)
=> (CP of 17 balls) - 720 = (CP of 5 balls)
=> (CP of 17 balls) - (CP of 5 balls) = 720
=> CP of 12 balls = 720
=> CP of 1 ball =
720 = 60 12
12. When an item is sold for Rs. 18,700, the owner loses 15%. At what price should that plot be sold to get a gain of 15%? A. Rs. 25100
B. Rs. 24200
C. Rs. 25300
D. Rs. 21200
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
When an item is sold for Rs. 18,700, the owner loses 15% ⇒ SP = 18700 and Loss = 15% ⇒ CP =
100 100 100 × SP = × 18700 = × 18700 100 − 15 85 100 − Loss%
To get a gain of 15%, SP =
100 + Gain% 100 + 15 115 × CP = × CP = × CP 100 100 100
115 100 115 23 × × 18700 = × 18700 = × 18700 100 85 85 17
=
= 23 × 1100 = 25300
13. 100 oranges were bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen. What is the percentage of profit or loss?
2 % Loss 7 2 C. 14 % Profit 7 A.
11
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
1 % Profit 7 2 D. 14 % Loss 7 B.
11
CP of 100 oranges = 350 ⇒ CP of 1 orange =
350 = 3.5 100
SP of 12 oranges = 48 ⇒ SP of 1 orange =
Profit% = =
48 =4 12
SP − CP 4 − 3.5 .5 × 100 = × 100 = × 100 CP 3.5 3.5
1 100 2 × 100 = = 14 7 7 7
14. A shopkeeper sells one radio for Rs. 840 at a gain of 20% and another for Rs. 960 at a loss of 4%. What is his total gain or loss percentage?
5% 12 C. 6 % 17
6% 15 D. 5 % 17
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
SP of 1st Radio = 840 Gain = 20% CP of 1st Radio =
100 (100 + Gain%)
× SP =
100 × 840 = 100 × 7 = 700 (100 + 20)
× SP =
100 × 960 = 100 × 10 = 1000 (100 − 4)
SP of 2nd Radio = 960 Loss = 4% CP of 2nd Radio =
100 (100 − Loss%)
Total CP = 700 + 1000 = 1700 Total SP = 840 + 960 = 1800 Total Gain = SP − CP = 1800 − 1700 = 100 Gain% =
Gain 100 100 15 × 100 = × 100 = =5 CP 1700 17 17
15. A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. What is his profit percentage? A. 6%
B. 5%
C. 4%
D. 7%
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option B Explanation :
CP of 1st variety rice = 20 CP of 2nd variety rice = 36 CP of the 56 kg rice mixture = (26 × 20 + 30 × 36) = 520 + 1080 = 1600
SP of the 1 kg rice mixture = 30 SP of the 56 kg rice mixture = 30 × 56 = 1680
Gain = SP − CP = 1680 − 1600 = 80 Gain% =
Gain 80 100 × 100 = × 100 = = 5% CP 1600 20
16. If a material is sold for Rs.34.80, there is a loss of 25%. Find out the cost price of the material? A. Rs.46.40
B. Rs.44
C. Rs.42
D. Rs.47.20
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
SP = 34.80 Loss = 25% CP =
=
100 (100 − Loss%)
× SP =
100 100 × 34.80 = × 34.80 75 (100 − 25)
4 × 34.80 = 4 × 11.60 = 46.40 3
17. A fruit seller sells apples at the rate of Rs.9 per kg and thereby loses 20%. At what price per kg, he should have sold them to make a profit of 5%? A. 11.32
B. 11
C. 12
D. 11.81
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
SP = 9 Loss = 20% CP =
=
100 (100 − Loss%)
× SP =
5 ×9 4
to make a profit of 5%, SP = =
100 100 ×9 = ×9 80 (100 − 20)
(100 + 5) 100 + Gain% × CP = × CP 100 100
105 5 105 5 21 5 21 1 189 × ×9 = × ×9 = × ×9 = × ×9 = = 11.81 100 4 100 4 20 4 4 4 16
18. A trader gives 12% additional discount on the discounted price, after giving an initial discount of 20% on the labeled price of an item. The final sale price of the item is Rs.704. Find out the labeled price? A. 1000
B. 2000
C. 1200
D. 920
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Let the labeled price = x SP = 704 Initial Discount = 20% Price after initial discount = x ×
80 100
Additional discount = 12% Price after additional discount = x ×
80 88 × 100 100
But Price after additional discount = SP = 704 ⇒ x×
80 88 × = 704 100 100
⇒ x×
4 22 × = 704 5 25
⇒ x = 704 ×
25 5 25 × = 176 × ×5 22 4 22
= 8 × 25 × 5 = 40 × 25 = 1000
19. A man sells two houses at the rate of Rs.1.995 lakhs each. On one he gains 5% and on the other, he loses 5%. What is his gain or loss percent in the whole transaction? A. 0.25%.
B. .3%
C. .4%
D. .5%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : If a person sells two items at the same price; one at a gain of x % and another at a loss of x %, then the seller always incurs a loss expressed as
Loss% = (
Common Loss and Gain% 10
)
2
So in this case, he will have a loss. Loss% = ( 1 =( ) 2
2
=(
x ) 10
5 ) 10
2
2
1 = ( ) = .25 4
20. John purchased a machine for Rs. 80,000. After spending Rs.5000 on repair and Rs.1000 on transport he sold it with 25% profit. What price did he sell the machine? A. Rs.107000.
B. Rs.107500.
C. Rs.108500.
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
CP = 80, 000 + 5000 + 1000 = 86000 P rofit = 25% SP = =
(100 + Gain%) 100
× CP =
(100 + 25) × 86000 100
125 5 × 86000 = × 86000 = 5 × 21500 = 107500 100 4
21. By selling an item for Rs.15, a trader loses one sixteenth of what it costs him. The cost price of the item is A. Rs.14
B. Rs.15
C. Rs.16
D. Rs.17
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
SP = 15 Loss =
CP 16
Loss = CP − SP = CP − 15 ⇒
CP = CP − 15 16
⇒
15 CP = 15 16
⇒
CP =1 16
⇒ CP = 16
22. A shopkeeper sells his goods at cost price but uses a weight of 800 gm instead of kilogram weight. What is his profit percentage? A. 18%
B. 40%
C. 25%
D. 20%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : If a trader professes to sell his goods at cost price, but uses false weights, then
Gain% = [
Error (True Value − Error)
So here profit percentage = [
=[
× 100] %
200 × 100] % (1000 − 200)
200 × 100] % = 25% 800
23. Prasanth bought a car and paid 10 % less than the original price. He sold it with 30% profit on the price he had paid. What percentage of profit did he earn on the original price? A. 17%
B. 16%
C. 18%
D. 14%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A
Explanation :
Let the original price = 100 Then the price at which he purchased (CP) = 90% of 100 = 90 P rofit = 30% SP = =
(100 + P rofit%) 100
× CP =
(100 + 30) × 90 100
130 × 90 = 13 × 9 = 117 100 (117 − 100) × 100 = 17% 100
Required % =
24. If a seller reduces the selling price of an item from Rs.400 to Rs.380, his loss increases by 2%. What is the cost price of the item? A. 1000
B. 800
C. 1200
D. 1100
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Initial Loss% =
CP − 400 × 100 CP
If the SP is reduced from 400 to 380, Loss% =
CP − 380 × 100 CP
It is given that If the SP is reduced from 400 to 380, Loss% increases by 2 ⇒
CP − 380 CP − 400 × 100 − × 100 = 2 CP CP
⇒ (CP − 380) − (CP − 400) = ⇒ 20 =
2 × CP 100
2 × CP 100
⇒ CP =
20 × 100 = 1000 2
25. A trader keeps the marked price of an item 35% above its cost price. The percentage of discount allowed to gain 8% is A. None
B. 30%
C. 25%
D. 20%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Let the Cost Price (CP) = 100 Then Market Price =
100 × 135 = 135 100
If he wants to gain 8%, SP = Discount % =
(100 + Gain%) 100
× CP =
(100 + 8) × 100 = 108 100
(135 − 108) 2700 × 100 = = 20 135 135
26. Arun bought a computer with 15% discount on the labeled price. He sold the computer for Rs.2880 with 20% profit on the labeled price. At what price did he buy the computer? A. Rs.3000
B. Rs.2080
C. Rs.2040
D. Rs.2000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
SP = 2880
Arun sold the computer with 20% profit on the Labeled Price ⇒ 2880 =
120 × Labeled Price 100
⇒ Labeled Price =
(2880 × 100) 28800 9600 = = = 2400 120 12 4
Arun bought a computer with 15% discount on the labeled price Labeled Price × 85 100
=> CP =
=
2400 × 85 = 24 × 85 = 2040 100
27. An item was sold for Rs.27.50 with a profit of 10%. If it was sold for Rs.25.75, what would have been the percentage of profit or loss? A. 3%
B. 2%
C. 4%
D. 5%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : SP = 27.5 Profit = 10%
CP =
100 (100 + P rofit%)
× SP =
100 2750 × 27.5 = = 25 110 (100 + 10)
If the item was sold for Rs.25.75,
Gain% =
(25.75 − 25) .75 × 100 = × 100 = 3% 25 25
28. If selling price of an article is Rs. 250, profit percentage is 25%. Find the ratio of the cost price and the selling price A. 5: 3
B. 3 : 5
C. 4 : 5
D. 5 : 4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : SP = 250 Profit = 25%
CP =
100 (100 + Profit%)
× SP =
100 100 × 250 == × 250 = 200 125 (100 + 25)
Required Ratio = 200 : 250 = 4:5
29. A material is purchased for Rs. 600. If one fourth of the material is sold at a loss of 20% and the remaining at a gain of 10%, Find out the overall gain or loss percentage?
1 % 2 C. 3% A.
1 % 2 D. 2%
3
B.
2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : CP = 600
Price Received by selling one fourth of the material at a loss of 20%
=
1 80 × 600 × = 120 4 100
Price Received by remaining material at a gain of 10%
=
3 110 × 600 × = 495 4 100
SP = 120 + 495 = 615
Profit% =
(SP − CP ) (615 − 600) 15 15 5 1 × 100 = × 100 = × 100 = = =2 % CP 600 600 6 2 2
30. A shopkeeper buys pencils at 9 for Rs. 16 and sells them at 11 for Rs. 22. Find out his loss or gain percentage?
1 % 2
A.
12
C.
14%
B.
12%
D.
11
2 % 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
16 9 22 SP of 1 pencil = =2 11
CP of 1 pencil =
(SP − CP ) Profit% = × 100 = CP
(2 −
16 ) 9 16 9
2 100 1 9 × 100 = × 100 = = 12 % 16 8 2 9
31. A reduction of 10% in the price of a pen enabled a trader to purchase 9 more for Rs.540. What is the reduced price of the pen? A. 8
B. 6
C. 5
D. 4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Let price of a pen = x Number of pens purchased for Rs.540 =
540 x
Price of a pen after the reduction of 10% =
90x 9x = 100 10
Number of pens purchased for Rs.540 with reduced price =
Additional pens purchased =
540 600 = 9x x 10
600 540 60 − = x x x
Given that Additional pens purchased = 9 ⇒
60 =9 x
x=
60 9
Reduced price of the pen =
60 90 × =6 9 100
32. Sunil purchases two books at Rs.300 each. He sold one book 10% gain and other at 10% loss. What is the total loss or gain in percentage? A. 10% gain
B. 1% loss
C. No loss or no gain
D. 1% gain
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : CP of two books = 300 + 300 = 600
SP of two books = 300 ×
110 90 300 + 300 × = × (110 + 90) = 3 × 200 = 600 100 100 100
Total SP = Total CP Hence No loss or no gain
I mportan t Formu las - Races an d Games of Skill 1. Race A race is a contest of speed in running, riding, driving, sailing, rowing etc over a particular distance. 2. Race Course Race course is the ground or path on which contests are conducted. 3. Starting Point Starting Point is the point from which a race starts. 4. Winning Point (or Goal) Winning Point (or Goal) is the point where a race finishes. 5. Dead-heat Race A race is said to be a dead-heat race if all the persons contesting the race reach the winning point (goal) exactly at the same time. 6. Winner Winner is the person who first reaches the winning point. 7. Let A and B be tw o contestants in a race. Lets ex am ine som e of the general statem ents and their m athem atical interpretations. statem ents
m athem atical interpretations
A beats B by t seconds
A finishes the race t seconds before B finishes.
A gives B a start of t seconds
A starts t seconds after B starts from the same starting point.
A gives B a start of x metres
While A starts from the starting point, B starts x meters ahead from the same starting point at the same time. To cover a race of 100 metres in this case, A will have to cover 100 metres while B will have to cover only (100 - x) metres.
Game of 100
A game in which the participant who scores 100 points first wins.
In a game of 100, A can give B 20 points
While A scores 100 points, B scores only 100-20=80 points.
If A is n times as fast as B and A gives B a start of x meters, then the length of the race course , so 8.
that A and B reaches the winning post at the same time
= x(
n ) n−1
metres
x metre race in t 1 seconds and B in t 2seconds, where t 1 < t 2, then A beats B by a x distance (t 2 − t 1 ) metres t2 If A can run
9.
1. A is 2 1⁄3 times as fast as B. If A gives B a start of 80 m, how long should the race course be so that both of them reach at the same time? A. 170 metre
B. 150 metre
C. 140 metre
D. 160 metre
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : -------------------------------------------------------------------Solution 1 -------------------------------------------------------------------Speed of A : Speed of B = 7⁄3 : 1 = 7 : 3 It means, in a race of 7 m, A gains (7-3)=4 metre
If A needs to gain 80 metre, race should be of
7 × 80 = 140 metre 4
-------------------------------------------------------------------Solution 2 --------------------------------------------------------------------
If A is n times as fast as B and A gives B a start of x meters, then the length of the race course , so that A and B reaches the winning post at the same time = x (
n ) n−1
metres
[Read more ...]
Here, n = 2
1 7 = , x = 80 3 3
Length of the race course = x ( . = 80
.
7 3
7 −1 3
= 80 (
n ) n−1
7 7 ) = 80 × ( ) = 20 × 7 = 140 metre 7−3 4
2. A can run 224 metre in 28 seconds and B in 32 seconds. By what distance A beat B? A. 36 metre
B. 24 metre
C. 32 metre
D. 28 metre
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : -------------------------------------------------------------------------------Solution 1 -------------------------------------------------------------------------------Clearly, A beats B by 4 seconds Now find out how much B will run in these 4 seconds
Distance 224 28 = = = 7 m/s Time taken by B 32 4 Distance covered by B in 4 seconds = Speed × time = 7 × 4 = 28 metre
Speed of B =
i.e., A beat B by 28 metre -------------------------------------------------------------------------------Solution 2 --------------------------------------------------------------------------------
If A can run where
t1
x metre race in t 1 seconds and B in t 2 seconds,
In a 100 m race, while A starts from the starting point, B starts 10 meters ahead from the same starting point at the same time. "A can give C 28 m" => In a 100 m race, while A starts from the starting point, C starts 28 meters ahead from the same starting point at the same time. i.e., in a 100-10=90 m race, B can give C 28-10=18 m
=> In a 100 m race, B can give C
18 × 100 = 20 m 90
----------------------------------------------------------------Solution 2 ----------------------------------------------------------------"A can give B 10 m". => In a 100 m race, A will have to cover 100 m while B will have to cover only 100-10=90 m. Similarly, "A can give C 28 m" => In a 100 m race, A will have to cover 100 m while C will have to cover only 100-28 = 72 m. => When B covers 90 metre, C covers 72 m
=> when B covers 100 metre, C covers
72 × 100 = 80 metre 90
i.e.,in a 100 m race, B can give C 100-80=20 metre
4. At a game of billiards, A can give B 15 points in 60 and A can give C to 20 points in 60. How many
points can B give C in a game of 90? A. 22 points
B. 20 points
C. 12 points
D. 10 points
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : -----------------------------------------------------Solution 1 ------------------------------------------------------"A can give 15 points to B in 60". => In a game of 60, A starts with 0 points where B starts with 15 points "A can give 20 points to C in 60". => In a game of 60, A starts with 0 points where C starts with 20 points ie, in a game of 60-15-45, B can give C 20-15=5 points => ie, in a game of 90, B can give C 5×2=10 points -----------------------------------------------------Solution 2 ------------------------------------------------------"A can give 15 points to B in 60". => In a game of 60, A will have to get 60 points while B will have to get only 60-15=45 points "A can give 20 points to C in 60". => In a game of 60, A will have to get 60 points while C will have to get only 60-20=40 points i.e., when B scores 45 points, C scores 40 points..
when B scores 90 points, C scores
40 × 90 = 80 points 45
i.e.,in a game of 90, B can give C 90-80=10 points
5. In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m. Then, A wins by: A. 60 m
B. 20 m
C. 40 m
D. 10 m
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option B Explanation : To reach the winning point , A will have to cover a distance of 500 - 140= 360 metre ratio of the speeds of two contestants A and B is 3 : 4 i.e., when A covers 3 metre, B covers 4 metre.
=> When A covers 360 metre, B covers
4 × 360 = 480 metre. 3
Remaining distance B have to cover = 500-480= 20 metre => A wins by 20 metre
6. In a race of 200 m, A can beat B by 31 m and C by 18 m. In a race of 350 m, C will beat B by: A. 20.25 m
B. 21.5 m
C. 22.75 m
D. 25 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : It means, when A convers 200 m, B covers only (200-31)=169 m and C covers only (200-18)=182 m => When C covers 182 m, B covers 169 m
=> When C covers 350 m, B covers
169 × 350 = 325 m 182
Hence, C beats B by 350-325 = 25 metre
7. A can run 22.5 m while B runs 25 m. In a kilometre race B beats A by: A. 102 m
B. 100 m
C. 75 m
D. 112.5 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B
Explanation : when B runs 25 m, A runs 22.5 m
=> When B runs 1000 metre, A runs
22.5 × 1000 = 900 m 25
=> When B runs 1 kilometre, A runs 900 m Hence, in a kilometre race, B beats A by (1 kilometre - 900 metre) = (1000 metre - 900 metre) = 100 metre
8. In a 100 m race, A can beat B by 25 m and B can beat C by 4 m. In the same race, A can beat C by: A. 21 m
B. 28 m
C. 26 m
D. 29 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : When A covers 100 m, B covers (100-25)=75 m When B covers 100 m, C covers (100-4)=96 m
=> When B covers 75 metre, C covers
96 × 75 = 72 m 100
i.e., When A cov ers 100 m , B covers 75 m and C cov ers 72 m => In a 100 m race, A beat C by (100-72)=28 m
9. In a 300 m race A beats B by 22.5 m or 6 seconds. B's time over the course is: A. None of these
B. 80 sec
C. 76 sec
D. 86 sec
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : B runs 22.5 m in 6 seconds
=> B runs 300 m in
6 × 300 = 80 seconds 22.5
i.e., B's time over the course = 80 seconds
10. A and B take part in 100 m race. A runs at 5 kmph. A gives B a start of 8 m and still beats him by 8 seconds. The speed of B is: A. 4.4 kmph
B. 4.25 kmph
C. 4.14 kmph
D. 5.15 kmph
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
5 25 = m/s 18 18 distance 100 Time taken by A to cover 100 m = = speed 25 ( ) 18
Speed of A = 5 kmph = 5 ×
= 72 seconds
It is given that, A gives B a start of 8 m and still beats him by 8 seconds. => B takes (72+8)=80 seconds to cover (100-8)=92 metre
Speed of B =
distance 92 92 18 = m/s = × kmph = 4.14 kmph time 80 80 5
11. A runs 1 2⁄3 times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time? A. 200 m
B. 270 m
C. 300 m
D. 160 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Speed of A : Speed of B = 5⁄3 : 1 = 5 : 3 i.e., In a race of 5 m, A runs 5 m and B runs 3 m i.e., A gains 2 m over B in a race of 5 m
=> A gains 80 m over B in a race of
5 × 80 = 200 m 2
Hence, winning point should be 200 m away from the starting point.
12. In a game of 100 points, A can give B 20 points and C 28 points. Then, B can give C: A. 40 points
B. 10 points
C. 14 points
D. 8 points
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : In a game of 100 points, A scores 100 points, B scores (10020)=80 points and C scores (100-28)=72 points i.e., when B scores 80 points, C scores 72 points
=> When B scores 100 points, C scores
72 × 100 = 90 points 80
i.e., In a game of 100 points, B can give C (100-90)=10 points
13. In 100 m race, A covers the distance in 36 seconds and B in 45 seconds. In this race A beats B by: A. 9 m
B. 25 m
C. 22.5 m
D. 20 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : In 100 m race, A covers the distance in 36 seconds and B in 45 seconds. Clearly, A beats B by (45-36)=9 seconds
Speed of B =
Distance 100 = m/s Time 45
Distance Covered by B in 9 seconds = Speed × Time =
100 × 9 = 20 metre 45
i.e., A beats B by 20 metre
14. In a 100 m race, A beats B by 10 m and C by 13 m. In a race of 180 m, B will beat C by: A. 5.4 m
B. 5 m
C. 4.5 m
D. 6 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : While A runs 100 m, B runs (100-10)=90 m and C runs (10013)=87 m i.e., when B runs 90 m, C runs 87 m
=> when B runs 180 m, C runs
87 × 180= 174 m 90
Hence, in a 180 m race, B will beat C by (180-174)=6 m
15. In a 200 metres race A beats B by 35 m or 7 seconds. A's time over the course is: A. 33 sec
B. 40 sec
C. 47 sec
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : B runs 35 m in 7 sec.
=> B runs 200 m in
7 × 200 = 40 seconds 35
Since A beats B by 7 seconds, A runs 200 m in (40-7) = 33 seconds Hence, A's time over the course = 33 seconds
16. A, B and C are the three contestants in one km race. If A can give B a start of 40 metres and A can give C a start of 64 metres. How many metres start can B give C? A. None of these
B. 20 m
C. 25 m
D. 35 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : While A covers 1000 m, B covers (1000-40)=960 m and C covers (1000-64)=936 m i.e., when B covers 960 m, C covers 936 m
When B covers 1000 m, C covers
936 × 1000 = 975 m 960
i.e., B can give C a start of (1000-975) = 25 m
17. In a game of 90 points A can give B 15 points and C 30 points. How many points can B give C in a game of 100 points? A. 140
B. 20
C. 300
D. 50
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : While A scores 90 points, B scores (90-15)=75 points and C scores (90-30)= 60 points i.e., when B scores 75 points, C scores 60 points
=> When B scores 100 points, C scores
60 × 100 = 80 points 75
i.e., in a game of 100 points, B can give C (100-80)=20 points
18. In a 100 metres race. A runs at a speed of 2 metres per seconds. If A gives B a start of 4 metres and still beats him by 10 seconds, find the speed of B. A. 1.6 m/sec.
B. 4 m/sec.
C. 2.6 m/sec.
D. 1 m/sec.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A
Explanation : Speed of A = 2 m/s
Time taken by A to run 100 m
distance 100 = = 50 seconds speed 2
A gives B a start of 4 metres and still A beats him by 10 seconds => B runs (100-4)=96 m in (50+10)=60 seconds
Speed of B =
distance 96 = = 1.6 m/s time 60
19. P runs 1 km in 3 minutes and Q in 4 minutes 10 secs. How many metres start can P give Q in 1 kilometre race, so that the race may end in a dead heat? A. 210 metre
B. 180 metre
C. 220 metre
D. 280 metre
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : P run 1 km in 3 minutes Q run 1 km in 4 minutes 10 secs
25 minutes 6 6 18 => Q runs (1 × × 3) = = 0.72 km in 3 minutes 25 25
=> Q runs 1 km in
Hence, in a 1 km race, P can give Q (1 - 0.72)=0.28 km = 280 metre
20. A runs 1 3⁄8 times As fast as B. If A gives B a start of 90 m and they reach the goal at the same time. The goal is at a distance of A. 330 m
B. 440 m
C. 120 m
D. 280 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A
Explanation : Speed of A : Speed of B = 11⁄8: 1 = 11 : 8 Let Speed of A = 11k and Speed of B = 8k Let the distance to the goal = x
Time taken by A = time taken by B ⇒
Distance travelled by A Speed of A
⇒
x x − 90 = 11k 8k
⇒
x x − 90 = 11 8
=
Distance travelled by B Speed of B
⇒ 8x = 11x − 990 ⇒ 3x = 990 ⇒ x = 330 Distance to the goal = 330 metre
21. Two boys A and B run at 4 1⁄5 and 8 km an hour respectively. A having 150 m start and the course being 1 km, B wins by a distance of A. 325 m
B. 60 m
C. 120 m
D. 275 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : A has a start of 150 m. So A has to run 1000-150=850 metre while B has to run 1000 metre.
Speed of A = 4
1 21 21 5 21 7 km/hr = km/hr = × = = m/s 5 5 5 18 18 6
Speed of B = 8 km/hour = 8 ×
5 20 = m/s 18 9
Time taken by B to travel 1000 metre =
distance 1000 = = 450 second Speed of B 20 ( ) 9
Distance travelled by A by this time = time × speed of A = 450 ×
7 = 525 metre 6
Hence, A win by 850-525=325 metre
22. In a 100 metres race, A can beat B by 10 metres and B can beat C by 5 metres. In the same race, A can beat C by A. 14.5 metre
B. 14 metre
C. 15.5 metre
D. 15 metre
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : While A runs 100 metre, B runs 100-10=90 metre While B runs 100 metre, C runs (100-5)=95 metre
=> While B runs 90 metre, C run
95 95 × 9 × 90 = = 85.5 metre 100 10
ie, when A run 100 metre, B run 90 metre and C run 85.5 metre Hence, A beat C by (100-85.5)= 14.5 metre
23. X, Y and Z are the three contestants in one km race. If X can give Y a start of 52 metres and X can also give Z a start of 83 metres, how many metres start Y can give Z? A. 33.3 m
B. 33 m
C. 32 m
D. 32.7 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
While X runs 1000 metre, Y runs (1000-52)=948 metre and Z runs (1000-83)=917 metre i.e., when Y runs 948 metre, Z runs 917 metre
=> When Y runs 1000 metre, Z runs
917 × 1000 = 967.30 metre 948
i.e., Y can give Z (1000-967.30) = 32.7 metre
24. In a race of 600 m. A can beat B by 60 m and in a race of 500 m. B can beat C by 50 m. By how many metres will A beat C in a race of 400 m? A. 76 metre
B. 74 metre
C. 72 metre
D. 78 metre
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : While A runs 600 m, B runs (600-60)=540 metre
=>While A runs 400 m, B runs
540 × 400 = 360 metre 600
While B runs 500 m, C runs (500-50)=450 metre
=>While B runs 360 m, C runs
450 × 360 = 324 metre 500
i.e., When A runs 400 metre, B runs 360 metre and C runs 324 metre. Hence, A beats C by (400-324)=76 metre in a race of 400 m
25. A can run 220 metres in 41 seconds and B in 44 seconds. By how many seconds will B win if he has 30 metres start? A. 8 sec
B. 4 sec
C. 2.5 sc
D. 3 sec
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : B has a start of 30 metre
=> A has to run 220 metre and B has to run (220-30)=190 metre Given that A takes 41 seconds to cover this 220 metre B takes 44 seconds to cover 220 metre
44 seconds to cover 1 metre 220 44 4 => B takes × 190 = × 190 = 38 seconds to cover 190 metre 220 20
=> B takes
i.e., A beats B by (41-38)= 3 seconds
26. In one km race A beats B by 4 seconds or 40 metres. How long does B take to run the kilometer? A. 112 sec
B. 110 sec
C. 101 sec
D. 100 sec
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : This means, B takes 4 seconds to run 40 metres
4 1 = seconds to run 1 metre 40 10 1 => B takes × 1000 = 100 seconds to run 1000 metre 10
=> B takes
27. In a game, A can give B 20 points, A can give C 32 points and B can give C 15 points. How many points make the game? A. 120 points
B. 90 points
C. 80 points
D. 100 points
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let x points make the game A can give B 20 points, A can give C 32 points => When A scores x points, B scores (x -20) points and C scores (x -32) points
B can give C 15 points => When B scores x points, C scores (x-15) points
(x − 15) points x (x − 20)(x − 15) => When B scores (x-20) point, C scores x
=> When B scores 1 point, C scores
i.e., (x − 32) =
points
(x − 20)(x − 15) x
⇒ x(x − 32) = (x − 20)(x − 15) ⇒ x 2 − 32x = x 2 − 35x + 300 ⇒ −32x = −35x + 300 ⇒ 3x = 300 ⇒ x = 100 i.e., 100 points make the game
28. In a game A can give B 20 points in 60 and C 18 points in 90. How many points can C give B in a game of 120? A. 20 points
B. 22 points
C. 18 points
D. 40 points
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : A can give C 18 points in 90 => While A scores 90 points, C scores (90-18)=72 points
90 points, C scores 1 point. 72 90 90 => While A scores × 120 = × 10 = 150 points, C scores 120 point. 72 6
=> While A scores
A can give B 20 points in 60 => While A scores 60 points, B scores (60-20)=40 points
40 2 = points 60 3 2 => While A scores 150 points, B scores × 150 = 100 points 3
=> While A scores 1 point, B scores
i.e., while C scores 120 points, B scores 100 points Hence, in a 120 race, C can give B (120-100)=20 points
29. In a km race A can beat B by 100 m and B can beat C by 60 m. In the same race A can beat C by A. 144 m
B. 164 m
C. 144 m
D. 154 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : In a km race A can beat B by 100 m => While A run 1000 m, B run (1000-100)=900 m In a km race B can beat C by 60 m => While B runs 1000 m, C runs (1000-60)=940 m
940 m 1000 940 => While B runs 900 m, C runs × 900 = 94 × 9 = 846 m 1000
=> While B run 1 m, C run
i.e., while A run 1000 m, C run 846 m Hence, A can beat C by (1000-846)= 154 m
30. In a flat race, A beats B by 15 metres and C by 29 metres. When B and C run over the course together, B wins by 15 metres. Find the length of the course A. 225 m
B. 125 m
C. 220 m
D. 256 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let x be the length of the course
A beats B by 15 metres and C by 29 metres => When A r uns x m etre, B runs (x-15) m etre and C runs (x-29) m etre When B and C run over the course together, B wins by 15 metres => when B runs x metre, C run (x-15) metre
=> when B runs 1 metre, C run
x − 15 metre x
(x − 15) x − 15 => when B runs (x-15) metre, C run × (x − 15) = x x (x − 15) i.e., (x − 29) = x
2
⇒ x 2 − 29x = x 2 − 30x + 225 ⇒ −29x = −30x + 225 ⇒ x = 225 Page 1
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Page 2
2
metre
31. A and B run a km race. If A gives B a start of 50 m, A wins by 14 sec and if A gives B a start of 22 sec, B wins by 20 m. The time taken by A to run a km is A. 125 sec
B. 120 sec
C. 105 sec
D. 100 sec
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let time taken by A to run 1000 m = a and time taken by B to run 1000 m = b If A gives B a start of 50 m, A wins by 14 sec distance travelled by A = 1000 distance travelled by B = (1000-50)=950 Time taken by B to run 950 m - Time taken by A to run 1000 m = 14 sec
950 b − a = 14 1000 95 b − a = 14---(Equation 1) 100 If A gives B a start of 22 sec, B wins by 20 m i.e., A starts 22 seconds after B starts from the same starting point. i.e., here B runs for b seconds where A runs for (b-22) seconds => Distance Covered by B in b seconds - Distance Covered by A in (b-22) seconds = 20
=> 1000 −
1000 × (b − 22) = 20 a
1000 × (b − 22) = 980 a 1000b − 22000 = 980a 100b − 2200 = 98a 50b − 1100 = 49a 50b − 49a = 1100---(Equation 2) (Equation 1) × 49 ⇒
95 × 49 b − 49a = 14 × 49 100
⇒ 46.55b − 49a = 686---(Equation 3) (Equation 2) - (Equation 3) ⇒ 3.45b = 414 ⇒b=
414 = 120 seconds 3.45
Substituting this value of b in Equation 1 ⇒
95 × 120 − a = 14 100
⇒ 114 − a = 14 ⇒ a = 114 − 14 = 100 i.e., The time taken by A to run a km = 100 seconds
32. A can run a kilometre in 4 min 54 sec and B in 5 min. How many metres start can A give B in a km race so that the race may end in a dead heat? A. 25 m
B. 20 m
C. 15 m
D. 10 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : i.e., in this race, A will win by (5 min - 4 min 54 sec) = (300 sec - 294 sec) = 6 sec B covers 1000 m in 5 min
=> B covers 1000 m in 300 sec
1000 10 = m in 1 sec 300 3 10 => B covers × 6 = 20 m in 6 sec 3
=> B covers
i.e., A can give B a start of 20 metre so that the race will end in a dead heat.
33. A runs 1 1⁄3 as fast as B. If A gives B a start of 30 metres. How far must be the wining post, so that the race ends in a dead heat? A. 100 m
B. 110 m
C. 140 m
D. 120 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the distance to the winning point = x Speed of A : Speed of B = 4/3 : 1 Let Speed of A = 4k/3 Speed of B = k A gives B a start of 30 metres => A run x metres and B run (x-30)metres
Time taken by A =
Distance run by A Speed of A
Time taken by B =
Distance run by B Speed of B
x 4k ( ) 3 x − 30 = k
=
For the race to be a a dead-heat race, Time taken by A = Time taken by B
⇒
⇒
x 4k ( ) 3 x 4 ( ) 3
=
x − 30 k
= x − 30
⇒ 3x = 4x − 120 ⇒ x = 120 i.e., the distance to the winning point = 120 metre
34. A can run 15 metres while B runs 20 metres. In a km race B beats A by A. 200 metres
B. 100 metres
C. 220 metres
D. 250 metres
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : A can run 15 metres while B runs 20 metres
15 metres while B runs 1 metre 20 15 A can run × 1000 = 750 metres while B runs 1000 metres 20
A can run
i.e., in a 1 km race, B beats A by (1000-750)=250 metres
35. A and B run a 5 km race on a round course of 400 m. If their speeds be in the ratio 5 : 4, how often does the winner pass the other? A. 5 times
B. 4 1⁄2 times
C. 2 1⁄2 times
D. 4 times
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : speeds of A and B are in the ratio 5 : 4
=> when A covers 5 metre, B covers 4 metre i.e., when A covers 5 metre, A gains 1 metre over B => when A covers 5 ×400 = 2000 metre, A gains 400 metre over B. i.e., when A covers 2000 metre, the winner pass the other 1 time.
=> when A covers 5000 metre, the winner pass the other
5000 1 = 2 times 2000 2
36. In a 100 m race, A runs at 6 km/hr. A gives B a start of 4 m and still beats him by 4 sec. Find the speed of B. A. 4.6 km/hr
B. 5.6 km/hr
C. 4.4 km/hr
D. 5.4 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
5 5 = m/s 18 3 distance 100 Time taken by A = = = 60 sec speed 5 ( ) 3
Speed of A = 6 km/hr = 6 ×
A gives B a start of 4 m i.e., B has to cover (100-4)=96 metre A beats B by 4 sec => Time taken by B to run the race = 60 + 4 = 64 sec
Speed of B = Page 1
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Distance 96 96 18 = m/s = × = 5.4 km/hr time 64 64 5 Page 2
1. Insert the missing number. 12, 25, 49, 99, 197, 395, (...) A. 789
B. 1579
C. 722
D. 812
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Each number is twice the previous one with 1 added or subtracted alternatively.
2.Insert the missing number. 34, 7, 37, 14, 40, 28, 43, (...) A. 31
B. 42
C. 46
D. 56
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : We have two series here 34, 37, 40, 43, ... (Increase by 3) 7, 14, 28, ... (Multiply by 2) Hence, next term is 28 × 2 = 56
3. Find the missing number. 1, 4, 9, 16, 25, 36, 49, (....) A. 64
B. 54
C. 56
D. 81
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : The series is 12, 22, 32, 42, 52, 62, 72, ... Hence, next term = 82 = 64
4. Insert the missing number. 2, 7, 10, 22, 18, 37, 26
A. 42
B. 52
C. 46
D. 62
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : There are two series here 2, 10, 18, 26, ... (Increase by 8) 7, 22, 37, ... (Increase by 15) Hence, next term is 37+15 = 52
5.
4, 12, 48, 240, 1440, (...)
A. 7620
B. 10080
C. 6200
D. 10020
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Go on multiplying the given numbers by 3, 4, 5, 6, 7
6.
2, 3, 6, 0, 10, -3, 14, (...)
A. 6
B. 2
C. -2
D. -6
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : There are two series 2, 6, 10, 14, ... (Adding 4) 3, 0, -3, ... (Subtracting 3) Hence, next term is -3 - 3 = -6
7.
5, 10, 13, 26, 29, 58, 61, (....)
A. 128
B. 64
C. 125
D. 122
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Numbers are alternatively multiplied by 2 and increased by 3 5 × 2 = 10 10 + 3 = 13 13 × 2 = 26 26 + 3 = 29 29 × 2 = 58 58 + 3 = 61 61 × 2 = 122
8.
1, 8, 27, 64, 125, 216, (....)
A. 354
B. 392
C. 343
D. 245
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : The pattern is 13, 23, 33, 43, 53, 63, etc Hence, next number is 73 = 343
9.
4, 5, 7, 11, 19, (...)
A. 22
B. 35
C. 27
D. 32
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 4 4 × 2 - 3 =5 5 × 2 - 3 =7 7 × 2 - 3 = 11
11 × 2 - 3 = 19 19 × 2 - 3 = 35
10.
23, 29, 31, 37, 41, 43, (...)
A. 53
B. 47
C. 48
D. 59
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : All are prime numbers in their order, starting from 23 Hence, next number is 47
11.
8, 24, 12, 36, 18, 54, (....)
A. 27
B. 68
C. 108
D. 72
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 8 × 3 = 24 24 ÷ 2 = 12 12 × 3 = 36 36 ÷ 2 = 18 18 × 3 = 54 54 ÷ 2 = 27
12.
7, 26, 63, 124, 215, 342, (....)
A. 481
B. 391
C. 511
D. 421
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : The series is (23 - 1), (33 - 1), (43 - 1), (53 - 1), (63 - 1), (73-
1), ... Hence, next number is (83 - 1) = 511
13.
2, -6, 18, -54, 162, (...)
A. 486
B. -486
C. 422
D. -428
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 2 × -3 = -6 -6 × -3 = 18 18 × -3 = -54 -54 × -3 = 162 162 × -3 = -486
14.
2, 6, 12, 20, 30, 42, 56, (...)
A. 61
B. 72
C. 64
D. 70
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : The pattern is 1×2, 2×3, 3×4, 4×5, 5×6, 6×7, 7×8. Hence, next number is 8×9 = 72
15.
8, 12, 18, 27, 40.5, (...)
A. 60.75
B. 62
C. 58
D. 60.5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 8
(8×3)÷ 2 = 12 (12×3)÷ 2 = 18 (18×3)÷ 2 = 27 (27×3)÷ 2 = 40.5 (40.5×3)÷ 2 = 60.75
16.
196, 144, 100, 64, 36, ?
A. 16
B. 8
C. 18
D. 12
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : The patter is 142, 122, 102, 82, 62, ... So next number is 42 = 16
17.
2, 3, 5, 16, 231, ?
A. 62005
B. 52120
C. 422
D. 53105
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : 2 3 32 - 22 = 5 52 - 32 = 16 162 - 52 = 231 2312 - 162 = 53105
18.
279936, 46656, 7776, 1296, 216, ?
A. 60
B. 46
C. 36
D. 66
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option C Explanation : Go on dividing by 6 to the next number
19.
220, 218, 214, 208, 200, 190, ?
A. 166
B. 178
C. 182
D. 190
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 220 220 218 214 208 200 190
20.
-
2 = 218 4 = 214 6 = 208 8 = 200 10 = 190 12 = 178
12, 38, 116, 350, 1052, ?
A. 1800
B. 2200
C. 2800
D. 3158
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : 12 12 × 3 + 2 = 38 38 × 3 + 2 = 116 116 × 3 + 2 = 350 350 × 3 + 2 = 1052 1052 × 3 + 2 = 3158
21.
9, 35, 91, 189, 341, ?
A. 502
B. 521
C. 559
D. 611
Hide Answer | Notebook | Discuss
Here is the answer and explanation Answer : Option C Explanation : 13 + 23 = 9 23 + 33 = 35 33 + 43 = 91 43 + 53 = 189 53 + 63 = 341 63 + 73 = 559
22.
78, 64, 48, 30, 10, (...)
A. -12
B. -14
C. 2
D. 8
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 78 64 48 30 10
23.
-
14 16 18 20 22
= 64 = 48 = 30 = 10 = -12
16384, 8192, 2048, 256, 16, ?
A. 1
B. 8
C. 2
D. 0.5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Terms are continuously divided by 2,4,8,16, and so on 16384 16384 ÷ 2 = 8192 8192 ÷ 4 = 2048 2048 ÷ 8 = 256 256 ÷ 16 = 16 16 ÷ 32 = 0.5
24.
46080, 3840, 384, 48, 8, 2, ?
A. 1
B. 1⁄64
C. 1⁄8
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 46080 ÷ 12 = 3840 3840 ÷ 10 = 384 384 ÷ 8 = 48 48 ÷ 6 = 8 8 ÷4 =2 2 ÷2 =1
25.
7, 21, 63, 189, 567, ?
A. 1521
B. 1609
C. 1307
D. 1701
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Multiply a term with 3 to get the next term
26.
1, 0.5, 0.25, 0.125, 0.0625, ?
A. 0.015625
B. 0.03125
C. 0.0078125
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Divide by 2 to get the next term
27.
Complete the series 20, 19, 17, ...., 10, 5
A. 15
B. 16
C. 13
D. 14
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : 20 19 17 14 10
-
28.
1 2 3 4 5
= 19 = 17 = 14 = 10 =5
Which fraction will come next,
9 32 9 C. 28
9 30 11 D. 32
A.
1 3 5 7 , , , ,? 2 4 8 16
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Numerators are getting increased by 2. i.e., numerators goes like 1,3,5,7,... Hence, next numerator = 7+2 = 9 Denominators are getting multiplied by 2. i.e., denominators goes like 2,4,8,16,... Hence, next denominator = 16×2 = 32
So, the next fraction is
29.
9 32
4, 10, (?), 82, 244, 730
A. 26
B. 28
C. 40
D. 48
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option B Explanation : 4 × 3 - 2 = 10 10 × 3 - 2 = 28 28 × 3 - 2 = 82 82 × 3 - 2 = 244 244 × 3 - 2 = 730
30.
Complete the series 95, 115.5, 138, ..., 189
A. 160.5
B. 162.5
C. 164.5
D. 166.5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 95 + 20.5 = 115.5 115.5 + 22.5 = 138 138 + 24.5 = 162.5 162.5 + 26.5 = 189
1. Find the odd man out. 1, 3, 9, 12, 19, 29 A. 12
B. 9
C. 1
D. 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 12 is an even number. All other given numbers are odd
2. Find the odd man out. 1, 8, 27, 64, 125, 196, 216, 343 A. 64
B. 196
C. 216
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : The pattern is 13, 23, 33, 43, 53, 63, 73. 196 is not a perfect cube
3. Find the odd man out. 15, 25, 30, 51, 85, 90, 115 A. 15
B. 25
C. 51
D. 90
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : All except 51 are multiples of 5
4. Find the odd man out. 24,36,52,72,96 A. 72
B. 52
C. 36
D. 24
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option B Explanation : All except 52 are multiples of 6
5. Find the odd man out. 187, 264, 386, 473, 682, 781 A. 386
B. 187
C. 781
D. 682
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : In all numbers except 386, the middle digit is the sum of other two digits.
6. Find the odd man out. 12, 24, 34, 48, 64, 84 A. 48
B. 34
C. 24
D. 12
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : All numbers except 34 are multiples of 4
7. Find the odd man out. 362, 482, 551, 263, 344, 284 A. 362
B. 482
C. 551
D. 344
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : In all numbers except 344, the product of first and third digits is the middle digit.
8. Find the odd man out. 742, 743, 633, 853, 871, 990, 532 A. 532
B. 990
C. 633
D. 742
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : In all numbers except 742, the difference of third and first digit is the middle digit.
9. Find the odd man out. 1, 5, 11, 17, 23, 29 A. 29
B. 17
C. 11
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : All given numbers except 1 are prime numbers. One is not a prime number because it does not have two factors. It is divisible by only 1
10. Find the odd man out. 7,13,19,25,29,37,43 A. 19
B. 29
C. 25
D. 43
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : All given numbers except 25 are prime numbers.
11. Find the odd man out. 1, 9, 16, 51, 121, 169, 225 A. 169
B. 51
C. 16
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option B Explanation : Each of the given numbers except 51 is a perfect square
12. Find the odd man out. 1, 4, 9, 17, 25, 36, 49 A. 1
B. 9
C. 17
D. 49
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : The patter is 12, 22, 32, 42, 52, 62, 72 But, instead of 42, 17 is given
13. Find the odd man out. 2, 5, 10, 17, 26, 38, 50, 65 A. 50
B. 38
C. 26
D. 65
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : The pattern is (1×1)+1 , (2×2)+1 , (3×3)+1 , (4×4)+1 , (5×5)+1 , (6×6)+1 , (7×7)+1 , (8×8)+1 Hence, in place of 38, the right number was (6×6)+1 = 37
14. Find the odd man out. 18, 16, 12, 24, 11, 34, 46 A. 16
B. 46
C. 16
D. 11
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : 11 is the only odd number in the given series
15. Find the odd man out. 1, 27, 216, 512, 1024, 1331 A. 1024
B. 512
C. 27
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : All given numbers except 1024 are perfect cubes
16. Find the odd man out. 1, 16, 81, 255, 625, 1296 A. 255
B. 1296
C. 81
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : The patter is 14, 24, 34, 44, 54, 64 Hence, in place of 255, the right digit is 44 = 256
17. Find the odd man out. 6, 13, 18, 25, 30, 37, 40 A. 40
B. 30
C. 37
D. 25
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : The difference between two beginning are 7, 5, 7, 5, 7, 5
successive
terms
Hence, in place of 40, right number is 37+5=42
18. Find the odd man out. 445, 221, 109, 46, 25, 11, 4
from the
A. 25
B. 109
C. 46
D. 221
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : To obtain next number, subtract 3 from the previous number and divide the result by 2 445 (445-3)/2 = 221 (221-3)/2 = 109 (109-3)/2 = 53 (53-3)/2 = 25 (25-3)/2 = 11 (11-3)/2 = 4 Clearly, 53 should have come in place of 46
19. Find the odd man out. 1050, 510, 242, 106, 46, 16, 3 A. 46
B. 106
C. 510
D. 1050
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 1050 (1050 - 30)/2 = 510 (510 - 26)/2 = 242 (242 - 22)/2 = 110 (110 - 18)/2 = 46 (46- 14)/2 = 16
(16- 10)/2 = 3 Hence, 110 should have come in place of 106
20. Find the odd man out. 2, 3, 5, 9, 12, 17, 23 A. 12
B. 9
C. 23
D. 2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 2 2 +1 =3 3 +2 =5 5 +3 =8 8 + 4 = 12 12 + 5 = 17 17 + 6 = 23 ie, 8 should have come in place of 9
21. Find the odd man out. 3, 8, 18, 38, 78, 158, 316 A. 38
B. 158
C. 316
D. 8
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : 3 3 × 2 +2 =8 8 × 2 + 2 = 18 18 × 2 + 2 = 38
38 × 2 + 2 = 78 78 × 2 + 2 = 158 158 × 2 + 2 = 318 Hence, 316 is wrong and 318 should have come in place of that
22. Find the odd man out. 5, 6, 14, 45, 185, 925, 5556 A. 5556
B. 925
C. 185
D. 6
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : 5 × 1 +1 =6 6 × 2 + 2 = 14 14 × 3 + 3 = 45 45 × 4 + 4 = 184 184 × 5 + 5 = 925 925 × 6 + 6 = 5556 Hence, it is clear that 184 should have come instead of 185
23. Find the odd man out. 23, 27, 36, 52, 77, 111, 162 A. 162
B. 111
C. 52
D. 27
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 23 + 22 = 27 27 + 32 = 36 36 + 42 = 52
52 + 52 = 77 77 + 62 = 113 113 + 72 = 162 Hence, 113 should have come in place of 111
24. Find the odd man out. 241, 263, 248, 271, 255, 277, 262 A. 277
B. 271
C. 263
D. 241
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Alternatively 22 is added and 15 is subtracted from the terms. Hence, 271 is wrong 241 241 + 22 = 263 263 - 15 = 248 248 + 22 = 270 270 - 15 = 255 255 + 22 = 277 277 - 15 = 262
25. Find the odd man out. 125, 127, 130, 135, 142, 153, 165 A. 165
B. 142
C. 153
D. 130
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Prime numbers 2, 3, 5, 7, 11, 13 are added successively.
Hence, 165 is wrong
26. Find the odd man out. 5, 10, 40, 81, 320, 640, 2560 A. 40
B. 81
C. 320
D. 2560
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Alternatively 2 and 4 are multiplied with the previous terms 5 5 × 2 = 10 10 × 4 = 40 40 × 2 = 80 80 × 4 = 320 320 × 2 = 640 640 × 4 = 2560 Hence, 81 is wrong. 80 should have come in place of 81.
27. Find the odd man out. 12, 21, 32, 45, 60, 77, 95 A. 95
B. 45
C. 32
D. 21
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 12 + 9 = 21 21 + 11 = 32 32 + 13 = 45 45 + 15 = 60
60 + 17 = 77 77 + 19 = 96 Hence, 95 is wrong. 96 should have come in place of 95
28. Find the odd man out. 3, 5, 15, 75, 1120, 84375 A. 84375
B. 1120
C. 15
D. 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Pattern : 1st * 2nd = 3rd, 2nd * 3rd = 4th, etc. 3 5 3 × 5 = 15 5 × 15 = 75 15 × 75 = 1125 75 × 1125 = 84375 Hence, 1120 is wrong. 1125 should have come in place of 1120
29. Find the odd man out. 3576, 1784, 888, 440, 216, 105, 48 A. 105
B. 216
C. 888
D. 1784
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 3576 (3576-8)/2 = 1784 (1784-8)/2 = 888
(888-8)/2 = 440 (440-8)/2 = 216 (216-8)/2 = 104 (104-8)/2 = 48 Hence, 105 is wrong. 104 should have come in place of 105
30. Find the odd man out. 30, -5, -45, -90, -145, -195, -255 A. -5
B. -145
C. -255
D. -195
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 30 30 - 35 = -5 -5 - 40 = -45 -45 - 45 = -90 -90 - 50 = -140 -140 - 55 = -195 -195 - 60 = -255 Hence, -145 is wrong. -140 should have come in place of -145
I mportan t Formu las - Simple I n terest 1. Introduction Money is not free and it costs to borrow the money. Normally, the borrower has to pay an extra amount in addition to the amount he had borrowed. i.e, To repay the loan, the borrower has to pay the sum borrowed and the interest. 2. Lender and Borrow er The person giving the money is called the lender and the person taking the money is the borrower. 3. Principal (sum ) Principal (or the sum) is the money borrowed or lent out for a certain period. It is denoted by P. 4. Interest Interest is the extra money paid by the borrower to the owner (lender) as a form of compensation for the use of the money borrowed. 5. Sim ple Interest (SI) If the interest on a sum borrowed for certain period is calculated uniformly, it is called simple interest(SI). 6. Am ount (A) The total of the sum borrowed and the interest is called the amount and is denoted by A
7.
The statement "rate of interest 10% per annum" means that the interest for one year on a sum of Rs.100 is Rs.10. If not stated explicitly, rate of interest is assumed to be for one year.
Let Principal = P, Rate = R% per annum and Time = T years. Then 8.
Simple Interest,
SI =
PRT 100
From the above formula , we can derive the followings
9.
P=
100 × SI RT
R=
100 × SI PT
T=
100 × SI PR
1. How much time will it take for an amount of Rs. 900 to yield Rs. 81 as interest at 4.5% per annum of simple interest? A. 2 years
B. 3 years
C. 1 year
D. 4 years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : P = Rs.900 SI = Rs.81 T =? R = 4.5%
T=
100 × SI 100 × 81 = = 2 years PR 900 × 4.5
2. Arun took a loan of Rs. 1400 with simple interest for as many years as the rate of interest. If he paid Rs.686 as interest at the end of the loan period, what was the rate of interest? A. 8%
B. 6%
C. 4%
D. 7%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let rate = R% Then, Time, T = R years P = Rs.1400 SI = Rs.686
SI=
PRT 100
⇒ 686 =
1400 × R × R 100
⇒ 686 = 14 R
2
⇒ 49 = R 2 ⇒R=7 i.e.,Rate of Interest was 7%
3. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is : A. Rs. 700
B. Rs. 690
C. Rs. 650
D. Rs. 698
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Simple Interest (SI) for 1 year = 854-815 = 39 Simple Interest (SI) for 3 years = 39 × 3 = 117 Principal = 815 - 117 = Rs.698
4. A sum fetched a total simple interest of Rs. 929.20 at the rate of 8 p.c.p.a. in 5 years. What is the sum? A. Rs. 2323
B. Rs. 1223
C. Rs. 2563
D. Rs. 2353
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : SI = Rs.929.20 P =? T = 5 years
R = 8%
P=
100 × SI 100 × 929.20 = = Rs.2323 RT 8×5
5. Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B? A. Rs. 6400
B. Rs. 7200
C. Rs. 6500
D. Rs. 7500
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the investment in scheme A be Rs.x and the investment in scheme B be Rs.(13900 - x)
We know that SI =
PRT 100
Simple Interest for Rs.x in 2 years at 14% p.a. =
x × 14 × 2 28x = 100 100
Simple Interest for Rs.(13900 - x) in 2 years at 11% p.a. = Total interest = Rs.3508 28x 22(13900 − x) + = 3508 100 100 28x + 305800 − 22x = 350800 6x = 45000 x=
45000 = 7500 6
Investment in scheme B = 13900 - 7500 = Rs.6400
(13900 − x) × 11 × 2 100
=
22(13900 − x) 100
46. A four years NSC certificate was purchased for Rs. 500 with Rs. 1000 being the maturity value. Find the rate of S.I. A. 24%
B. 22%
C. 16%
D. 25%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : P = Rs. 500 SI = Rs.500 T =4 R=?
R =
100 × SI 100 × 500 100 = = = 25% PT 500 × 4 4
47. If simple interest on a certain sum of money for 8 years at 4% per annum is same as the simple interest on Rs. 560 for 8 years at the rate of 12% per annum then the sum of money is A. Rs.1820
B. Rs.1040
C. Rs.1120
D. Rs.1680
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the sum of money be x then
x×4×8 100
=
560 × 12 × 8 100
x × 4 × 8 = 560 × 12 × 8 x × 4 = 560 × 12 x = 560 × 3 = 1680
48. If a sum of money trebles itself in 40 years,what is the rate of interest?
A. 5%
B. 6%
C. 4%
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : ---------------------------------------------------------------------------------Solution 1 ---------------------------------------------------------------------------------Let the sum of money be Rs.x After 40 years, this becomes 3x Simple Interest = 3x - x = 2x
Simple Interest = 2x = 2=
PRT 100
x × R × 40 100
R × 40 100
200 = 40R R = 5% ---------------------------------------------------------------------------------Solution 2 ----------------------------------------------------------------------------------
If a sum of money becomes n times in T years at simple interest, then the rate of interest per annum can be given be
R=
100(n − 1) % T
[Read more ...] n =3 T = 40
R=
100(n − 1) 100(3 − 1) 100 × 2 100 = = = = 5% T 40 40 20
6. A person borrows Rs.5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at 61⁄4% p.a for 2 years. Find his gain in the transaction per year. A. Rs. 167.50
B. Rs. 150
C. Rs.225
D. Rs. 112.50
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : ----------------------------------------------------------------------------Solution 1 ----------------------------------------------------------------------------The person borrows Rs. 5000 for 2 years at 4% p.a. simple interest
Simple interest that he needs to pay =
PRT 5000 × 4 × 2 = = 400 100 100
He also lends it at 61⁄4% p.a for 2 years
25 5000 × ×2 PRT 4 Simple interest that he gets = = 100 100
= 625
His overall gain in 2 years = Rs.625 - Rs.400 = Rs.225
His overall gain in 1 year =
225 = Rs.112.5 2
----------------------------------------------------------------------------Solution 2 ----------------------------------------------------------------------------The person borrows Rs. 5000 for 2 years at 4% p.a. simple interest He also lends it at 61⁄4% p.a for 2 years
6
1 1 % − 4% = 2 % 4 4
So his gain in the transaction for 1 year = The simple interest he gets for Rs.5000 for 1 year at 2 1⁄4% per annum
=
PRT = 100
9 ×1 4 100
5000 ×
= 112.5
7. What will be the ratio of simple interest earned by certain amount at the same rate of interest for 5 years and that for 15 years? A. 3 : 2
B. 1 : 3
C. 2 : 3
D. 3 : 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : ----------------------------------------------------------------------------------Solution 1 ----------------------------------------------------------------------------------Let Principal = P Rate of Interest = R%
PR × 5 ) 100 Required Ratio = PR × 15 ( ) 100 (
=
5 1 = = 1 :3 15 3
----------------------------------------------------------------------------------Solution 2 -----------------------------------------------------------------------------------
Simple Interest =
PRT 100
Here Principal(P) and Rate of Interest (R) are constants Hence, Simple Interest ∝ T
Required Ratio =
Simple Interest for 5 years Simple Interest for 15 years
=
T1 5 1 = = = 1 :3 T2 15 3
8. A sum of money amounts to Rs.9800 after 5 years and Rs.12005 after 8 years at the same rate of simple interest. The rate of interest per annum is
A. 15%
B. 12%
C. 8%
D. 5%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Simple Interest for 3 years = (Rs.12005 - Rs.9800) = Rs.2205
Simple Interest for 5 years =
2205 × 5 = Rs.3675 3
Principal(P) = (Rs.9800 - Rs.3675) = Rs.6125
R=
100 × SI 100 × 3675 = = 12% PT 6125 × 5
9. A certain amount earns simple interest of Rs. 1200 after 10 years. Had the interest been 2% more, how much more interest would it have earned? A. Rs. 25
B. None of these
C. Rs. 120
D. Cannot be determined
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Since we do not have the principal and rate of interest, we can not find out the required details.
10. A man took loan from a bank at the rate of 8% p.a. simple interest. After 4 years he had to pay Rs. 6200 interest only for the period. The principal amount borrowed by him was: A. Rs.17322
B. Rs.20245
C. Rs.18230
D. Rs.19375
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Principal(P) = ? Time(T) = 4 years
Simple Interest(SI) = Rs.6200 R = 8%
P=
100 × SI 100 × 6200 = = Rs.19375 RT 8×4
11. A sum of Rs. 14,000 amounts to Rs. 22,400 in 12 years at the rate of simple interest. What is the rate of interest? A. 7%
B. 6%
C. 5%
D. 4%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Simple Interest for 4 years (SI) = (22400 - 14000) = Rs.8400 R=? T = 12 years P = Rs. 14000
R=
100 × SI 100 × 8400 = = 5% PT 14000 × 12
12. A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest? A. 3.46%
B. None of these
C. 4.5%
D. 5%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the sum of Rs.725 is lent out at rate R% for 1 year Then, at the end of 8 months, ad additional sum of 362.50 more is lent out at rate 2R% for remaining 4 months(1/3 year) Total Simple Interest = 33.50
362.50 × 2R ×
725 × R × 1 => 100
+
=>
725 × R × 1 100
+
=>
725R 725R + = 33.50 100 300
100
1 3
= 33.50
362.50 × 2R = 33.50 300
=> 725R (
1 1 ) = 33.50 + 100 300
=> 725R (
4 ) = 33.50 300
=> 725R × 4 = 10050 => 725R = 2512.5 => 145R = 502.5 => R =
502.5 = 3.46% 145
13. An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes: A. None of these
B. 10.25%
C. 10.5%
D. 10%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the automobile financier lends Rs.100
1 100 × 10 × PRT 2 Simple Interest for first 6 months = = 100 100
= Rs. 5
After 6 months, he adds the simple interest to principal i.e., after 6 months, principal becomes Rs.100 + Rs.5 = Rs.105
1 105 × 10 × PRT 2 Simple Interest for next 6 months = = 100 100
= Rs. 5.25
Amount at the end of 1 year = Rs.105 + Rs. 5.25 = Rs.110.25 i.e., Effective Simple Interest he gets for Rs.100 for 1 year = 110.25 - 100 = 10.25 i.e, the Effective Rate of Interest = 10.25%
(∵ R =
100 × SI 100 × 10.25 = = 10.25%) PT 100 × 1
14. A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in
all from both of them as interest. The rate of interest per annum is: A. 5%
B. 10%
C. 7%
D. 8%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the rate of interest per annum be R%
Simple Interest for Rs. 5000 for 2 years at rate R% per annum + Simple Interest for Rs. 3000 for 4 years at rate R% per annum = Rs.2200 5000 × R × 2 3000 × R × 4 + 100 100 ⇒ 100R + 120R = 2200 ⇒ 220R = 2200 ⇒ R = 10
⇒
= 2200
i.e, Rate = 10%.
15. What annual payment will discharge a debt of Rs. 6450 due in 4 years at 5% per annum? A. Rs.1500
B. Rs.1400
C. Rs.1800
D. Rs.1600
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : ------------------------------------------------------------------------Solution 1 ------------------------------------------------------------------------The amount needs to be repaid in 4 years = Rs.6450 Suppose Rs.x is paid annually to repay this debt Then, amount paid after 1st year = Rs.x
Interest for this Rs.x for the remaining 3 years =
x×5×3 15x = 100 100
Then, amount paid after 2nd year = Rs.x
Interest for this Rs.x for the remaining 2 years =
x×5×2 10x = 100 100
Amount paid after 3rd year = Rs.x
Interest for this Rs.x for the remaining 1 year =
x×5×1 5x = 100 100
Amount paid after 4th year = Rs.x and this closes the entire debt
=> x +
15x 10x 5x +x+ +x+ + x = 6450 100 100 100
=> 4x +
30x = 6450 100
=> 4x +
3x = 6450 10
=> 40x + 3x = 64500 => 43x = 64500 => x = 1500 ------------------------------------------------------------------------Solution 2 (using form ula) -------------------------------------------------------------------------
The annual instalment which will discharge a debt of D due in T years at R% simple interest per annum
=
100D RT(T-1) 100T + 2
[Read more ...]
100D RT(T-1) 100T + 2 645000 645000 = = = 1500 400 + 30 430
As per the above formula, the annual payment =
=
100 × 6450 5 × 4 × (4 − 1) 100 × 4 + 2
16. A lends Rs. 1500 to B and a certain sum to C at the same time at 8% per annum simple interest. If after 4 years, A altogether receives Rs. 1400 as interest from B and C, then the sum lent to C is A. Rs.2875
B. Rs.1885
C. Rs.2245
D. Rs.2615
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the sum lent to C be Rs. x Simple Interest for Rs.1500 at 8% per annum for 4 years + Simple Interest for Rs.x at 8% per annum for 4 years = Rs.1400
1500 × 8 × 4 x×8×4 + 100 100 32x ⇒ 480 + = 1400 100 32x ⇒ = 920 100 920 × 100 ⇒x= = 2875 32
⇒
= 1400
17. A sum of Rs. 10 is given as a loan to be returned in 6 monthly installments at Rs.3. What is the rate of interest? A. 820%
B. 620%
C. 780%
D. 640%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Amount borrowed = Rs.10 Let rate of interest = R%
Simple Interest for Rs.10 for 6 months at R% = i.e, 10 +
R is due in 6 months 20
Payment after 1st month = Rs.3
10 × R × 100
1 2
=
R 20
Interest for this Rs.3 for the remaining 5 months =
5 12 100
3 × R×
Payment after 2nd month = Rs.3
Interest for this Rs.3 for the remaining 4 months =
4 12 100
3 × R×
... Payment after 5th month = Rs.3
Interest for this Rs.3 for the remaining 1 month =
1 12 100
3 × R×
Payment after 6th month = Rs.3 and this closes the loan
=> (3 + 3 + 3 + 3 + 3 + 3) +
5 12 100
. 3 × R× .
+
4 12 100
3 × R×
+. . . +
1 12 100
3 × R×
= 10 +
R 20
3R (5 + 4+. . . +1) R 12 18 + = 10 + 100 20 15R R 18 + = 10 + 400 20 R 15R 5R R 8= − = = 20 400 400 80 R = 640%
18. If the simple interest on a certain sum of money after 3 1⁄8years is 1⁄4 of the principal, what is the rate of interest per annum? A. 6%
B. 4%
C. 8%
D. 12%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the the sum of money(P) be Rs.x Time(T) = 3 1⁄8 Years = 25⁄8 Years
Simple interest (SI) = x ⁄4
x 100 × SI 100 × x × 8 4 Rate of interest per annum(R) = = = PT 25 4 × x × 25 x× 8 100 ×
= 8%
19. If a sum of Rs. 9 is lent to be paid back in 10 equal monthly installments of re. 1 each, then the rate of interest is A. 11.33%
B. 11%
C. 266.67%
D. 26.67%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Amount borrowed = Rs.9 Let rate of interest = R%
Simple Interest for Rs.9 for 10 months at R% = i.e., 9 +
10 12 100
9×R×
90R is due in 10 months 1200
=
Payment after 1st month = Rs.1
Interest for this Rs.1 for the remaining 9 months =
9 12 100
1×R×
Payment after 2nd month = Rs.1
Interest for this Rs.1 for the remaining 8 months = ⋯
8 12 100
1×R×
Payment after 9th month = Rs.1
Interest for this Rs.1 for the remaining 1 month =
1 12 100
1×R×
Payment after 10th month = Rs.1 and this closes the loan
90R 1200
9 12 100
1×R×
9+
90R = 10 × 1 + 1200
9+
90R R = 10 + (9 + 8 + ⋯ + 1) 1200 1200
9+
90R R 9 × 10 ( ) = 10 + 1200 1200 2
9+
90R 45R = 10 + 1200 1200
+
8 12 100
1×R×
+⋯+
1 12 100
1×R×
45R =1 1200 R=
1200 = 26.67% 45
20. Divide Rs. 2379 into 3 parts so that their amount after 2,3 and 4 years respectively may be equal, the rate of interest being 5% per annum at simple interest. The first part is A. Rs. 828
B. Rs. 746
C. Rs. 248
D. Rs. 1024
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the parts be x, y and z R = 5% x + interest on x for 2 years = y + interest on y for 3 years = z + interest on z for 4 years
(x +
x×5×2 y× 5 × 3 z× 5 × 4 ) = (y + ) = (z + ) 100 100 100
(x +
x 3y z ) = (y + ) = (z + ) 10 20 5
11x 23y 6z = = 10 20 5
Let
11x 23y 6z = = = k 10 20 5
Then, x =
10k , 11
y=
20k , 23
(where k is a constant) z=
5k 6
we know that x + y + z = 2379 10k 20k 5k + + = 2379 11 23 6 10k × 23 × 6 + 20k × 11 × 6 + 5k × 11 × 23 = 2379 × 11 × 23 × 6 1380k + 1320k + 1265k = 2379 × 11 × 23 × 6 3965k = 2379 × 11 × 23 × 6 k=
2379 × 11 × 23 × 6 3965
First part, x = =
10k 10 2379 × 11 × 23 × 6 = × 11 11 3965
2 × 2379 × 23 × 6 793
= 2 × 3 × 23 × 6 = 828
=
10 × 2379 × 23 × 6 3965
21. In how many years, Rs. 150 will produce the same interest at 6% as Rs. 800 produce in 2 years at 4½ %? A. 4 years
B. 6 years
C. 8 years
D. 9 years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let Simple Interest for Rs.150 at 6% for n years = Simple Interest for Rs.800 at 4½ % for 2 years
150 × 6 × n 100
=
9 ×2 2 100
800 ×
150 × 6 × n = 800 ×
9 ×2 2
150 × 6 × n = 800 × 9 3 × 6 × n = 16 × 9 6 × n = 16 × 3 2 × n = 16 n = 8 years
22. A sum of Rs. 2500 amounts to Rs. 3875 in 4 years at the rate of simple interest. What is the rate of interest? A. 12.25%
B. 12%
C. 6%
D. 13.75%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Simple Interest, SI = (3875 - 2500) = Rs.1375 Principal, P = Rs. 2500 Time, T = 4 years
R=?
R=
100 × SI 100 × 1375 100 × 1375 = = = 13.75% PT 2500 × 4 10000
23. What is the interest due after 40 days for Rs. 3200 at 10% A. 35.07
B. 36.21
C. 35.52
D. 34
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Simple Interest, SI =
PRT = 100
3200 × 10 × 100
40 365
= 35.07
24. What is the rate of interest at which Rs.150 becomes Rs. 220 in 10 years. A.
11 % 3
C. 12%
B.
14 % 3
D. 14%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Simple Interest, SI = 220 - 150 = Rs.70
R=
100 × SI 100 × 70 2×7 14 = = = % PT 150 × 10 3 3
25. A person invested in all Rs. 2600 at 4%, 6% and 8% per annum simple interest. At the end of the year, he got the same interest in all three cases. The money invested at 4% is: A. Rs.2200
B. Rs.800
C. Rs.1600
D. Rs.1200
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Let x , y and x be his investments at 4%, 6% and 8% respectively Simple Interest on x at 4% for 1 year = Simple Interest on y at 6% for 1 year = Simple Interest on z at 8% for 1 year
x×4×1 100
=
y× 6 × 1 100
=
z× 8 × 1 100
⇒ 4x = 6y = 8z ⇒ 2x = 3y = 4z Hence, we have, y =
2x 2x x and z = = 3 4 2
we know that x + y + z = 2600 ⇒ x+
2x x + = 2600 3 2
⇒ 6x + 4x + 3x = 2600 × 6 ⇒ 13x = 2600 × 6 ⇒x=
2600 × 6 = 200 × 6 = 1200 13
i.e., Money invested at 4% = Rs.1200
26. David invested certain amount in three different schemes A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in Scheme C was 150% of the amount invested in Scheme A and 240% of the amount invested in Scheme B, what was the amount invested in Scheme B? A. Rs.5000
B. Rs.2000
C. Rs.6000
D. Rs.3000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let x , y and x be his investments in A, B and C respectively. Then Then, Interest on x at 10% for 1 year + Interest on y at 12% for 1 year + Interest on z at 15% for 1 year = 3200
x × 10 × 1 y × 12 × 1 z × 15 × 1 + + = 3200 100 100 100 ⇒ 10x + 12y + 15z = 320000 − − − (1) Amount invested in Scheme C was 240% of the amount invested in Scheme B
=> z =
240y 60y 12y = = − − − (2) 100 25 5
Amount invested in Scheme C was 150% of the amount invested in Scheme A
150x 3x = 100 2 2z 2 12y 8y => x = = × = − − − (3) 3 3 5 5
=> z =
From(1),(2) and (3), 10x + 12y + 15z = 320000
8y 12y ) + 12y + 15 ( ) = 320000 5 5 16y + 12y + 36y = 320000 64y = 320000 320000 10000 y= = = 5000 64 2
10 (
i.e.,Amount invested in Scheme B = Rs.5000
27. A sum of Rs. 1550 was lent partly at 5% and partly at 8% p.a. simple interest. The total interest received after 3 years was Rs. 300. The ratio of the money lent at 5% to that lent at 8% is: A. 16 : 15
B. 15 : 16
C. 15 : 8
D. 8 : 15
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the partial amount at 5% be x and the partial amount at 8% be (1550-x) Interest on x at 5% for 3 years + interest on (1550-x) at 8% for 3 years = 300
x× 5 × 3 100
+
(1550-x) × 8 × 3 100
= 300
(1550-x) × 8 x× 5 + = 100 100 100 5x + 8(1550 − x) = 10000 5x + 12400 − 8x = 10000 3x = 2400 x = 800 Required Ratio = x : (1550-x) = 800 : (1550-800) = 800 : 750 = 16 : 15
28. A sum of money doubles in 12 years. In how many years, it will treble at S.I. A. 12 years
B. 8 years
C. 6 years
D. 24 years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Simple Interest, SI =
PRT 100
i.e, SI ∝ T when rate(R) and principal (P) are constants Let x be the sum of money and which will treble in n years (Please note that when the money doubles, simple interest is 2x - x=x and when the money trebles, simple interest is 3x - x = 2x) (2x-x) ∝ 12 => x ∝ 12 -------------(1) (3x-x) ∝ n => 2x ∝ n -------------(2) From (1) and (2),
x 12 = 2x n 1 12 = 2 n ⇒ n = 24 years i.e, in 24 years, the money will treble.
29. A man invests a certain sum of money at 6% per annum simple interest and another sum at 7% per annum simple interest. His income from interest after 2 years was Rs. 354. One-forth of the first sum is equal to one-fifth of the second sum. The total sum invested was : A. Rs.3100
B. Rs.2700
C. Rs.2200
D. Rs.1800
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the man invests Rs.x at 6% and Rs.y at 7%
Simple Interest on Rs.x at 6% for 2 years + Simple Interest on Rs.y at 7% for 2 years = Rs.354
x× 6 × 2 100
+
y× 7 × 2 100
= 354
x × 6 × 2 + y × 7 × 2 = 354 × 100 x × 6 + y × 7 = 177 × 100 6x + 7y = 17700
⋯ (1)
One-forth of the first sum is equal to one-fifth of the second sum
x y = 4 5 4y => x = 5
=>
⋯ (2)
Solving (1) and (2),
6x + 7y = 17700 6(
4y ) + 7y = 17700 5
24y + 35y = 17700 × 5 59y = 17700 × 5 y = 300 × 5 = 1500 x=
4y 4 × 1500 = = 4 × 300 = 1200 5 5
total sum invested = x + y = 1500 + 1200 = 2700
30. The interest on a certain deposit at 5% per annum is Rs. 101.20 in one year. How much will the additional interest in one year be on the same deposit at 6% per annum? A. Rs.20.8
B. Rs.19.74
C. Rs.20.24
D. Rs.19.5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C
Explanation : -------------------------------------------------------------------------------Solution 1 --------------------------------------------------------------------------------
100 × SI 100 × 101.20 = = 20 × 101.20 = Rs. 2024 RT 5×1 2024 × 6 × 1 Simple Interest for Rs.2024 at 6% per annum for 1 year, SI = = 121.44 100
Principal, P =
Additional Interest = Rs.121.44 - Rs.101.20 = Rs.20.24 -------------------------------------------------------------------------------Solution 2 --------------------------------------------------------------------------------
Principal, P =
100 × SI 100 × 101.20 = = 20 × 101.20 = Rs. 2024 RT 5×1
All parameters remains same except the increase in interest rate. and additional Interest Rate = 6% - 5% = 1%
Hence, Additional Interest = Simple Interest for Rs.2024 at 1% per annum for 1 year 2024 × 1 × 1 = = 20.24 100
31. A sum was put a simple interest at a certain rate for 2 years. Had it been put at 4% higher rate, it would have fetched Rs. 60 more. The sum is: A. Rs.750
B. Rs.700
C. Rs.820
D. Rs.940
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the sub be Rs.x and the initial rate be R%.Then
x × (R+4) × 2 100
−
⇒
x× 4 × 2 100
⇒
x× 2 = 15 100
x× R× 2 100
= 60
= 60
⇒ 2x = 1500 ⇒ x = 750
32. The simple interest at x% for x years will be Rs. x on a sum of: A. Rs. x C.
100 x
Rs.
B. Rs. x2 D.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : P =? R = x% Simple Interest,SI = x T = x years
P=
100 × SI 100 × x 100 = = RT x×x x
Rs.
100 x2
33. A sum of money becomes 7⁄5 of itself in 4 years at a certain rate of simple interest. The rate per annum is : A. 9%
B. 10%
C. 11%
D. 12%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the sum of money be Rs.x
Amount after 4 years =
7x 5
T = 4 years R=?
7x − x) = 5 2x 100 × 100 × SI 5 R= = = PT x×4
Simple Interest, SI = (
2x 5 40 = 10% 4
34. If the simple interest on Rs. 2000 is less than the Simple Interest on Rs. 3000 at 5% by Rs. 50, find the time. A. 2.5 year
B. 1 year
C. 1.5 year
D. 2 year
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : -----------------------------------------------------------------------------------Solution 1 ------------------------------------------------------------------------------------
3000 × 5 × T 100
−
(3000 − 2000)
5×T = 50 100
1000 ×
2000 × 5 × T 100
= 50
5×T = 50 100
50T = 50 T = 1 year -----------------------------------------------------------------------------------Solution 2 -----------------------------------------------------------------------------------Difference in principal = 3000-2000 = 1000 i.e., Simple Interest on Rs.1000 will be Rs.50
1000 × 5 × T 100 50T = 50 T = 1 year
= 50
35. A sum of Rs. 7700 is to be divided among three brothers Vikas, Vijay and Viraj in such a way that simple interest on each part at 5% per annum after 1, 2 and 3 years respectively remains equal. The Share of Vikas is more than that of Viraj by A. Rs.1200
B. Rs.1400
C. Rs.2200
D. Rs.2800
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : ------------------------------------------------------------------------------------Solution 1 ------------------------------------------------------------------------------------Let Vikas, Vijay and Viraj gets Rs.x, Rs.y and Rs.z respectively. Simple Interest on x at 5% for 1 year = Simple Interest on y at 5% for 2 year
= Simple Interest on z at 5% for 3 year
⇒
x×5×1 100
=
y× 5 × 2 100
=
z× 5 × 3 100
⇒ 5x = 10y = 15z ⇒ x = 2y = 3z ⇒ y=
x x and z = 2 3
x + y + z = 7700 ⇒ x+
--- (1)
(∵ the total amount is Rs. 7700)
x x + = 7700 (∵ substituted the values of y and z from from equation 1) 2 3
⇒
11x = 7700 6
⇒
x = 7000 6
⇒ x = 4200 z=
x 4200 = = 1400 3 3
i.e, Vikas gets Rs.4200 and Viraj gets Rs.1400 Share of Vikas is more than that of Viraj by (4200 - 1400) = 2800 ------------------------------------------------------------------------------------Solution 2 -------------------------------------------------------------------------------------
If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R1, R2, ... , Rnrespectively
and
time
periods
are
T 1,
T 2,
...
,
T n respectively, then the ratio in which the sum will be divided in n parts can be given by
1 1 1 : :⋯ R 1 T 1 R2 T 2 RnT n [Read more ...] T 1 = 1 , T 2 = 2, T 3 = 3 R1 = 5 , R2 = 5, R3 = 5 Share of Vikas : Share of Vijay : Share of Viraj
1 1 1 : : 5×1 5×2 5×3 1 1 1 = : : 1 2 3 = 6 :3 :2
=
Total amount is Rs. 7700
6 = 700 × 6 = 4200 11 2 Share of Viraj = 7700 × = 700 × 2 = 1400 11
Share of Vikas = 7700 ×
Share of Vikas is greater than Share of Viraj by (4200 - 1400) = Rs. 2800
36. The simple interest on Rs.500 at 6% per annum from May 3rdto July 15th in the same year is A. Rs. 8
B. Rs. 6
C. Rs. 4
D. Rs. 9
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Time from May 3rd to July 15th = 28 days of May + 30 days of June and 15 days of July
= 73 days=
73 1 years = years 365 5
Simple interest =
PRT = 100
500 × 6 × 100
1 5
=6
37. Find the Simple Interest on Rs. 7500 at 11% for 2 Years and 5 months. A. Rs.1994.25
B. Rs.1991.25
C. Rs.1992.75
D. Rs.1993.75
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Time,T = 2 year 5 months = 29 months = SI =
PRT = 100
7500 × 11 × 100
29 12
29 year 12
= 75 × 11 ×
29 = 1993.75 12
38. Arun borrowed a sum for 4 years on S.I. at 12%. The total interest paid was Rs. 360. Find the Principal. A. Rs.700
B. Rs.650
C. Rs.800
D. Rs.750
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
P=
100 × SI 100 × 360 100 × 30 = = = 25 × 30 = 750 RT 12 × 4 4
39. The simple interest on Rs. 1820 from March 9, 2003 to May 21, 2003 at 7 1⁄2% rate is A. Rs. 27.30
B. Rs. 22.50
C. Rs. 28.80
D. Rs. 29
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Time, T = (22 + 30 + 21)days = 73 days = 1 15 %= % 2 2 15 1 1820 × × PRT 2 5 SI = = 100 100
73 1 year = year 365 5
Rate, R = 7
=
1820 × 100
3 2
=
910 × 3 2730 = = 27.30 100 100
40. At what rate percent of simple interest will a sum of money double itself in 20 years? A. 4%
B. 5%
C. 6%
D. 8%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let sum = x Time = 20 years Simple Interest = x
R=
100 × SI 100 × x 100 = = = 5% PT x × 20 20
41. If the simple interest on a certain sum of money for 4 years is one–fifth of the sum, then the rate of interest per annum is A. 4%
B. 7%
C. 6%
D. 5%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the Principal(P) be x Then, Simple Interest(SI) = x/5 Time(T) = 4 years
Rate of interest per annum(R) =
100 × SI = PT
100 × x×4
x 5
=
20 = 5% 4
42. A certain sum in invested for T years. It amounts to Rs. 400 at 10% per annum. But when invested at 4% per annum, it amounts to Rs. 200. Find the time (T). A. 45 years
B. 60 years
C. 40 years
D. 50 Years
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : ------------------------------------------------------------------------------Solution 1 ------------------------------------------------------------------------------Let the principal = Rs.x and time = y years Principal,x amounts to Rs.400 at 10% per annum in y years Simple Interest = (400-x)
Simple Interest =
PRT 100
⇒ (400 − x) =
x × 10 × y 100
⇒ (400 − x) =
xy 10
--- (equation 1)
Principal,x amounts to Rs.200 at 4% per annum in y years Simple Interest = (200-x)
Simple Interest =
PRT 100
⇒ (200 − x) =
x×4×y 100
⇒ (200 − x) =
xy 25
--- (equation 2)
(
xy ) 10
(equation 1) 400 − x ⇒ = 200 − x (equation 2) xy ( ) 25 ⇒
400 − x 25 = 200 − x 10
⇒
400 − x 5 = 200 − x 2
⇒ 800 − 2x = 1000 − 5x ⇒ 200 = 3x ⇒x=
200 3
200 )y 3 200 )= Substituting this value of x in Equation 1, we get, (400 − 3 10 (
⇒ (400 −
200 20y )= 3 3
⇒ 1200 − 200 = 20y ⇒ 1000 = 20y y=
1000 = 50 years 20
------------------------------------------------------------------------------Solution 2 -------------------------------------------------------------------------------
If a certain sum of money P lent out for a certain time T amounts to P1 at R1% per annum and to P2 at R2% per annum, then
P=
P2 R1 − P 1 R2 R1 − R2
T=
P1 − P2 × 100 years P2 R1 − P 1 R2
[Read more ...] R1 = 10%, R2 = 4% P1 = 400, P2 = 200
P1 − P2 400 − 200 × 100 = P 2R 1 − P 1 R 2 200 × 10 − 400 × 4 200 1 = × 100 = × 100 = 50 years 400 2
T=
× 100 =
200 × 100 2000 − 1600
43. A sum of money is lent at S.I. for 6 years. If the same amount is paid at 4% higher, Arun would have got Rs. 120 more. Find the principal A. Rs. 200
B. Rs. 600
C. Rs. 400
D. Rs. 500
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : This means, simple interest at 4% for that principal is Rs.120
P=
100 × SI 100 × 120 100 × 30 = = = 100 × 5 = 500 RT 4×6 6
44. Mr. Mani invested an amount of Rs. 12000 at the simple interest rate of 10% per annum and another amount at the simple interest rate of 20% per annum. The total interest earned at the end of one year on the total amount invested became 14% per annum. Find the total amount invested. A. Rs. 25000
B. Rs. 15000
C. Rs. 10000
D. Rs. 20000
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option D Explanation : -----------------------------------------------------------------------------------Solution 1 -----------------------------------------------------------------------------------Let his investments be Rs.12000 and Rs.x Rs. 12000 is invested at the simple interest rate of 10% per annum for 1 year
Simple Interest =
PRT 12000 × 10 × 1 = = Rs. 1200 100 100
Rs. x is invested at the simple interest rate of 20% per annum for 1 year
Simple Interest =
PRT x × 20 × 1 = 100 100
Total interest = Rs. (1200 +
i.e., Rs. (1200 +
⇒ (1200 +
= Rs.
x 5
x ) 5
x ) is the simple interest for Rs.(12000 + x) at 14% per annum for 1 year 5
(12000 + x) × 14 × 1 x )= 5 100
⇒ 120000 + 20x = 14 × 12000 + 14x ⇒ 6x = 14 × 12000 − 120000 = 48000 ⇒ x = 8000 Total amount invested = 12000 + x = 12000 + 8000 = Rs. 20000 -----------------------------------------------------------------------------------Solution 2 ------------------------------------------------------------------------------------
If an amount P1 is lent out at simple interest of R1% per annum and another amount P2 at simple interest rate of R2% per annum, then the rate of interest for the whole sum can be given by
P1 R1 + P 2 R2 P1 + P2
R=
[Read more ...] P1 = Rs. 12000, R1 = 10% P2 = ?, R2 = 20% R = 14%
14 =
12000 × 10 + P 2 × 20 12000 + P 2
12000 × 14 + 14P 2 = 120000 + 20P 2 6P 2 = 14 × 12000 − 120000 = 48000 ⇒ P 2 = 8000 Total amount invested = (P1 + P2) = (12000 + 8000) = Rs. 20000
45. A lent Rs. 4000 to B for 2 years and Rs. 2000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is : A. 14%
B. 15%
C. 12%
D. 13.75%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the rate of interest per annum be R% Simple Interest for Rs.4000 for 2 years at R% + Simple Interest for Rs.2000 for 4 years at R% = 2200
4000 × R × 2 100
+
2000 × R × 4 100
80R + 80R = 2200 160R = 2200 16R = 220 4R = 55 R=
55 = 13.75% 4
= 2200
Basics Con cepts an d Formu las - Simplification 1.
Order of Operations - BODMAS Rule
What is BODMAS Rule? BODMAS rule defines the correct sequence in which operations are to be performed in a given mathematical expression to find its value. In BODMAS, B = Bracket, O = Order (Powers, Square Roots, etc.) DM = Division and Multiplication (left-to-right) AS = Addition and Subtraction (left-to-right) This m eans, to sim plify an expression, the follow ing order m ust be follow ed. i. Do operations in Brackets first, strictly in the order (), {} and [] ii. Evaluate exponents (Powers, Roots, etc.) iii. Perform division and multiplication, working from left to right. (division and multiplication rank equally and done left to right). iv. Perform addition and subtraction, working from left to right. (addition and subtraction rank equally and done left to right). In India, the term BODMAS is used whereas in USA the acronym PEMDAS is used. The full form of PEMDAS is "Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction". Both are correct and only difference is that these are used in different parts of the world. Some other variations which are used across the world to represent the same concept are BIDMAS, ERDMAS, PERDMAS and BPODMAS. Exam ples a. 12 + 22 ÷ 11 × (18 ÷ 3)2 - 10 = 12 = 12 = 12 = 84
+ 22 ÷ 11 × 62 - 10 (∵ Brackets first) + 22 ÷ 11 × 36 - 10 (∵ exponents) + 2 × 36 - 10 = 12 + 72 - 10 (∵ division and multiplication, left to right) - 10 = 74 (∵ Addition and Subtraction, left to right)
b. 4 + 10 - 3 × 6 / 3 + 4 = 4 + 10 - 18/3 + 4 = 4 + 10 - 6 + 4 (∵ division and multiplication, left to right) = 14 - 6 + 4 = 8 + 4 = 12 (∵ Addition and Subtraction, left to right)
2.
Modulus of a Real Num ber
For any real number
|x| = {
x −x
x , the absolute value or modulus of x is denoted by |x| and is defined as
if x ≥ 0 if x < 0
Hence, the Modulus(absolute value) of Exam ple |8| = |-8| = 8
x is always either positive or zero, but never negative
1. The price of 80 apples is equal to that of 120 oranges. The price of 60 apples and 75 oranges together is Rs.1320. The total price of 25 apples and 40 oranges is A. Rs. 660
B. Rs. 620
C. Rs. 820
D. Rs. 780
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let the price of one apple = a and price of one orange = b The price of 80 apples is equal to that of 120 oranges 80a = 120b => 2a = 3b
2a -----(Equation 1) 3
⇒b=
price of 60 apples and 75 oranges together is Rs.1320=>60a + 75b = 1320 => 4a + 5b = 88
5(2a) = 88 (∵ Substituted the value of b from equation 1) 3
⇒ 4a +
=> 12a + 10a = 88 × 3 => 6a + 5a = 44 × 3 => 11a = 44 × 3=> a = 4 × 3 = 12
b=
2a 2 × 12 = =8 3 3
Total price of 25 apples and 40 oranges = 25a + 40b = (25 × 12) + (40 × 8) = 300 + 320 = 620
2. The price of 24 apples is equal to that of 28 oranges. The price of 45 apples and 60 oranges together is Rs.1350. The total price of 30 apples and 40 oranges is A. Rs.920
B. Rs.940
C. Rs.880
D. Rs.900
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : -------------------------------------------------------------------------------------Solution 1 -------------------------------------------------------------------------------------Let the price of one apple = a and price of one orange = b Price of 24 apples is equal to that of 28 oranges 24a = 28b => 6a = 7b
⇒b=
6a -----(Equation 1) 7
price of 45 apples and 60 oranges together is Rs.1350 => 45a + 60b = 1350 => 3a + 4b = 90
⇒ 3a + ⇒ a+
4(6a) = 90 (∵ Substituted the value of b from equation 1) 7
4(2a) = 30 7
=> 7a + 8a = 30 × 7 = 210 => 15a = 210
⇒a=
b=
210 42 = = 14 15 3
6a 6 × 14 = = 6 × 2 = 12 7 7
Total price Price of 45 Price of 15 Price of 30
of 30 apples and 40 oranges= 30a + 45b = (30 × 14) + (40 × 12) = 420 + 480 = 900-----------------------------------------------------------------------------apples and 60 oranges = Rs.1350 apples and 20 oranges = Rs.1350/3 = Rs.450 (∵ Divided LHS and RHS by 15) apples and 40 oranges = Rs.450 × 2 = Rs.900 (∵ Multiplied LHS and RHS by 2)
3. There are two buildings P and Q. If 15 persons are sent from P to Q, then the number of persons in each building is the same. If 20 persons are sent from Q to P, then the number of persons in P is double the number of persons in Q. How many persons are there in building P? A. 80
B. 140
C. 120
D. 90
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the number of persons in building P = p and the number of persons in building Q = q If 15 persons are sent from P to Q, then the number of persons in each building is the same => p-15 = q+15 => p - q = 30 ----(Equation 1) If 20 persons are sent from Q to P, then the number of persons in P is double the number of persons in Q => 2(q - 20) = (p + 20) => 2q - 40 = p + 20 => 2q - p = 60 ----(Equation 2) (Equation 1) + (Equation 2)=> q = 90 From Equation 1, p = 30 + q = 30 + 90 = 120 i.e., Building P has 120 persons
4. The price of 3 tables and 4 chairs is Rs. 3300. With the same money one can buy 2 tables and 10 chairs. If one wants to buy 1 table and 1 chair, how much does he need to pay? A. Rs.940
B. Rs.1050
C. Rs.1040
D. Rs.950
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let price of a table = t and price of a chair = c 3t + 4c = 3300 ---(Equation 1) 2t + 10c = 3300 => t + 5c = 1650 ---(Equation 2) (Equation 2) × 3 => 3t + 15c = 4950 ---(Equation 3) (Equation 3) - (Equation 1) => 11c = 1650 => c = 150 Substituting the value of c in equation 1, we get 3t + (4 × 150) = 3300 => 3t = 3300 – 600 = 2700 =>t = 2700/3 = 900 Cost of 1 table and 1 chair = c + t = 150 + 900 = 1050
5. There are 6 working days in a regular week and for each day, the working hours are 10. A man earns Rs. 2.10 per hour for regular work and Rs. 4.20 per hour for overtime. If he earns Rs.525 in 4 weeks, how many hours did he work? A. 245
B. 285
C. 275
D. 255
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Regular working hours in 4 weeks = (4 × 6 × 10) = 240 hours Amount earned by working in these regular working hours = 240 × 2.10 = Rs.504 Additional amount he earned = 525 - 504 = Rs.21
Hours he worked overtime = 21/4.2 = 210/42 = 5 hours Total hours he worked = 240 + 5 = 245 hours
6. A man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, then the number of hens will be A. 22
B. 24
C. 26
D. 20
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let number of hens = h and number of cows = c number of heads = 48 => h + c = 48 ---(Equation 1) number of feet = 140 => 2h + 4c = 140 => h + 2c = 70 ---(Equation 2) (Equation 2) - (Equation 1) gives 2c - c = 70 - 48 => c = 22 Substituting the value of c in Equation 1, we get h + 22 = 48 => h = 48 - 22 = 26 i.e., number of hens = 26
7. A sum of Rs.2200 has been divided among A, B and C such that A gets 1/4 of what B gets and B gets 1/5 of what C gets. What is B's share? A. Rs.341
B. Rs.364
C. Rs.372
D. Rs.352
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Let C's share = Rs. x Then B's share =
x 5
A's share =
x 1 x × = 5 4 20
Hence, x +
x x + = 2200 5 20
⇒ x (1 + ⇒ x(
1 1 ) = 2200 + 5 20
25 ) = 2200 20
5 ⇒ x ( ) = 2200 4 ⇒x=
2200 × 4 = 440 × 4 = Rs.1760 5
B's share =
x 1760 = = Rs.352 5 5
8. A fires 5 shots to B's 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed: A. 30 birds
B. 22 birds
C. 18 birds
D. 38 birds
Hide Answer | Notebook
| Discuss
Here is the answer and explanation Answer : Option A Explanation :
Let the total number of shots = x
Shots fired by A =
5x 8
Shots fired by B =
3x 8
Killing shots by A = Missing shots by B =
5x 1 5x × = 8 3 24 3x 1 3x × = 8 2 16
B has missed 27 times ⇒ ⇒x=
3x = 27 16
27 × 16 = 144 3
Hence, killing shots by A =
5x 5 × 144 5 × 12 = = = 30 24 24 2
i.e., A has killed 30 birds
9. If p - q = 6 and p2 + q2 = 116, what is the value of pq? A. 30
B. 40
C. 20
D. 50
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
2
(a − b) = a 2 − 2ab + b2 [Read more ...] (p - q)2 = p2 - 2pq + q2 (p - q)2 = (p2 + q2)- 2pq 62 = 116 - 2pq 36 = 116 - 2pq 2pq = 80 pq = 40
10. To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two-fifth of its present? A. 63
B. 64.5
C. 62.5
D. 60.5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
---------------------------------------------------------------------------------------Solution 1 (Chain Rule) ---------------------------------------------------------------------------------------Let x buckets of water be required ,if the capacity of the bucket is reduced to two-fifth More buckets, less capacity (Indirect proportion) Hence we can write as
Capacity ⇒ 1 × 25 =
1:
2 } :: x : 25 5
2 ×x 5
2x 5 25 × 5 ⇒x= = 62.5 2 ⇒ 25 =
i.e., 62.5 buckets are needed ----------------------------------------------------------------------------------------Solution 2 -----------------------------------------------------------------------------------------Let capacity of 1 bucket = x Capacity of the tank = 25x New capacity of the bucket = 2x/5
Hence, number of buckets needed =
25x 2x ( ) 5
=
25 × 5 = 62.5 2
----------------------------------------------------------------------------------------Solution 3 ---------------------------------------------------------------------------------------Or more simply, you can assume as capacity of 1 bucket = 1 Then, capacity of the tank = 25 New capacity of the bucket = 2/5
Hence, number of buckets needed =
25 2 ( ) 5
= 62.5
11. John gets on the elevator at the 14th floor of a building and rides up at the rate of 84 floors per minute. At the same time, Vinod gets on an elevator at the 58th floor of the same building and rides down at the rate of 92 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross? A. 38
B. 36
C. 32
D. 35
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let their paths cross after x minutes Then, 84x + 92x = 58 - 14 => 176x = 44
⇒ x=
44 1 = 176 4
Number of floors covered by John in these
1 1 minute = 84 × = 21 4 4
Hence, their path cross at 14 + 21 = 35, i.e., at 35th floor
12. A man has Rs. 312 in the denominations of one-rupee notes, five-rupee notes and twenty-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ? A. 36
B. 24
C. 28
D. 32
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A
Explanation : Let the number of notes of each denomination be x Then, x + 5x + 20x = 312 => 26x = 312 => x = 312/26 = 12 the total number of notes that he has = 3x = 3 × 12 = 36
13. Free notebooks were distributed equally among children of a class. The number of notebooks each child got was one-eighth of the number of children. Had the number of children been half, each child would have got 16 notebooks. Total how many notebooks were distributed? A. 602
B. 528
C. 423
D. 512
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : ---------------------------------------------------------------------------------------Solution 1 ---------------------------------------------------------------------------------------Let number of children be n If the number children = n, number of books each child will get = n/8
total books distributed = n ×
n n2 = 8 8
If the number children = n/2, number of books each child will get = 16
total books distributed =
n × 16 = 8n 2
n2 Since total books are same, we can write as = 8n 8 n ⇒ =8 8 ⇒ n = 64 Total number of notebooks those were distributed =
n2 64 × 64 = = 64 × 8 = 512 8 8
----------------------------------------------------------------------------------------Solution 2 (Chain Rule) ----------------------------------------------------------------------------------------Let n be the total number of children. More children, less notebooks(Indirect proportion) If the number children = n, number of books each child will get = n/8 If the number children = n/2, number of books each child will get = 16 Hence we can write as
children
n:
n n } :: 16 : 2 8
n n = × 16 8 2 n 16 ⇒ = =8 8 2 ⇒ n = 8 × 8 = 64
⇒ n×
Then, total number of notebooks those were distributed = n ×
n 64 × 64 = = 64 × 8 = 512 8 8
14. Eight people are planning to share equally the cost of a rental car. If one person withdraws from the arrangement and the others share equally the entire cost of the car, then the share of each of the remaining persons increased by:
2 5 2 C. 7 A.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
1 8 1 D. 7 B.
Let x be the the cost of the rental car When 8 persons share equally, share of one person =
x 8
When 1 person withdraws and other 7 persons share equally, share of one person = Increase in the share =
x x x − = 7 8 56
x ) 56 Required fraction = x ( ) 8 (
=
1 7
(723 + 1992) 2 − (723 − 1992)2 723 × 1992
15.
x 7
A. 4
B. 33
C. 6
D. 1
=?
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
2 (a + b) = a 2 + 2ab + b2 2 (a − b) = a 2 − 2ab + b2
[Read more ...]
(723 + 1992)2 − (723 − 1992) 2 723 × 1992 =
= =
(a + b)2 − (a − b) 2 ab
(where a = 723 and b = 1992)
a2 + 2ab + b2 − (a 2 − 2ab + b2 ) ab 4ab =4 4
16. One-third of Rahul's savings in National Savings Certificate is equal to one-half of his savings in Public Provident Fund. If he has Rs. 1,80,000 as total savings, how much has he saved in Public Provident Fund? A. Rs. 72000
B. Rs. 44000
C. Rs. 58000
D. Rs. 92000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let savings in National Savings Certificate = x and savings in Public Provident Fund = (180000 - x)
1 1 x = (180000 − x) 3 2 ⇒ 2x = 3(180000 − x) ⇒ 2x = 540000 − 3x ⇒ 5x = 540000 ⇒x=
540000 = 108000 5
Savings in Public Provident Fund = (180000 - 108000) = 72000
17. 8 / 4 / 2 = ? A. 4
B. 1
C. 0
D. 2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 8 / 4 / 2 = (8 / 4) / 2 = 2/2 = 1
18. 20 + 20 × 2 = ? A. 40
B. 50
C. 60
D. 70
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : 20 + 20 × 2 = 20 + 40 = 60
19. 25 / 5 × 5 = ? A. 25
B. 15
C. 20
D. 30
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 25 / 5 × 5 = 5 × 5 = 25
20. 5 × 5 / 5 = ? A. 5
B. 1
C. 10
D. 25
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 5 × 5 / 5 = 25 / 5 = 5
21. b - [b -(a+b) - {b - (b - a+b)} + 2a] = ? A. 0
B. 4a
C. a
D. -2a
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : b - [b -(a+b) - {b - (b - a+b)} + 2a] = b - [b -a -b - {b -b +a -b} + 2a] = b - [b -a -b - {a -b} + 2a] = b - [b -a -b - a +b + 2a] = b - [b] =b - b =0
1 1 1 +3 +4 = ? 3 2 4 1 1 B. 10 A. 10 6 12 1 1 C. 10 D. 10 4 2 22.
2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
2
1 1 1 +3 +4 3 2 4
=
7 7 17 + + 3 2 4
=
(28 + 42 + 51) 12
=
121 12
= 10
1 12
23. If a * b = 2a - 4b + 2ab, then 2*3 + 3*2 = ? A. 2
B. 0
C. 14
D. 12
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : a * b = 2a - 4b + 2ab Hence, 2*3 = 2(2) - 4(3) + 2(2 × 3) = 4 - 12 + 12 = 4 3*2 = 2(3) - 4(2) + 2(3 × 2) = 6 - 8 + 12 = 10 ∵ 2*3 + 3*2 = 4 + 10 = 14
1
24.
4+
1 4+
=? 1
4+
24 305 72 C. 305
1 4
1 64 81 D. 320
A.
B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
1 4+
1 4+
= 1
1 4+ 4
1 4+
1
4+
1 17 ( ) 4
=
1 4+
1 4+
4 17
=
1 1 4+ 72 ( ) 17
=
1 1 = 17 305 4+ ( ) 72 72
=
72 305
25. If the number of boys in a class are 8 times the number of girls, which value can never be the total number of students? A. 27
B. 45
C. 81
D. 42
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option D Explanation : Let the number of girls = x and the number of boys = 8x Then, total number of students = x + 8x = 9x i.e., the total number of students must be a multiple of 9 From the given choices, 42 cannot be a multiple of 9. Hence, 42 cannot be the total number of students.
26.
What fraction of
A.
5 4
C.
2
1 4
3 1 needs to be added to itself to become 2 5 4 3 B. 2 4 3 D. 4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Let x be the required fraction 3 3 1 x+ = 2 5 5 4 3 3 9 x+ = 5 5 4 ⇒
3 9 3 (45 − 12) 33 x= − = = 5 4 5 20 20
⇒x=
33 5 11 3 × = =2 20 3 4 4
27. An organization decided to raise Rs. 6 lakh by collecting equal contribution from each of its employees. If each of them had contributed Rs. 60 extra, the contribution would have been Rs. 6.24 lakh. How many employees are there in that organization? A. 400
B. 300
C. 200
D. 100
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Required number of employees =
(624000 - 600000) 24000 = = 400 60 60
28. In a group of ducks and cows, the total number of legs are 28 more than twice the number of heads. Find the total number of cows. A. 14
B. 12
C. 16
D. 8
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let the number of ducks be d and number of cows be c Then, total number of legs = 2d + 4c = 2(d + 2c) total number of heads = c + d Given that total number of legs are 28 more than twice the number of heads => 2(d + 2c) = 28 + 2(c + d) => d + 2c = 14 + c + d => 2c = 14 + c => c = 14
i.e., total number of cows = 14
29. If a - b = 6 and a2 + b2 = 116, then what is the value of ab? A. 20
B. 40
C. 60
D. 80
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
2
(a − b) = a 2 − 2ab + b2 [Read more ...] => (a - b)2 = a2 - 2ab + b2 => 62 = 116 - 2ab => 36 = 116 - 2ab => 2ab = 116 - 36 = 80 => ab = 40
30. A room has equal number of men and women. Eight women left the room, leaving twice as many men as women in the room . What was the total number of men and women present in the room initially? A. 32
B. 34
C. 28
D. 30
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Assume that initial number of men = initial number of women = x 2(x-8) = x => 2x - 16 = x => x = 16 Total number of men and women = 2x = 2 × 16 = 32
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31. From a group of boys and girls, 15 girls leave. They are then left 2 boys for each girl. After this, 45 boys leave. There are then 5 girls for each boy. Find the number of girls in the beginning A. 40
B. 20
C. 32
D. 60
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : -------------------------------------------------------------------------------------Solution 1 -------------------------------------------------------------------------------------Assume that initial number of boys = b initial number of girls = g 15 girls leave and they are then left 2 boys for each girl => 2(g - 15) = b => 2g - b = 30 ---(Equation 1) After this, 45 boys leave. There are then 5 girls for each boy 5(b - 45) = (g - 15) 5b - g = 210---(Equation 2) (Equation 2) × 2 => 10b - 2g = 420---(Equation 3) (Equation 1) + (Equation 3) => 9b = 450 => b = 450/9 = 50 Substituting this value of b in Equation 1, we get 2g - 50 = 30 => 2g = 80 => g = 40 i.e., number of girls in the beginning = 40 -------------------------------------------------------------------------------------Solution 2 -------------------------------------------------------------------------------------Assume the number of boys at present = x Then, the number of girls at present = 5x Before the boys left, the number of boys were (x+45) and number of girls were 5x
Hence, 2(5x) = x + 45 => 10x = x + 45 => x = 5 Number of girls in the beginning = (5x + 15) = (5×5 + 15) = 40
32. How many pieces of 85 cm length can be cut from a rod of 42.5 meters long? A. 15
B. 20
C. 2
D. 50
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Number of pieces =
4250 850 = = 50 85 17
33. In a garden, there are 10 rows and 12 columns of mango trees. The distance between the two trees is 2 metres and a distance of one metre is left from all sides of the boundary of the garden. What is the length of the garden? A. 30 m
B. 28 m
C. 26 m
D. 24 m
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Between the 12 mango trees, there are 11 gaps and each gap has 2 meter length Also, 1 meter is left from all sides of the boundary of the garden. Hence, length of the garden = (11 × 2) + 1 + 1 = 24 meter
34. In a garden, 26 trees are planted at equal distances along a yard 300 metres long, one tree being at each end of the yard. What is the distance between two consecutive trees? A. 10
B. 20
C. 14
D. 12
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option D Explanation : 26 trees have 25 gaps between them. Length of each gap = 300/25 = 12 i.e., distance between two consecutive trees = 12
35. A boy was asked to multiply a number by 22. He instead multiplied the number by 44 and got the answer 308 more than the correct answer. What was the number to be multiplied? A. 16
B. 10
C. 14
D. 12
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the number be x 22x + 308 = 44x => 44x - 22x = 308 => 22x = 308 => x = 308/22 = 154/11 = 14
36. In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If he attempts all 80 questions and secures 120 marks, How many questions does he answer correctly? A. 30
B. 60
C. 50
D. 40
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the number of correct answers be x Then, number of wrong answers = (80 - x) 4x – (80 - x) = 120 => 4x – 80 + x = 120 => 5x = 200 => x = 200/5 = 40
i.e., he does 40 questions correctly
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I mportan t Formu las - Squ are Root an d Cu be Root 1. Square Root
If a2 = b,
√ b= a
(i.e., square root of b is a.)
Examples
√4=2
22 = 4 and 52 = 25 and
√ −− 25 = 5
2. Cube Root
If a3 = b,
√3 b = a
(i.e., cube root of b is a.)
Examples
23 = 8 and 53 = 125 and
√3 8 = 2 3 −−− √ 125 = 5
3. Im portant Properties 1.
2.
− ×√− −− m n =√− mn √− − −− − m √− m = √ − n √n
3.
1 −−−−−−−−−−−−−−− − 1− n −−−−−−−−−−−− − −−−−−−− − √ a. √ a. √ a ⋯ n times = a 2
4.
−−−−−−−−−−− −−−−−−−− a−−−− ⋯ ∞− = a √ a. √ a. √ −
5.
if
−−−−−−−−−−−−−− − −−−−−−−−− a−−−− ⋯ ∞−− = p, then p(p − 1) = a √ a +√ a +√−
−−− x √ 648 1. If −−− = x , find the value of x. √ 512 A. 24
B. 12
C. 48
D. 36
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
−−− x √ 648 −−− = x √ 512 −−− −−− −−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−− ⇒ x 2 = √ 512 × √ 648 = √ 512 × 648 = √ 2 × 2 × 2 × 64 × 2 × 2 × 2 × 81 = 2×2×2×8×9 − −−−−−−−−−−−− − x = √ 2 × 2 × 2 × 8 × 9 = 2 × 4 × 3 = 24
2.
−−−−− √ 5.4756 = ?
A. 2.24
B. 1.24
C. 1.34
D. 2.34
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
−−−−− √ 5.4756 = 2.34
3.
−−− −− If 3√ 5 + √ 125 = 17.88, then what will be the value of √ 80 + 16√ 5?
A. 21.66
B. 13.41
C. 22.35
D. 44.7
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
−−− 3√ 5 + √ 125 = 17.88 −−− ⇒ 3√ 5 + √ − 25 × 5− = 17.88 ⇒ 3√ 5 + 5√ 5 = 17.88 ⇒ 8√ 5 = 17.88 ⇒√5=
17.88 = 2.235 8
−−− 80 + 16√ 5 = √ − 16 × 5− + 16 √ 5 √ −− = 4√ 5 + 16√ 5 = 20 √ 5 = 20 × 2.235 = 44.7
4. The cube root of 0.000729 is A. 0.09
B. 0.9
C. 0.21
D. 0.11
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
(0.000729)
1/3
=(
729 106
1/3
)
=
9 102
=
9 = .09 100
5. What is the least perfect square which is divisible by each of 21, 36 and 66? A. 213444
B. 214434
C. 214344
D. 231444
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : LCM of 21, 36, 66 = 2772 ie, all multiples of 2772 are divisible by 21, 36 and 66 Prime factorization of 2772 is,
2772 = 2 × 2 × 3 × 3 × 7 × 11 ie, to make it a perfect square, we have to multiply it by 7 and 11 Hence, required number = 2772 × 7 × 11 = 213444
−− √− 144 11 15 × −−− × −−− is equal to: 6. 11 √ 225 √ 196 A. 0.85
B. 0.72
C. 2.8
D. 0.4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
−− √− 144 11 15 × −−− × −−− 11 √ 225 √ 196 12 11 15 = × × 11 15 14 12 = 14 6 = 7 = 0.85
1 ) 7. (√ 7 − √7 36 A. √7 36 C. 7
2
simplifies to: 7 36 7 D. −− √ 36 B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
1 (√ 7 − ) √7
2
1 1 ) = (√ 7 ) − 2 × √ 7 × +( √7 √7 1 1 36 = 7−2+ = 5+ = 7 7 7 2
2
8. The square root of 16641 is A. 129
B. 121
C. 211
D. 229
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
−−−−− √ 16641 = 129
9.
−−−−−− √− 0.0576 × ?− = 0.24.
A. None of these
B. 10
C. 1
D. 0.1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
−−−−−−−− √ 0.0576 × x = 0.24. ⇒ 0.0576 × x = (0.24) 2 ⇒ 0.0576 × x = 0.0576 ⇒x=1
10.
−−−−−−−− √− 0.000256 × ?− = 1.6.
A. 0.1
B. 10
C. 10000
D. 1000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
−−−−−−−−−− √ 0.000256 × x = 1.6 2
⇒ 0.000256 × x = (1.6) ⇒ 0.000256 × x = 2.56 2.56 2560000 ⇒x= = = 10000 0.000256 256
11. How many two-digit numbers satisfy this property : The last digit (units digit) of the square of the two-digit number is 8 ? A. 1
B. 2
C. 3
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : A number ending with 8 can never become a perfect square Let's examine this in detail 1 × 1 =1 Hence, if the unit digit of a number is 1, unit digit of its square is 1 2 × 2 =4 Hence, if the unit digit of a number is 2, unit digit of its square is 4 3 × 3 =9 Hence, if the unit digit of a number is 3, unit digit of its square is 9 4 × 4 = 16 Hence, if the unit digit of a number is 4, unit digit of its square is 6 5 × 5 = 25 Hence, if the unit digit of a number is 5, unit digit of its square is 5 6 × 6 = 36 Hence, if the unit digit of a number is 6, unit digit of its square is 6 7 × 7 = 49 Hence, if the unit digit of a number is 7, unit digit of its square is 9 8 × 8 = 64 Hence, if the unit digit of a number is 8, unit digit of its square is 4 9 × 9 = 81 Hence, if the unit digit of a number is 9, unit digit of its square is 1 0 × 0 =0 Hence, if the unit digit of a number is 0, unit digit of its square is 0
12.
−−−−−−−−−− if a = 0.2917, then the value of √ 4a 2 − 4a + 1 + 3a is :
A. 0.5834
B. 0.2917
C. 1.2917
D. 2.2917 | Discuss
Hide Answer | Notebook Here is the answer and explanation Answer : Option C Explanation :
−−−−−−−−−−−−−−−−−−−− −−−−−−− −−−−−−−−−− √ 4a 2 − 4a + 1 + 3a = √ (1) 2 − 2 × 1 × 2a + (2a ) 2 + 3a = √ (1 − 2a) 2 + 3a = 1 − 2a + 3a = 1 + a = 1 + 0.2917 = 1.2917
13. A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 98.01, the number of the member is the group is: A. 101
B. 98
C. 99
D. 88
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the number of members = n Then n2 = Total amount of money collected = 9801 paise
−−−− ie, n = √ 9801 = 99 i.e., number of members = 99
14.
if √ 7 = 2.645, then find the value of
√ 7 10 −−− − + √ 175 2 √7
A. 7.22
B. 8.92
C. 6.72
D. 10.77
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
√ 7 10 −− 175 − +√− 2 √7 =
√ 7 10 −−−−− − + √ 7 × 25 2 √7
=
√ 7 10 − + 5√ 7 2 √7 2
=
(√ 7) − (2 × 10) + (5 √ 7 × 2√ 7) 2√ 7
=
7 − 20 + 70 57 = 2√ 7 2√ 7
=
28.5 28.5 28500 = = = 10.77 2.645 2645 √7
Please note that
57
57 can be solved further in the below lines as well 2√ 7
57√ 7 14 2√ 7 2√ 7 × √ 7 57 × 2.645 150.765 = = = 10.77 14 14
15.
=
if x =
57 × √ 7
=
√ 3+1 √ 3−1 and y = , what is the value of (x 2 + y 2 ) √ 3−1 √ 3+1
A. 15
B. 14
C. 13
D. 10
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
x=
(√ 3 + 1) (√ 3 − 1)
=
(√ 3 + 1) (√ 3 + 1) (√ 3 − 1) (√ 3 + 1)
(√ 3 − 1) (√ 3 − 1) √ 3−1 y= = √ 3+1 (√ 3 + 1) (√ 3 − 1) x 2 + y 2 = (2 + √ 3)
16.
=
2
+ (2 − √ 3)
2
=
(√ 3 + 1)
2
=
3−1
(√ 3 − 1) 3−1
2
=
3 + 2√ 3 + 1 2
3 − 2√ 3 + 1 2
=
=
4 − 2√ 3 = 2−√ 3 2
= (4 + 4 √ 3 + 3) + (4 − 4√ 3 + 3) = 2(4 + 3) = 14
−− −− The square root of (14 + 2√ 13 )(14 − 2 √ 13 ) is
A. 8
B. 9
C. 10
D. 12
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
−−−−−−−−−−−−2− −−−−−−−−−−−−−−−−−−− − −−−−−−−−−− √ (14 + 2√ −− 13 )(14 − 2√ −− 13 ) = √ (14) 2 − (2√ −− 13 ) = √ − 196 − (4 × 13) −−−−−−− −−− = √ 196 − 52 = √ 144 = 12
17.
−−−−−−−− −−− √ 248 + √ 64 =?
A. 21
B. 14
C. 12
D. 16
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
−−−−−−−− −−− −−− −−−− √ 248 + √ 64 = √ − 248 + 8− = √ 256 = 16
−−−9− 18. √ 1 =? 16 1 A. 1 B. 1 6 1 C. 1 D. 1 2
4 + 2√ 3 = 2+√ 3 2
1 3 1 4
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option D Explanation :
−−−9− −25 −− −− 1 √ 25 5 √ 1 √ = = −− = = 1 16 16 √ 16 4 4
−−−−−−−−−−−−−−−−− − −−−−−−−−−−−− −−−−−−−− 19. √ 41 − √ 21 + √ 19 − √ 9 =? A. 3
B. 4
C. 5
D. 6
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
−−−−−−−−−−−−−−−−− − −−−−−−−−−−−− −−−−−−−−−−−−−−−− −−−−−−−−−−−−− −−−−−−−− −−−−−−−−−− −− −−−−−−−−− −−− √ 41 − √ 21 + √ 19 − √ 9 = √ 41 − √ 21 + √ − 19 − 3− = √ 41 − √ 21 + √ −− 16 −−−−−−−−−− − −−−−−−−− −−− −−− 21 + 4− = √ 41 − √ −− 25 = √ − 41 − 5− = √ −− 36 = 6 = √ 41 − √ −
−−−−− 12 =? 20. √ 4 125 3
A. 1 C.
1
2 5 4 D. 1 5
B.
3 5
1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
−−−− −512 −−− −2−−−−−−−−−−−−−−−−−−−−−−−−− − 2×2×2 12− ×2×2×2×2×2×2×2×2 8 3 3 3 √3 4 √ √ = = = = =1 125 125 5×5×5 5 5 5
21.
−−−−−−− −−−−−− √3 √ − 0.000064 =?
A. 0.2
B. 0.4
C. 0.1
D. 0.3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A
Explanation :
−−−− − −−−− −−− − −−− −−−− 3 −−− 64 8− 8− 2 √ 64 √8 −−−−−−− − −−−−− − 3 3 3 3 √ √ 0.000064 = √ √ =√ =√ =√ = 3 −− = = 0.2 − − − 6 3 1000 √ 10 10 10 10 √ 106 3
22. What is the smallest number by which 3600 be divided to make it a perfect cube? A. 110
B. 210
C. 420
D. 450
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : 3600 = 24 × 32 × 52 ie, the smallest number by which 3600 be divided to make it a perfect cube = 2 × 32 × 52 = 2 × 9 × 25 = 450
23. 140√? + 315 = 1015 A. 25
B. 15
C. 5
D. 50
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 140√x + 315 = 1015 => 140√x = 1015 - 315 = 700 = 140 × 5 => √x = 5 => x = 52 = 25
24.
√
−−−−−− 0.289− 0.00121
+ √ 9 =?
A. 18.45
B. 15.45
C. 12.45
D. 15 | Discuss
Hide Answer | Notebook Here is the answer and explanation Answer : Option A Explanation :
−−−−−−− −−−−− −−−−− 0.289 28900 170 √ 28900 √ +√ 9= √ +3 = + 3 = + 3 = 15.45 + 3 = 18.45 −− 0.00121 121 11 √− 121
25. A man plants 49284 apple trees in his garden and arranges them so that there are as many rows as there are apples trees in each row. The number of rows is A. 182
B. 202
C. 222
D. 122
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let n be the number of rows Then n × n = 49284
−−−−− ie, n = √ 49284 = 222
26.
−− −− What is the difference between (√ 18 + √ 3) and (2 + √ 12 )
A. 3√2 - 2√3
B. √2 - 2√3
C. 2√2 - √3
D. 2√2 + √3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
−− −− Difference = (√ 18 + √ 3) − (√ 2 + √ 12 ) −−−− −−−− = (√ 2 × 9 + √ 3) − (√ 2 + √ 3 × 4 ) = (3√ 2 + √ 3) − (√ 2 + 2√ 3) = 3√ 2 + √ 3 − √ 2 − 2√ 3 = 2√ 2 − √ 3
27.
−− if a*b = a + b − √ ab , then find the value of 16 ∗ 9
A. 14
B. 13
C. 12
D. 11
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
−−−−− 16 ∗ 9 = 16 + 9 − √ 16 × 9 −−−−− = 25 − √ 16 × 9 = 25 − (√ −− 16 × √ 9) = 25 − (4 × 3) = 25 − 12 = 13
1
−
1 √ −− 11
√9 10 + √ −− 99 1 28. + = 1 1 ? 2 + −− √ 9 √ 11 A. 1
B. 2
C. 3
D. 4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
1 1 − −− √ 9 √ 11 1 1 . √ 9 + √ −− 11 .
−− 10 + √ 99 1 ]= +[ x 2
√ −− 11 − √ 9 10 + √ −− 99 1 ]+ [ ]= ⇒ [ −− x 2 √ 11 + √ 9 −− −− (√ 11 − √ 9) (√ 11 − √ 9 ) ⇒ [ −− −− (√ 11 + √ 9) (√ 11 − √ 9 ) ⇒
−− (√ 11 − √ 9)
2
11 − 9
−− 10 + √ 99 1 ]+ [ ]= x 2
−− 10 + √ 99 1 ]= +[ x 2
−− −− 11 − 2√ 11 √ 9 + 9 10 + √ 99 1 ]+ [ ]= ⇒[ 2 x 2 −− −− 20 − 2√ 99 10 + √ 99 1 ]+ [ ]= ⇒[ 2 x 2 10 + √ −− 99 1 −− ]= ⇒ (10 − √ 99 ) + [ x 2 ⇒
−− −− (10 − √ 99 ) (10 + √ 99 ) x
⇒
(100 − 99) 1 = x 2
⇒
1 1 = x 2
=
1 2
⇒x=2
29.
−− What is the square root of (8 + 2√ 15 )?
A. 2√5 + 2√3
B. √5 + √3
C. √2 + √6
D. 2√2 + 2√6
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B
Explanation :
8 + 2√ −− 15 = 5 + 3 + 2 × √ 5 × √ 3 = (√ 5) 2 + (√ 3)2 + (2 × √ 5 × √ 3) = (√ 5 + √ 3)
2
−−−−−−−− −− − √ −−−−−−−−2− (8 √ 15 ) = (√ 5 + √ 3) = √ 5 + √ 3 + 2 Hence, √
30.
1 (√ 9 − √ 8)
−
1 (√ 8 − √ 7)
+
1 (√ 7 − √ 6)
−
1 (√ 6 − √ 5)
A. 1.5
B. .25
C. 0.5
D. 5
+
1 (√ 5 − √ 4)
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
(√ 9 + √ 8) 1 = (√ 9 − √ 8) (√ 9 − √ 8)(√ 9 + √ 8)
=
(√ 9 + √ 8) = √9+√ 8 9−8
Similarly all other terms can be rewritten. Thus, 1 1 1 1 1 − + − + (√ 9 − √ 8) (√ 8 − √ 7) (√ 7 − √ 6) (√ 6 − √ 5) (√ 5 − √ 4) = (√ 9 + √ 8) − (√ 8 + √ 7) + (√ 7 + √ 6) − (√ 6 + √ 5) + (√ 5 + √ 4) = √ 9+√ 4= 3+2 = 5
=?
I mportan t Formu las - Stocks an d Sh ares 1. Introduction To start a big business or an industry, a large amount of money is required. This may be beyond the capacity of one or two individuals. Hence, a number of individuals join hands to form a company called Joint Stock Company. 2. Stock Capital The total amount of money required by the company is called the stock Capital. 3. Shares or Stock The whole capital of the company is divided into equal units. Each unit is called a share or a stock. 4. Shareholder or Stockholder Each individual who purchases one or more shares is called a shareholder (stockholder) of the company. The company issues share certificates to each of its shareholders indicating the number of shares allocated and the value of each share. 5. F ace v alue Face value of a share is the value printed on the share certificate. It is also called nominal value or par value The face value of a share always remains the same. 6. Market v alue The stocks of different companies can be traded (bought or sold) in the market through brokers at stock exchanges. The price at which a stock is traded in the market is called its market value. a. If the market value of a stock is higher than its face value, the stock is traded at a premium or above Par b. If the market value of a stock is same as its face value, the stock is traded at par c. If the market value of a stock is less than its face value, the stock is traded at a discount or below Par The market value (trading price) of a share can vary time to time.
Let's consider an example . Assume that the face value of a company X is Rs.10 and it is now traded at a premium of Rs.2. Then its market value now is (Rs.10 + Rs.2) = Rs.12. Similarly, if the company X having face value of Rs.10 is now traded at a discount of Rs.2, it means the market value of X now is (Rs.10 – Rs.2) is Rs.8 7. Dividend The annual profit of a company is distributed among its shareholders. The distributed profit is called the dividend. Dividends are declared annually, semi-annually and quarterly as per the company regulations. Dividend on a share is normally expressed as a certain percentage of its face value. Sometimes, it is also expressed as a certain amount per share. 8. Brokerage As we have seen earlier, stocks of different companies can be traded (bought or sold) in the market through brokers at stock exchanges. The brokers charge is called brokerage Brokerage is added to the cost price when the stock is purchased. Brokerage is subtracted from the selling price when the stock is sold Number of shares held by a person 9.
=
Total Investment of the person Investment in 1 share
=
Total Income Income from one share
=
Total Face Value Face value of 1 share
10. What does the statem ent "Rs.100 , 8% stock at Rs.110" m ean? It means, The face value of the stock = Rs.100 Market value (MV) of the stock = Rs.110 Annual dividend per share = 8% of the face value =
100 × 8 = Rs.8 100
Ie, here , an investment of Rs.110 gives an annual income of Rs.8 Rate of interest per annum = Annual income from an investment of Rs.100 =
8 × 100 = 7.27% 110
Please note that generally investors invest in shares not merely because of this annual return. They also will have a capital gain if the market value the share goes higher.
1. A man buys Rs. 20 shares paying 9% dividend. The man wants to have an interest of 12% on his money. What is the market value of each share? A. Rs.12
B. Rs.18
C. Rs.15
D. Rs.21
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Face value of each share = Rs.20 Dividend per share = 9% of 20 =
9 × 20 9 = 100 5
He needs to have an interest of 12% on his money
ie, Money Paid for a share × Money Paid for a share =
12 9 = 100 5
9 100 × = 15 5 12
ie, Market Value of the share = Rs.15
2. A man invested Rs.1552 in a stock at 97 to obtain an income of Rs.128. What is the dividend from the stock? A. None of these
B. 9.7%
C. 7.5%
D. 8%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : By investing Rs.1552, income = Rs.128
By investing Rs.97, income =
128 × 97 =8 1552
ie, dividend = 8%
3. The cost price of a Rs. 100 stock at 4 discount, when brokerage is A. Rs. 96.25
B. Rs. 96.2
C. Rs. 97.25
D. Rs. 97.5
1 % is 5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Cost Price = 100 − 4 +
1 = 96.2 5
4. In order to obtain an income of Rs. 650 from 10% stock at Rs. 96, one must make an investment of A. Rs. 3100
B. Rs. 6500
C. Rs. 6240
D. Rs. 9600
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Market Value = Rs. 96. Required Income = Rs. 650. Here face value is not given. Take face value as Rs.100 if it is not given in the question To obtain Rs.10 (ie,10% of the face value 100), investment = Rs.96
To obtain Rs.650, investment =
5. By investing in
16
96 × 650 = 6240 10
2 % stock at 64, one earns Rs. 1500. The investment made is 3
A. Rs. 9600
B. Rs. 7500
C. Rs. 5640
D. Rs. 5760
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Market Value = Rs.64 Face Value is not given and hence take it as Rs.100
16
2 50 % of the face value = 3 3
ie, to earn
50 , investment = Rs.64 3
Hence, to earn Rs.1500, investment needed =
64 × 3 × 1500 = 5760 50
6. A man invested Rs. 4940 in Rs. 10 shares quoted at Rs. 9.50. If the rate of dividend be 14%, his annual income is A. Rs.728
B. Rs.648
C. Rs.720
D. Rs.622
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Market Value of a share = Rs.9.50 Investment = Rs.4940 Number of shares = 4940/9.50 = 520 Face Value of a share = Rs.10 dividend = 14%
dividend per share =
(10 × 14) = Rs. 1.4 100
His annual income = 520 × 1.4 = Rs.728
7. A man invests some money partly in 12% stock at 105 and partly in 8% stock at 88. To obtain equal dividends from both, he must invest the money in the ratio: A. 31 : 44
B. 31 : 27
C. 16 : 15
D. 35 : 44
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : ---------------------------------------------Solution 1 ----------------------------------------------
In case of stock1, if he invest Rs.105, he will get a dividend of Rs.12 (assume face value = 100) In case of stock2, if he invest Rs.88, he will get a dividend of Rs.8 (assume face value = 100) ie, if he invest Rs.(88*12)/8, he will get a dividend of Rs.12 Required ratio = 105 : (88 × 12)/8 = 105 : (11 × 12) = 35 : (11 × 4) = 35 : 44 ---------------------------------------------Solution 2 ---------------------------------------------Let the amount of money he invest in 12% stock = P Then, Number of shares = P/105 (where 105 is the market value per share) Total dividend = (P/105) × 12 (Assume face value = Rs.100 and hence dividend per share = Rs.12) Let the amount of money he invest in 8% stock = Q Then, Number of shares = Q/88 (where 88 is the market value per share) Total dividend = (Q/88)× 8 (Assume face value = Rs.100 and hence dividend per share = Rs.8) If both needs to give equal dividends,
P Q × 12 = ×8 105 88 P 8 × 105 105 35 35 = = = = Q 88 × 12 11 × 12 11 × 4 44 P:Q = 35:44
8. A man bought 40 shares of Rs. 60 at 5 discount, the rate of dividend being obtained is A. 13.64%
B. 15.5%
C. 14%
D. 14.25%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Face Value of a share = Rs.60
12
1 % The rate of interest 2
He bought each share at Rs.60 - Rs.5 = Rs.55 Number of shares = 40
Dividend = 12
1 25 %= % 2 2
Dividend per share =
60 × 25 = Rs. 7.5 2 × 100
Total dividend = (40 × 7.5) ie, He got a dividend of (40 × 7.5) for an investment of Rs.(40 × 55)
Interest obtained =
40 × 7.5 × 100 = 13.64% 40 × 55
9. By investing Rs. 1800 in 9% stock, Syam earns Rs. 120. The stock is then quoted at A. Rs.135
B. Rs. 96
C. Rs. 85
D. Rs. 122
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Assume that face value = Rs.100. Dividend per share = Rs.9 (as it is a 9% stock) By investing Rs. 1800, he earns Rs.120
Investment needed to earn Rs.9 =
1800 × 9 = Rs.135 120
ie, stock is then quoted (then market value) = Rs.135
10. The market value of a 10.5% stock, in which an income of Rs. 756 is derived by investing Rs. 9000, brokerage being 1⁄4 %, is: A. Rs. 124.75
B. Rs. 108.25
C. Rs. 125.25
D. Rs. 112.20
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A
Explanation : Assume that face value = Rs.100 Dividend per share = Rs.10.5 (as it is 10.5% stock) By investing Rs.9000, earnings = Rs.756
To get an earning of Rs.10.5, investment required =
9000 × 10.5 = Rs.125 756
ie,market value of Rs. 100 stock = Rs.(125- 1/4) = Rs.124.75
11. A 14% stock yielding 8% is quoted at: A. Rs. 125
B. Rs. 83.33
C. Rs. 120
D. Rs. 175
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Assume that face value = Rs.100 as it is not given To earn Rs.8, money invested = Rs.100
To earn Rs.14, money invested =
100 × 14 = Rs.175 8
ie, market value of the stock = Rs.175
12. Which is better investment: 11% stock at 143 or A.
9
3 % stock at 120 4
C. 11% stock at 143
9
3 % stock at 120? 4
B. Cannot be compared, as the total amount of investment is not given. D. Both are equally good
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Assume that face value of both stocks are Rs.100 as it is not given. Hence, dividend per share in 1st case is Rs.11 and dividend per share in 2nd case is Rs.39/4
Assume that investment in each case is (Rs.143 × Rs.120)
Number of shares once can purchase in 1st case =
143 × 120 = 120 143
Income in 1st case = Rs.(120 × 11) = Rs.1320
Number of shares once can purchase in 2nd case =
143 × 120 = 143 120
Income in 2nd case = 143 × 39/4 = Rs.1394.25
Hence, 9
3 % stock at 120 is better 4
13. Sakshi invests a part of Rs. 12,000 in 12% stock at Rs. 120 and the remainder in 15% stock at Rs. 125. If his total dividend per annum is Rs. 1360, how much does he invest in 12% stock at Rs. 120? A. Rs. 6000
B. Rs. 4500
C. Rs. 5500
D. Rs. 4000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let Sakshi invests Rs.x in 12% stock and he invests (12000-x) in 15% stock Assume that face value of both stocks are Rs.100 as it is not given. Hence, dividend per share in 1st case is Rs.12 and dividend per share in 2nd case is Rs.15 Number of 12% shares he can purchase with Rs.x = x/120
Total dividend from 12% shares =
x × 12 120
Number of 15% shares he can purchase with Rs.(12000-x) = (12000-x)/125
Total dividend from 15% shares = Total dividend = Rs.1360
(12000 − x) × 15 125
(12000 − x) × 15 x × 12 + 120 125 (12000 − x) × 3 x + 10 25
= 1360
= 1360
5x + 6(12000 − x) = 1360 × 50 5x + 72000 − 6x = 68000 x = 4000
14. Rs. 9800 are invested partly in 9% stock at 75 and 10% stock at 80 to have equal amount of incomes. The investment in 9% stock is: A. Rs. 4800
B. Rs. 5400
C. Rs. 5000
D. Rs. 5600
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Assume that face value of both stocks are Rs.100 as it is not given. Hence, dividend per share in 1st case is Rs.9 and dividend per share in 2nd case is Rs.10 Let investment in 9% stock = Rs.x and investment in 10% stock = Rs.(9800-x) Market value of a 9% stock= Rs.75 Number of 9% stocks he can purchase with Rs.x = x/75
Total dividend from 9% stocks =
x×9 75
Market value of a 10% stock = Rs.80 Number of 10% stocks he can purchase with Rs.(9800-x) = (9800-x)/80
Total dividend from 10% stocks =
(9800 − x) × 10 80
incomes fom both of the investments is same
(9800 − x) × 10 x×9 = 75 80 (9800 − x) x×3 = 25 8 24x = 25 × 9800 − 25x 49x = 25 × 9800 x=
25 × 9800 = 25 × 200 = 5000 49
ie, investment in 9% stock = Rs.5000
15. A 6% stock yields 8%. The market value of the stock is: A. Rs. 133.33
B. Rs. 96
C. Rs. 75
D. Rs. 48
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Assume that face value = Rs.100 as it is not given To earn Rs.8, money invested = Rs.100
To earn Rs.6, money invested =
100 × 6 100 × 3 = = 25 × 3 = 75 8 4
ie, market value of the stock = Rs.75
16. 12500 shares, of par value Rs. 20 each, are purchased from Ram by Mohan at a price of Rs. 25 each. Find the amount required to purchase the shares. A. 312500
B. 311500
C. 313500
D. 314500
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Face value of each share = Rs.20
Market value of each share = Rs.25 Number of shares = 12500 Amount required to purchase the shares = 12500 × 25 = 312500
17. 12500 shares, of par value Rs. 20 each, are purchased from Ram by Mohan at a price of Rs. 25 each. If Mohan further sells the shares at a premium of Rs. 11 each, find his gain in the transaction. A. Rs. 75000
B. Rs. 70000
C. Rs. 85000
D. Rs. 65000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Face value of each share = Rs.20 Market value of each share = Rs.25 Number of shares = 12500 Amount required to purchase the shares = 12500 × 25 = 312500 Mohan further sells the shares at a premium of Rs. 11 each ie, Mohan further sells the shares at Rs.(20+11) = Rs.31 per share total amount he gets by selling all the shares = 12500 × 31 = 387500 His gain = 387500 - 312500 = Rs.75000
18. Find the annual dividend received by Nishita from 1200 preferred shares and 3000 common shares both of par value Rs. 50 each if the dividend paid on preferred shares is 10% and semi-annual dividend of 3½ % is declared on common shares. A. Rs. 18500
B. Rs. 16500
C. Rs. 14500
D. Rs. 19500
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Total number of preferred shares = 1200 Face value = Rs.50
dividend paid on preferred shares is 10%
Dividend per share =
50 × 10 = Rs.5 100
Total Dividend = 1200 × 5 = 6000 Total number of common shares = 3000 Face value = Rs.50 Semi-annual dividend of 3½ % is declared on common shares.
semi-annual dividend per share = Total semi-annual dividend =
50 × 7 7 = Rs. 2 × 100 4
7 × 3000 = Rs.5250 4
annual dividend = Rs.5250 × 2 = Rs.10500 Total dividend on all all shares(preferred and common) = 6000 + 10500 = Rs.16500
19. A man invested Rs. 26000 in 5% stock at 104. He sold the stock when the price rose to Rs. 120 and invested the sale proceeds in 6% stock. By doing this his income increased by Rs. 2500. At what price did he purchase the second stock? A. Rs. 125
B. Rs. 48
C. Rs. 24
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Assuming that face value of the first stock = Rs.100 as it is not given in the question Since it is a 5% stock, we can take the dividend per stock = Rs.5 Market Value of the first stock = Rs.104 Investment on the first stock = Rs.26000 Number of stocks purchases = 26000/104 = 250 His total income from all these stocks = Rs.250 × 5 = Rs.1250
He sells each of this stock at Rs.120 ie, amount he earns = Rs.120 × 250 = Rs.30000 He invest this Rs.30000 in 6% stock (here also face value is not given and hence take it as Rs.100) His new income = Rs.(1250 + 2500) = Rs.3750 ie, By Rs.30000 of investment , he earns an income of Rs.3750
To get an income of Rs.6, investment needed =
30000 × 6 = Rs.48 3750
This is the market value of the second stock
20. Anu invested Rs. 32400 in 8% stock at 90. She sold out Rs. 18000 stock when the price rose to Rs 95 and the remaining stock at Rs.98. She invested the total sale proceeds in 10% stock at 96½. Find the change in income of Anu. A. Rs.2220
B. Rs.120
C. Rs.2720
D. Rs.720
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Since the face value of the stock is not given, we can take it as Rs.100 it is an 8% stock. ie, dividend per stock = Rs.8 Total investment = Rs.32400 Market Value = Rs.90
Amount of stock she purchased =
32400 × 100 = Rs.36000 90
Number of shares she purchased = 32400/90 = 360 Total income = 360 × 8 = Rs.2880 She sold out Rs. 18000 stock when the price was Rs.95
Amount received by selling Rs.18000 stock at Rs.95 =
18000 × 95 = Rs.17100 100
Amount received by selling remaining Rs.18000 stock at Rs.98 =
18000 × 98 = Rs.17640 100
Total amount received = Rs. (17100 + 17640) = Rs. 34740 This Rs.34740 is invested in 10% stock at 96½
34740 = 360 193 ( ) 2
Number of shares =
Total income from these stocks = 360 × 10 = Rs.3600 Change in income = Rs.3600 - Rs.2880 = Rs.720
21. A man bought 20 shares of Rs. 50 at 5 discount, the rate of dividend being obtained is A. 12.5%
B. 15%
C. 15.5%
D. 17%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Face Value of a share = Rs.50 He bought each share at Rs.50 - Rs.5 = Rs.45 Number of shares = 20
Dividend = 13
1 27 %= % 2 2
Dividend per share =
50 × 27 = Rs. 6.75 2 × 100
Total dividend = (20 × 6.75) ie, He got a dividend of (20 × 6.75) for an investment of Rs.(20 × 45)
Interest obtained =
20 × 6.75 × 100 = 15% 20 × 45
22. Find the income on 7½ % stock of Rs. 2000 purchased at Rs. 80. A. Rs.6
B. Rs.160
C. Rs.148
D. Rs.150
13
1 % The rate of interest 2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Face Value of the stock = Rs.2000
Dividend is 7 Dividend =
1 15 %= % of the face value 2 2
2000 × 15 = Rs.150 2 × 100
23. John buys 100 shares of par value Rs. 5 each, of a company, which pays an annual dividend of 12% at such a price that he gets 10% on his investment. Find the market value of a share. A. Rs.6
B. Rs.12
C. Rs.4
D. Rs.8
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Face value of each share = Rs.5
Total dividend received by John = 100 × 5 ×
12 = Rs.60 100
Let market value of 100 shares = Rs.x
x×
10 = 60 100
x = 600 ie, Market value of 100 shares = Rs.600 Hence, Market value of each share = Rs.6
24. A man invested Rs. 5050 in 5% stock at 99 and sold it when the price rose to Rs. 101. He invested the sale proceeds in 8% stock at 88. Find the change in man's income if the Brokerage is Rs. 2. A. 0
B. Rs.160
C. Rs.180
D. Rs.190
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option D Explanation : Purchase price of the first stock = Rs.99 + Rs.2 = Rs.101 Number of stocks purchased in this case = 5050/101 = 50 Since the face value is not given, it can be taken as Rs.100. So, dividend per share = Rs.5 Income = 50 × 5 = Rs.250 Sale Price of the stock = 101-2 = 99 Amount received by selling the stock = 50 × 99 = 4950 Then he invests this Rs.4950 in 8% stock at 88 Purchase price of this stock = 88+2 = 90 Number of stocks purchased in this case = 4950/90 = 55 Since the face value is not given, it can be taken as Rs.100. So, dividend per share = Rs.8 Income = 55 × 8 = Rs.440 Change in income = 440-250 = Rs.190
25. To produce an annual income of Rs. 800 from a 8% stock at 90, the amount of stock needed is: A. Rs.10000
B. Rs.14400
C. Rs.10800
D. Rs.16000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Since face value is not given, take it as Rs.100. As it is an 8% stock, income (dividend) per stock = Rs.8 ie, For an income of Rs.8,amount of stock needed = Rs.100
For an income of Rs.800, amount of stock needed =
100 × 800 = 10000 8
26. A man sells 4000 common shares of a Company x (each of par value Rs. 10), which pays a dividend of 40% at Rs. 30 per share. He invests the sale proceeds in ordinary shares of Company Y (each of par value Rs. 25) that pays a dividend of 15%. If the market value of Company Y is Rs. 15, find the number of shares of Company Y purchased by the man A. 8000
B. 12000
C. 16000
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Just too much information is given in the question to confuse. This is a straight and simple question Market Value of Company X (his selling price) = Rs.30 Total shares sold = 4000 Amount he gets = Rs.(4000 × 30) He invests this amount in ordinary shares of Company Y Market Value of Company Y(His purchasing price) = 15
Number of shares of company Y which he purchases =
4000 × 30 = 8000 15
27. A company issued 20000 shares of par value Rs. 10 each. If the total divided declared by the company is Rs.24000, find the rate of dividend paid by the company. A. 12%
B. 12.5%
C. 14%
D. 14.5%
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Number of shares = 20000 Face value of each share = Rs.10
divided per share = 10 ×
R where R is the rate of interest 100
Total divided = 20000 × 10 ×
R 100
20000 × 10 × ⇒ R=
R = 24000 100
24000 = 12 2000
ie, divided = 12%
28. A company declared a semi-annual dividend of 12%. Find the annual dividend of Sam owing 2000 shares of the company having a par value of Rs. 10 each. A. Rs.4800
B. Rs.2400
C. Rs.3600
D. None of these
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
semi-annual dividend =
10 × 12 = Rs.1.2 100
Total semi-annual dividend = 2000 × 1.2 = Rs.2400 Total annual dividend = 2 × Rs.2400 = Rs.4800
29. A company has issued 500 preferred shares and 400 common shares both of par value Rs. 100 each. The dividend on a preferred share and a common share is 8% and 12%, respectively. The company had a total profit of 150000 rupees out of which some amount was kept in reverse fund and the remaining disturbed as dividend. Find the amount kept in reserve fund. A. Rs.141200
B. Rs.160000
C. Rs.7200
D. Rs.182200
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Face value of each preferred share = 100
Dividend per preferred share =
100 × 8 100
Total dividend in all preferred shares =
500 × 100 × 8 = Rs.4000 100
Face value of each common share = 100
Dividend per common share =
100 × 12 100
Total dividend in all common shares =
400 × 100 × 12 100
= Rs.4800
Total dividend = Rs.4000 + Rs.4800 = Rs.8800 amount kept in reserve fund = Rs.150000 - Rs.8800 = Rs.141200
30. Arun invested Rs. 333000 in 5½ % stocks at 110 .If brokerage is Rs.1, what is his annual income from his investment. A. Rs.16500
B. Rs.12500
C. Rs.16000
D. Rs.18000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Investment = Rs.333000 Since face value is not given, we can take it as Rs.100 and dividend per share = Rs.11/2 Market Value = 110 + 1 = 111 Number of shares purchased = 333000/111 = 3000
Total income = 3000 ×
11 = Rs.16500 2
31. Find the income on 8% stock of Rs.1200 purchased at Rs.120? A. Rs.96
B. Rs.66
C. Rs.90
D. Rs.88
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Face Value = Rs.1200 Market Value = Rs.120
Dividend (income)=
1200 × 8 = Rs.96 100
I mportant Concepts and Formulas - A lgebra (I nclusive of Surds and I ndices ) I.
Basic A lgebraic Formulas
1.
a(b + c) = ab + ac
2.
(a + b) = a 2 + 2ab + b2
3.
(a − b) = a 2 − 2ab + b2
4.
(a + b) + (a − b) = 2(a 2 + b2 )
5.
(a + b) = a 3 + 3a 2 b + 3a b2 + b3 = a 3 + 3ab(a + b) + b3
6.
(a − b) = a 3 − 3a 2 b + 3a b − b = a 3 − 3ab(a − b) − b
7.
a 2 − b2 = (a − b)(a + b)
8.
a 3 + b3 = (a + b)(a 2 − ab + b2 )
9.
a 3 − b3 = (a − b)(a 2 + ab + b2 )
(Distributive Law)
2
2
2
2
3
3
2
3
3
10.
a n − bn = (a − b)(a n−1 + a n−2 b + a n−3 b2 + ... + bn−1 )
11.
(a + b + c) = a 2 + b2 + c2 + 2(ab + bc + ca)
12.
(a + b)(a + c) = a 2 + (b + c)a + bc
2
1.
a n = a. a. a ... (n times)
2.
a m . a n . . . a p = a m+n+...+p
3.
a m . a n = a m+n
4.
am = a m−n an
5.
(a m ) = amn = (a n )
6.
(ab) n = an bn
7.
a ( ) b
8.
a −n =
9.
an =
n
n
=
m
an bn
1 an
1 a −n
10.
a 0 = 1 where a ∈ R, a ≠ 0
11.
q p a p/q = √ −− a
12.
if a m = a n where a ≠ 0 and a ≠ ±1, then m = n
13.
if a n = bn where n ≠ 0, then a = ±b
2
1.
(x + a ) = x 2 + 2ax + a 2
2.
(x − a ) = x 2 − 2ax + a 2
3.
(x + a)(x + b) = x 2 + (a + b)x + ab
4.
x 2 − a 2 = (x − a)(x + a)
5.
(x n + 1) is completely divisible by (x + 1) when n is odd
6.
(x n + a n ) is completely divisible by (x + a) when n is odd
7.
(x n − a n ) is completely divisible by (x − a) for every natural number n
8.
(x n − a n ) is completely divisible by (x + a) when n is even
2
Binom ial Theorem
If a and b are any real numbers and n is a positive integer,
(a + b)
n
= a n + na n−1 b +
n(n − 1) n−2 2 n(n − 1)(n − 2) a b + 2! 3!
a n−3 b3 + ⋯ +
n(n − 1)(n − 2) ⋯ (n − r + 1) r!
an−r br + ⋯ + bn
n n n n n n = ( ) a n + ( ) a n−1 b + ( ) a n−2 b2 + ( ) an−3 b3 + ⋯ + ( ) a n−r br + ⋯ + ( ) bn 0 1 2 3 r n n n n−r r =∑ ( )a b r r=0
n where ( ) is known as Binomial Coefficient and is defined by r n n! ( )= r (r!)(n − r)!
=
n(n − 1)(n − 2) ⋯ (n − r + 1) r!
n where r = 1,2,...,n and ( ) = 1 0
Binomial Coefficient also occurs in many other mathematical areas than algebra, especially in combinatorics where
( nr ) represents the number of combinations of n distinct things taking r at a time and is denoted by nCr or C(n,r). The properties of binomial coefficients have led to extending its meaning beyond the basic case where n and r are nonnegative integers with r ≤ n; such expressions are also called binomial coefficients.
Binom ial Ex pansions 0
(a + b) = 1 1
(a + b) = a + b 2
(a + b) = a 2 + 2ab + b2 3
(a + b) = a 3 + 3a 2 b + 3a b2 + b3 4
2
3
4
(a + b) = a 4 + 4a 3 b + 6 a2 b + 4a b + b ... and so on
Binom ial Series One of Newton's achievement was to extend the Binomial Theorem to the case where n can be any real number. If n is any real number and -1 < x < 1, then
(1 + x)
n
= 1 + nx +
n(n − 1) n(n − 1)(n − 2) x2 + 2! 3!
x3 + ⋯
n n n n = ( ) x 0 + ( ) x 1 + ( ) x2 + ( ) x 3 + ⋯ 0 1 2 3 ∞ n = ∑ ( ) xr r r=0
n(n − 1)(n − 2) ⋯ (n − r + 1) n where ( ) = r r!
II.
n where r ≥ 1 and ( ) = 1 0
Surds and I mportant P roperties i.
Surds Let
a be a rational number and n be a positive integer such that √n a is irrational. Then √n a is called a surd of order n.
Examples :
√ 3 is a surd of order 2 √ 7 is a surd of order 3 3
4√ 3 is a surd of order 2 Please note that numbers like
3 −− √ 9 , √ 27 etc are not surds because they are not irrational numbers
Every surd is an irrational number. But every irrational number is not a surd. (eg : are irrational numbers.)
ii.
Quadratic Surds
π , e etc are not surds though they
A surd of order 2 is called a quadratic surd Examples :
iii.
√ 2, √ 3, (3 + √ 5), etc.
Rules of Surds
1.
2.
−− − −−− √ m × √ n = √ mn − −− − m √− m =√ − n √n
3.
1 −−−−−−−−−−−−−−− 1− n −−−−−−−−−−−− −−−−−−−−−− √ a. √ a. √ a ⋯ n times = a 2
4.
−−−−−−−−−−− − −−−−−−−− −−−−−− √ a. √ a. √ a ⋯ ∞ = a
5.
if
1.
The conjugate surd of
2.
if √ b and √ d are quadratic surds and if then a = c and b = d
−−−−−−−−−−−−−− −−−−−−−−− −− a−−−− ⋯ ∞− = p, then p(p − 1) = a √ a +√ a +√−
√ a + √ b = ±(√ a − √ b) a +√ b= c+√ d
,
i.e., rational part on the left side = rational part on the right side. and irrational part on the left side = irrational part on the right side.
3.
III.
if √ b and √ d are quadratic surds and if then a = 0 and b = d
a +√ b= √ d ,
Quadratic Equations and How to Solve Quadratic Equations A . Quadratic Equations A quadratic equation is a second degree univariate polynomial equation.
A quadratic equation can be written as (general form or standard form of quadratic equation)
a x 2 + bx + c = 0
Example :
where x is a variable, a, b and c are constants and
x 2 + 5x + 6 = 0
(Please note that if a=0, equation becomes a linear equation) The solutions of a quadratic equation are called its roots.
B. How to Solve Quadratic Equations
a≠0
There are many methods to solve Quadratic equations. Quadratic equations can be solved by factoring, completing the square, graphing, Newton's method, and using the quadratic formula. We can go through some of the popular methods for solving quadratic equations here. 1.
How to Solve Quadratic Equations By Factoring
Step 1: Write the quadratic equation in standard form. i.e. , bring all terms on the left side of the equal sign and 0 on the right side of the equal sign. Step 2: Factor the left side of the equation Step 3: Equate each factor to 0 and solve the equations
Factoring is one of the fastest ways for solving quadratic equations. The concept will be clear from the following examples.
Example1 : Solve the Quadratic Equation
x 2 + 4x − 11 = x − 1
Step 1 : Write the quadratic equation in standard form.
x 2 + 4x − 11 = x − 1 ⇒ x 2 + 4x − 11 − x + 1 = 0 ⇒ x 2 + 3x − 10 = 0 Step 2 : Factor the left side of the equation. We know that Assume
x 2 + (a + b)x + ab = (x + a)(x + b)
. We will use the same concept for factoring.
x 2 + 3x − 10 = x 2 + (a + b)x + ab
We need to find out a and b such that a + b = 3 and ab = -10 a = +5 and b = -2 satisfies the above condition. Hence
x 2 + 3x − 10 = (x + 5)(x − 2)
x 2 + 3x − 10 = 0 ⇒ (x + 5)(x − 2) = 0 Step 3 :Equate each factor to 0 and solve the equations (x + 5)(x - 2) = 0 => (x + 5) = 0 or (x - 2) = 0 => x = -5 or 2 Hence, the solutions of the quadratic equation
x 2 + 4x − 11 = x − 1
are x = -5 and x = 2.
(In other words, x = -5 and x = 2 are the roots of the quadratic equation
Example2 : Solve the Quadratic Equation
x 2 − 7x + 10 = 0
This equation is already in the standard form. Hence let's go to step 2
x 2 + 4x − 11 = x − 1
Step 2 : Factor the left side of the equation. Here we need to find out a and b such that a + b = -7 and ab = +10 a = -5 and b = -2 satisfies the above condition. Hence
Hence,
x 2 − 7x + 10 = (x − 5)(x − 2) x 2 + 3x − 10 = 0
⇒ (x − 5)(x − 2) = 0 Step 3 :Equate each factor to 0 and solve the equations (x - 5)(x - 2) = 0 => (x - 5) = 0 or (x - 2) = 0 => x = 5 or 2 Hence, the solutions of the quadratic equation
x 2 − 7x + 10 = 0
are x = 5 and x = 2.
(In other words, x = 5 and x = 2 are the roots of the quadratic equation
Example3 : Solve the Quadratic Equation
x 2 − 7x + 10 = 0
)
3x 2 − 14x + 8 = 0
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation. Here we need to follow a slightly different approach for factoring because here the coefficient of x2 ≠ 1 (whereas In example 1 and example 2, the coefficient of x2 was 1) Hence, to factor
3x 2 − 14x + 8 , we need to do the followings
1. Product of the second degree term and the constant. i.e.,
3x 2 × 8 = 24x 2
2. We got product as 24x2 and have middle term as -14x. From this , it is clear that we can take -12x and -2x such that their sum is -14x and product is 24x2 Hence,
3x 2 − 14x + 8 = 0
Hence,
3x 2 − 14x + 8 = 0
= 3x
2
− 12x − 2x + 8 = 3x(x − 4) − 2(x − 4) = (x − 4)(3x − 2)
⇒ (x − 4)(3x − 2) = 0 Step 3 :Equate each factor to 0 and solve the equations (x - 5)(x - 2) = 0 => (x - 4) = 0 or (3x - 2) = 0 => x = 4 or
2 . 3
Hence, the solutions of the quadratic equation (In other words, x = 4 and x =
3x 2 − 14x + 8 = 0
are x = 4 and x =
2 3
2 2 are the roots of the quadratic equation 3x − 14x + 8 = 0 3
Example4 : Solve the Quadratic Equation
6x 2 − 17x + 12 = 0
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation. coefficient of x2 ≠ 1 Hence, to factor
6x 2 − 17x + 12 , we need to do the followings
1. Product of the second degree term and the constant. i.e.,
6x 2 × 12 = 72x 2
2. We got product as 72x2 and have middle term as -17x. From this , it is clear that we can take -8x and -9x such that their sum is -17x and product is 72x2 Hence,
6x 2 − 17x + 12 = 6x 2 − 8x − 9x + 12 = 2x(3x − 4) − 3(3x − 4) = (3x − 4)(2x − 3)
Hence,
6x 2 − 17x + 12 = 0
⇒ (3x − 4)(2x − 3) = 0 Step 3 :Equate each factor to 0 and solve the equations (3x - 4)(2x - 3) = 0 => (3x - 4) = 0 or (2x - 3) = 0 => x =
4 3 or . 3 2
Hence, the solutions of the quadratic equation (In other words, x =
6x 2 − 17x + 12 = 0
are x =
4 3 and x = . 3 2
4 3 2 and x = are the roots of the quadratic equation 6x − 17x + 12 = 0 3 2
(Special Factorization Examples) Example5 : Solve the Quadratic Equation
x2 − 9 = 0
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation. Here,x
2
− 9 is in the form a 2 − b2 where a = x and b = 3
We know that
a 2 − b2 = (a − b)(a + b)
We will use the same concept for factoring this.
x 2 − 9 = x 2 − 32 = (x − 3)(x + 3) Hence,
x2 − 9 = 0
⇒ (x − 3)(x + 3) = 0 Step 3 :Equate each factor to 0 and solve the equations (x - 3)(x + 3) = 0
=> x = 3 or -3 Hence, the solutions of the quadratic equation
x2 − 9 = 0
are x = 3 and x = -3
(In other words, x = 3 and x = -3 are the roots of the quadratic equation
Example6 : Solve the Quadratic Equation
x2 − 9 = 0
25x 2 − 16 = 0
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation. 2
25x 2 − 16 = (5x) − 42 = (5x − 4)(5x + 4) Hence,
25x 2 − 16 = 0
⇒ (5x − 4)(5x + 4) = 0 Step 3 : Equate each factor to 0 and solve the equations (5x - 4)(5x + 4) = 0 =>5x = 4 or -4
x=
4 −4 . or 5 5
Hence, the solutions of the quadratic equation (In other words, x =
25x 2 − 16 = 0
are x =
4 −4 and x = 5 5
4 −4 2 and x = are the roots of the quadratic equation 25x − 16 = 0 5 5
Example7 : Solve the Quadratic Equation
x 2 + 6x + 9 = 0
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation. Here,x 2
+ 6x + 9
We know that
is in the form a2
+ 2ab + b2
a 2 + 2ab + b2 = (a + b)
2
where a = x and b = 3
We will use the same concept for factoring this.
x 2 + 6x + 9 = x 2 + 2 × x × 3 + 32 = (x + 3 )2 Hence,
x 2 + 6x + 9 = 0
⇒ (x + 3) 2 = 0 Step 3 :
(x + 3) 2 = 0 => x + 3 = 0 => x = -3 Hence, the solution of the quadratic equation
x 2 + 6x + 9 = 0
is x = -3
(In other words, x = -3 is the root of the quadratic equation
Example8 : Solve the Quadratic Equation
x 2 + 6x + 9 = 0
x 2 − 6x + 9 = 0
This equation is already in the standard form. Hence let's go to step 2
Step 2 : Factor the left side of the equation. Here,x
2
− 6x + 9
We know that
2
is in the form a
− 2ab + b2
a 2 − 2ab + b2 = (a − b) 2
where a = x and b = 3
We will use the same concept for factoring this.
x 2 − 6x + 9 = x 2 − 2 × x × 3 + 32 = (x − 3 )2 Hence,
x 2 + 6x + 9 = 0 2
⇒ (x − 3) = 0 Step 3 : 2
(x − 3) = 0 => x - 3 = 0 => x = 3 Hence, the solution of the quadratic equation
x 2 − 6x + 9 = 0
(In other words, x = 3 is the root of the quadratic equation
2.
is x = 3
x 2 − 6x + 9 = 0
How to Solve Quadratic Equations By Quadratic Formula
Quadratic F orm ula Consider a quadratic equation Its solutions can be given by
a x 2 + bx + c = 0
where
a≠0
−−−−−−− −b ± √ b2 − 4ac x= 2a Example1 : Solve the Quadratic Equation
x 2 + 4x − 11 = x − 1 ⇒ x 2 + 4x − 11 − x + 1 = 0 ⇒ x 2 + 3x − 10 = 0
x 2 + 4x − 11 = x − 1
−−−−−−− −b ± √ b2 − 4ac x= 2a =
−3 ± 7 2
=
4 −10 or 2 2
=
−−−−−−−−−−−−−−− −3 ± √ 32 − 4 × 1 × (−10)
=
2×1
9−−− + 40− −3 ± √ − 2
= 2 or − 5 Hence, the solutions of the quadratic equation
x 2 + 4x − 11 = x − 1
are x = 2 and x = -5
(In other words, x = 2 and x = -5 are the roots of the quadratic equation
Example2 : Solve the Quadratic Equation
−−−−−−− −b ± √ b2 − 4ac x= 2a =
7±√ 9 7±3 = 2 2
=
10 4 or 2 2
=
x 2 + 4x − 11 = x − 1
x 2 − 7x + 10 = 0
−−−−−−−−−−−−−− − 2 7 ± √ (−7) − 4 × 1 × 10
=
2×1
−−−−−− 7 ± √ 49 − 40 2
= 5 or 2 Hence, the solutions of the quadratic equation
x 2 − 7x + 10 = 0
are x = 5 and x = 2.
(In other words, x = 5 and x = 2 are the roots of the quadratic equation
Example3 : Solve the Quadratic Equation
−−−−−− − 2 −b ± √ b − 4ac x= 2a
=
=
−−− 14 ± √ 100 14 ± 10 = 6 6
=
24 4 or 6 6
= 4 or
3x 2 − 14x + 8 = 0
−−−−−−−−−−−−−− − 2 14 ± √ (−14) − 4 × 3 × 8 2×3
x 2 − 7x + 10 = 0)
=
−−−−−−− 14 ± √ 196 − 96 6
2 3
Hence, the solutions of the quadratic equation (In other words, x = 4 and x =
3x 2 − 14x + 8 = 0
are x = 4 and x =
2 3
2 are the roots of the quadratic equation 3x 2 − 14x + 8 = 0 3
Example4 : Solve the Quadratic Equation
6x 2 − 17x + 12 = 0
−−−−−−− −b ± √ b2 − 4ac x= 2a =
17 ± √ 1 17 ± 1 = 12 12
=
18 16 or 12 12
=
3 4 or 2 3
=
−−−−−−−−−−−−−−− − 2 17 ± √ (−17) − 4 × 6 × 12 2×6
Hence, the solutions of the quadratic equation (In other words, x =
=
−− − −6 ± √ −4 2
=
−−−−−−−− 17 ± √ 289 − 288 12
6x 2 − 17x + 12 = 0
are x =
4 3 and x = . 3 2
4 3 2 and x = are the roots of the quadratic equation 6x − 17x + 12 = 0 3 2
Example5 : Solve the Quadratic Equation
−−−−−−− −b ± √ b2 − 4ac x= 2a
=
=
−−−−−−−−−−−−− − 2 −6 ± √ (6) − 4 × 1 × 10 2×1
x 2 + 6x + 10 = 0 −−−−−− −6 ± √ 36 − 40 = 2
−6 ± 2i 2
= −3 ± i = (−3 + i) or (−3 − i) Hence, the solutions of the quadratic equation
x 2 + 6x + 10 = 0
are x =(-3 + i) and x = (-3 - i)
(In other words, x =(-3 + i) and x = (-3 - i) are the roots of the quadratic equation
Example5 : Solve the Quadratic Equation
−−−−−−− −b ± √ b2 − 4ac x= 2a =
=
−−−−−−−−−−−−− − 2 −6 ± √ (−6) − 4 × 1 × 9 2×1
x 2 + 6x + 9 = 0 =
−−−−−−− −6 ± √ (36 − 36 2×1
−6 ± √ 0 −6 = 2 2
= −3 Hence, the solutions of the quadratic equation (In other words,
x 2 + 6x + 9 = 0
is
x = −3
x = −3 is the roots of the quadratic equation x 2 + 6x + 9 = 0
Example6 : Solve the Quadratic Equation
x2 − 9 = 0
x 2 + 6x + 10 = 0
−−−−−−− −b ± √ b2 − 4ac x= 2a =±
=
−−−−−−−−−−−−−−− 0 ± √ (0) 2 − 4 × 1 × (−9) 2×1
=
±√ −− 36 2
6 2
= 3 or − 3 Hence, the solutions of the quadratic equation
x2 − 9 = 0
are x = 3 and x = -3
(In other words, x = 3 and x = -3 are the roots of the quadratic equation
x2 − 9 = 0
C. How to find out the number of solutions(roots) of a quadratic equation quickly?
To find out the number of solutions (roots) of a quadratic equation 2
discriminant, D. The discriminant, D = b
– 4ac
a x 2 + bx + c = 0
quickly, we need to find out the
If the discriminant is positive, there will be two real solutions for the quadratic equation. If the discriminant is 0, there will be one real solution (it repeats itself) for the quadratic equation. If the discriminant is negative, there will be two complex solutions for the quadratic equation.
Ex am ple1 : Find out the number of roots for 2
Discriminant, D = b D > 0. Hence
x 2 − 7x + 12 = 0
– 4ac = (−72 ) − (4 × 1 × 12) = 49 − 48 = 1.
x 2 − 7x + 12 = 0
will have two real roots.
Let's verify if this is correct by solving the quadratic equation.
x 2 − 7x + 12 = 0 ⇒ (x − 3)(x − 4) = 0 ⇒ x = 3 or 4. Yes, two real roots we got.
Ex am ple2 : Find out the number of roots for 2
Discriminant, D = b D = 0. Hence
x 2 − 4x + 4 = 0
– 4ac = (−42 ) − (4 × 1 × 4) = 16 − 16 = 0.
x 2 − 4x + 4 = 0
will have one real root (it repeats itself)
Let's verify if this is correct by solving the quadratic equation.
x 2 − 4x + 4 = 0 ⇒ (x − 2) 2 = 0 ⇒ x = 2 Yes, real root (it repeats itself).
Ex am ple3 : Find out the number of roots for 2
Discriminant, D = b D < 0. Hence
x2 + 1 = 0
– 4ac = (02 ) − (4 × 1 × 1) = 0 − 4 = −4
x2 + 1 = 0
will have two complex roots
Let's verify if this is correct by solving the quadratic equation.
x2 + 1 = 0
−−−−−−− −b ± √ b2 − 4ac x= 2a
−−−−−−−−−−− −−−− − 0 ± √ 02 − 4 × 1 × 1 √ −− −4 ±2 i √ 0−4 = = = = =±i 2×1 2 2 2
Yes, we got two complex roots, +i and -i
D. Sum and P roducts of the roots of a quadratic equation
Let
x 1 and x 2 are the roots of a quadratic equation a x 2 + bx + c = 0 where a ≠ 0
Sum of the roots of the quadratic equation,
x1 + x2 =
Product of the roots of the quadratic equation,
−b a
x1 × x2 =
c a
Ex am ple 1: Find the sum and product of the roots of the quadratic equation sum of the roots=
−b 7 = =7 a 1
product of the roots=
c 10 = = 10 a 1
Ex am ple 2: Find the sum and product of the roots of the quadratic equation sum of the roots=
x 2 − 7x + 10 = 0
−b −3 = a 2
product of the roots=
c 1 = a 2
2x 2 + 3x + 1 = 0
1. (132)7 × (132)? = (132)11.5. A. 3
B. 3.5
C. 4
D. 4.5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
a m . a n = a m+n (132)7 × (132)x = (132)11.5 => 7 + x = 11.5 => x = 11.5 - 7 = 4.5
2.
a ( ) b
x−2
x−7
b =( ) a
. What is the value of x ?
A. 3
B. 3.5
C. 4
D. 4.5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
1 a −n
an =
a ( ) b
x−2
a ⇒( ) b
b =( ) a x−2
x−7
a =( ) b
−(x−7)
⇒ x − 2 = −(x − 7) ⇒ x − 2 = −x + 7 ⇒ x − 2 = −x + 7 ⇒ 2x = 9 ⇒x=
9 = 4.5 2
3. If 7(x - y) = 343 and 7(x + y) = 16807, what is the value of x? A. 4
B. 3
C. 2
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 7(x - y) = 343 = 73 => x - y = 3 ---------------------------(Equation 1) 7(x + y) = 16807 = 75 => x + y = 5 ---------------------------(Equation 2) (Equation 1)+ (Equation 2) => 2x = 3 + 5 = 8 => x = 8⁄2 = 4
4. (0.04)-2.5 = ? A. 125
B. 25
C. 3125
D. 625
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
a −n =
(0.04)
1 an
−2.5
. 2 = (5 )
.
=( 5 2
2.5
1 ) .04
=(
100 ) 4
2.5
5
= 5 = 3125
5. (6)6.5 × (36)4.5 ÷ (216)4.5 = (6)? A. 1
B. 2
C. 4
D. 6
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : (6)6.5 × (36)4.5 ÷ (216)4.5 = (6)6.5 × [(6)2]4.5 ÷ [(6)3]4.5 = (6)6.5 × (6)9 ÷ (6)13.5 = (6)(6.5 + 9 - 13.5) = (6)2
1
6.
1 + P (n−m)
A.
2
C.
1
+
1
1 + P (m−n) 1 B. 1+P 1 D. P
=?
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
= (25)
2.5
2 = (5 )
2.5
1 1+P =
=
=
1 Pn 1+ m P
+
Pm +P
P
m
P P
m m
1
+
(n−m)
n
1 + P (m−n) 1 Pm 1+ n P
+
P
n
Pn +P
m
n
+P +P
n
=1
7.
If x = (8 + 3√ 7) , what is the value of (√ x −
A.
√ −− 12
B.
C.
2
D.
4 √ −− 14
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
2
(a − b) = a 2 − 2ab + b2 a 2 − b2 = (a − b)(a + b)
(√ x −
1 ) √x
= x−2+ = x+
2
1 x
1 −2 x
= (8 + 3√ 7) +
= (8 + 3√ 7) +
= (8 + 3√ 7) +
= (8 + 3√ 7) +
= (8 + 3√ 7) +
1 −2 (8 + 3√ 7) (8 − 3√ 7) (8 + 3√ 7) (8 − 3√ 7) (8 − 3√ 7) 82 − (3√ 7) (8 − 3√ 7) 64 − 63 (8 − 3√ 7) 1
2
−2
−2
−2
= 8 + 3√ 7 + 8 − 3 √ 7 − 2 = 14 We got that (√ x −
Hence , (√ x −
1 ) √x
1 ) = √ −− 14 √x
2
= 14
−2
1 )? √x
8. if 6m = 46656, What is the value of 6m-2 A. 36
B. 7776
C. 216
D. 1296
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : ------------------------------------------------------------------------Solution 1 --------------------------------------------------------------------------
am = a m−n an Given that 6m = 46656
6m−2 =
6m 62
=
46656 62
=
46656 = 1296 36
-------------------------------------------------------------------------Solution 2--------------------------------------------------------------------------6 m = 46656 6 46656 6 7776 6 1296 6 216 6 36
6
Hence, 46656 = 66 Thus, we got that 6m = 66 => m = 6 6m-2 = 6(6-2) = 64 = 1296
9. 10222 ÷ 10220 = ? A. 10
B. 100
C. 1000
D. 10000
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
am = a m−n an 10222 ÷ 10220 = 10222 - 220 = 102 = 100
10. If m and n are whole numbers and mn = 196, what is the value of (m - 3)(n+1) ? A. 2744
B. 1
C. 121
D. 1331
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : mn = 196 We know that 142 = 196 Hence we can take m = 14 and n = 2 (m - 3)(n+1) = (14 - 3)(2+1) = 113 = 1331
(1024)
11.
n/5
× 42n+1
16 × 4n−1 n
A. 256
B. 64
C. 16
D. 9
=?
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
(1024)
n/5
2n+1
×4
n−1
n
16 × 4 n/5
=
=
(45 )
× 42n+1
2 n
n−1 (4 ) × 4
4n × 42n+1 42n × 4n−1
= 4(n+2n+1)−(2n)−(n−1) = 42 = 16
12.
x−4
a If ( ) b
b =( ) a
x−8
, what is the value of x?
A. 4
B. 6
C. 8
D. 10
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
1 a −n
an =
a ( ) b
x−4
a ⇒( ) b
b =( ) a x−4
x−8
a =( ) b
−(x−8)
⇒ x − 4 = −(x − 8) ⇒ x − 4 = −x + 8 ⇒ 2x = 12 ⇒x=
12 =6 2
13. If 1000.20 = x, 100.60 = y and xz = y2, what is the value of z? A. 3
B. 6
C. 4.2
D. 2.2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
(a m ) n = amn = (a n ) m x = (100)0.2
y = (10)0.6 xz = y2 => (1000.2)z = (100.6)2 => 1000.2z = 10(0.6 × 2) => (102)0.2z = 10(0.6 × 2) => 10(2 × 0.2z)= 10(0.6 × 2) => 2 × 0.2z = 0.6 × 2
(∵ powers are equal as base values are equal in both side)
=> 0.2z = 0.6
⇒ z=
14.
(
.6 6 = =3 .2 2
xq ) xr
(q+r−p)
.(
(r+p−q)
xr ) xp
.(
xp ) xq
A. x(a - b - c)
B. .5
C. 1
D. x(a + b + c)
(p+q−r)
=?
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
(
xq ) xr
(q+r−p)
= [x (q−r) ]
.(
xr ) xp
(q+r−p)
(r+p−q)
. [x (r−p) ]
.(
xp ) xq
(r+p−q)
(p+q−r)
. [x (p−q) ]
(p+q−r)
= x[(q - r)(q + r - p) + (r - p)(r + p - q) + (p - q)(p + q - r)] = x[(q - r)(q + r)- p(q - r) + (r - p)(r + p) - q(r - p) + (p - q)(p + q) - r(p - q)] 2 2 2 2 2 2 = x[(q - r ) - p(q - r) + (r - p ) - q(r - p) + (p - q ) - r(p - q)] 2 2 2 2 2 2 = x[(q - r + r - p + p - q ) - p(q - r) - q(r - p) - r(p - q)]
= x[0 - p(q - r) - q(r - p) - r(p - q)] = x[- p(q - r) - q(r - p) - r(p - q)] = x[-pq + pr -qr + pq - rp + rq ] = x[0] =1
15.
1 1 + x (q−p) + x (r−p)
+
1 1 + x (p−q) + x (r−q)
A. 0
B. 1
C. .5
D. 2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
+
1 1 + x (q−r) + x (p−r)
=?
1 1 + x (q−p)
=
+
+ x (r−p)
1 xq xr 1+ p + p x x
=
xp xp + xq + xr
=
xp + xq + xr xp + xq + xr
+ x (r−q)
1 xp xr 1+ q + q x x
+
+
1 1 + x (p−q)
xq xq + xp + xr
+
+
+
1 1 + x (q−r)
1 xq xp 1+ r + r x x
xr xr + xq + xp
=1
16.
65610.14 × 65610.11 = ?
A. 16
B. 9
C. 4
D. 1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
a m . a n = a m+n (a m ) n = amn = (a n ) m 65610.14 × 65610.11 = 6561(0.14 + 0.11)= 6561(0.25) = 6561(1/4) = (38)(1/4) = 38/4 = 32 = 9
17. What is the value of (2 × 4 × 5)5n A. 25n + 45n + 55n C. (40)
B. (405)n D. (40n)5
5n
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
(a m ) n = amn = (a n ) m (2 × 4 × 5)5n = (40)5n = (405)n = (40n)5
18.
If (√ 3)
n
= 6561, (n)
64 C. 16√ 3 A.
3/2
=?
64√ 3 D. 16 B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
+ x (p−r)
a p/q = √q −− ap (√ 3)
n
= 6561
⇒ (√ 3)
n
= (√ 3 )
16
⇒ n = 16
−−− (n) 3/2 = (16) 3/2 = √ 163 = 16√ −− 16 = 16 × 4 = 64
19. 5x × 23 = 36. 5(x+1) = ? A. 22
B. 21
C. 20.5
D. 22.5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
a m . a n = a m+n 5x × 23 = 36 ⇒ 5x =
36 23
5(x+1) = 5x × 5 =
20.
36 36 × 5 ×5 = 2×2×2 23
=
9×5 = 22.5 2
36120 = (36 × x )40 . What is the value of x?
A. 44
B. 44
C. 62
D. 64
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
(ab) n = an bn am = a m−n an (a m ) n = amn = (a n ) m 36120 = (36 × x ) 40 ⇒ 36120 = 3640 × x 40 ⇒ x 40 =
36120 = 36(120−40) = 3680 3640
⇒ (√ x 2 ) 40 = 3680 ⇒ (√ x ) 80 = 3680 ⇒ √ x = 36 ⇒ x = 362 = 64
21. (6561)(1/2) + (6561)(1/4) + (6561)(1/8) = ? A. 98
B. 86
C. 93
D. 81
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : 3 6561
3 2187
3 729
3 243
3 81
3 27
3 9
3
3 9
3
Hence, 6561 = 38
(6561)(1/2) + (6561)(1/4) + (6561)(1/8)
= (38)(1/2) + (38)(1/4) + (38)(1/8)
= (3)(8/2) + (3)(8/4) + (3)(8/8)
= (3)4 + (3)2 + (3)1
= 81 + 9 + 3
= 93
22.
√ −− 3n = 6561. 3√ n = ?
A. 81
B. 9
C. 16
D. 25
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : 3 6561
3 2187
3 729
3 243
3 81
3 27
Hence, 6561 = 38
−− Given that √ 3n = 6561 −− √ 3n = 38 3n = 316 ⇒ n = 16 3√ n = 3√ 16 = 34 = 81
23. If 5(a + b) = 5 × 25 × 125 , what is (a + b)2 A. 12
B. 16
C. 34
D. 36
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : 5(a + b) = 5 × 25 × 125 = 51 × 52 × 53 = 5(1 + 2 + 3) = 56 => (a + b) = 6 => (a + b)2 = 62 = 36
24. (7-1 - 11-1) + (7-1 + 11-1) = ? A. 2 × 7-1
B. 2 × 11-1
C. 14
D. 22
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option A Explanation : (7-1 - 11-1) + (7-1 + 11-1) = 7-1 + 7-1 - 11-1 + 11-1 = 7-1 + 7-1 = 2 × 7-1
25. (5)1.25 × (12)0.25 × (60)0.75 = ? A. 420
B. 260
C. 200
D. 300
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
(ab) n = an bn a m . a n = a m+n (5)1.25 × (12)0.25 × (60)0.75 =(5)1.25 × (12)0.25 × (12 × 5)0.75 = (5)1.25 × (12)0.25 × (12)0.75 × (5)0.75 = (5)(1.25 + 0.75) × (12)(0.25 + 0.75) = (5)2 × (12)1 = 25 × 12 = 300
26. 36 × 36 × 36 × 36 = 6? A. 10
B. 8
C. 4
D. 6
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : 36 × 36 × 36 × 36 = 62 × 62 × 62 × 62 = 6(2+2+2+2) = 68
27. What is the value of (7-14 - 7-15) ? A. 6 × 7-15
B. 6 × 7-14
C. 7 × 7-15
D. 7 × 7-14
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
a −n = an =
1 an
1 a −n
(7−14 − 7−15 ) =
=
7−1 715
=
6 715
1 1 7 1 − = − 714 715 715 715 = 6 × 7−15
28. If 3(n + 4) - 3(n + 2) = 8, What is the value of n? A. 0
B. -1
C. -2
D. 2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : 3(n + 4) - 3(n + 2) = 8 3(n + 2 + 2) - 3(n + 2) = 8 3(n + 2) × 3(2) - 3(n + 2) = 8 3(n + 2)[3(2) - 1] = 8 3(n + 2) × 8 = 8 3(n + 2) = 1 => n + 2 = 0 => n = -2
29.
7 −−−− If 2x = √ 1024 , what is the value of x ?
A. None of these
B. -7/10
C. 10/7
D. 7/10
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
a p/q = √q −− ap −−−− 2x = √7 1024 −− 7 − ⇒ 2x = √ 210 ⇒ 2x = 210/7 ⇒ x = 10/7
30.
[(3−2 − 5−2 )
17
15 4 7 C. 4 A.
÷ (3−2 − 5−2 )
18
4 15 4 D. 7 B.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
1/2
]
=?
17
[(3−2 − 5−2 )
−2
−2 −5 )
−2
−2 −5 )
= [(3
= [(3
÷ (3−2 − 5−2 ) (17−18)
(−1)
]
1/2
1/2
1
=
(3−2 − 5−2 ) 1/2
1
= (
1 32
−
1 52
) 1/2
32 × 52
=
(52 − 32 )
=[
32 × 52 ] 16
=
3×5 4
=
15 4
1/2
]
1/2
18
]
1/2
I mportan t Formu las - Time an d Distan ce Distance Time
1.
Speed =
2.
Distance = Speed × Time
3.
Time =
4.
Distance Speed
To convert Kilometers per Hour(km/hr) to Meters per Second(m/s)
x km/hr = x × 5.
5 m/s 18
To convert Meters per Second(m/s) to Kilometers per Hour(km/hr)
x m/s = x ×
18 km/hr 5
6.
If a car covers a certain distance at x kmph and an equal distance at y kmph, 2xy the average speed of the whole journey = kmph x+y
7.
Speed and time are inversely proportional (when distance is constant) 1 ⇒ Speed ∝ (when distance is constant) Time
8.
If the ratio of the speeds of A and B is a : b, then the ratio of the times taken by them to cover the same distance is
1 1 : or b : a a b
1. A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is A. 11 hrs
B. 8 hrs 45 min
C. 7 hrs 45 min
D. 9 hts 20 min
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : ---------------------------------------------------------------Solution 1 ---------------------------------------------------------------Given that time taken for riding both ways will be 2 hours lesser than the time needed for waking one way and riding back
From this, we can understand that time needed for riding one way = time needed for waking one way - 2 hours
Given that time taken in walking one way and riding back = 5 hours 45 min Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min
In fact, you can do all these calculations mentally and save a lot of time which will be a real benefit for you. ---------------------------------------------------------------Solution 2 -----------------------------------------------------------------
Let the distance be x km. Then
(Time taken to walk x km) + (Time taken to ride x km) = 5 hour 45 min 45 3 23 =5 hour = 5 hour = hour .......(Equation1) 60 4 4
(Time taken to ride 2x km) = 5 hour 45 min - 2 = 3 hour 45 min 45 3 15 =3 hour = 3 hour = hour .......(Equation2) 60 4 4
Solving equations 1 and 2 (Equation 1 ) × 2 ⇒ (Time taken to walk 2x km)+ (Time taken to ride 2x km) = 23 hour .......(Equation3) 2
(Equation 3 - Equation 2) ⇒ Time taken to walk 2x km = =
23 15 − 2 4
46 15 31 3 − = hours = 7 hours = 7 hours 45 minutes 4 4 4 4
2. A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour? A. 8.2
B. 4.2
C. 6.1
D. 7.2
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : distance = 600 meter time = 5 minutes = 5 x 60 seconds = 300 seconds
Speed = = 2×
distance 600 = = 2m/s time 300
18 36 km/hr = km/hr = 7.2 km/hr 5 5
3. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour? A. 12
B. 11
C. 10
D. 9
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : speed of the bus excluding stoppages = 54 kmph speed of the bus including stoppages = 45 kmph
Loss in speed when including stoppages = 54 - 45 = 9kmph => In 1 hour, bus covers 9 km less due to stoppages
Hence, time that the bus stop per hour = time taken to cover 9 km
=
distance 9 1 60 = hour = hour = min = 10 min speed 54 6 6
4. A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km. A. 121 km
B. 242 km
C. 224 km
D. 112 km
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : distance = speed x time
Let time taken to travel the first half = x hr then time taken to travel the second half = (10 - x) hr
Distance covered in the the first half = 21x Distance covered in the the second half = 24(10 - x)
But distance covered in the the first half = Distance covered in the the second half => 21x = 24(10 - x) => 21x = 240 - 24x => 45x = 240 => 9x = 48 => 3x = 16
⇒x=
16 3
Hence Distance covered in the the first half = 21x = 21 × Total distance = 2 × 112 = 224 km
16 = 7 × 16 = 112 km 3
5. A car traveling with 5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. What is the actual speed of the car? A. 30 km/hr
B. 35 km/hr
C. 25 km/hr
D. 40 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Time = 1 hr 40 min 48 sec = 1hr + distance = 42 km speed =
⇒
distance 42 = time 126 ( ) 75
=
40 48 2 1 126 hr + hr = 1 + + = hr 60 3600 3 75 75
42 × 75 126
5 42 × 75 of the actual speed = 7 126
⇒ actual speed =
42 × 75 7 42 × 15 7 × 15 × = = = 7 × 5 = 35 km/hr 126 5 18 3
6. A man covered a certain distance at some speed. If he had moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. What is the the distance in km? A. 36
B. 38
C. 40
D. 42
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the distance be x km , the speed in which he moved = v kmph
Time taken when moving at normal speed - time taken when moving 3 kmph faster = 40 minutes
x x 40 − = v v+3 60 1 1 2 ]= ⇒ x[ − v v+3 3 ⇒
⇒ x[
v+3−v 2 ]= 3 v (v + 3)
⇒ 2v(v + 3) = 9x................(Equation1)
Time taken when moving 2 kmph slower - Time taken when moving at normal speed = 40 minutes
x x 40 − = v−2 v 60 1 1 2 ⇒ x[ − ]= v−2 v 3 ⇒
⇒ x[
v−v+2 2 ]= 3 v (v − 2)
⇒ x[
2 2 ]= 3 v (v − 2)
⇒ x[
1 1 ]= 3 v (v − 2)
⇒ v(v − 2) = 3x................(Equation2)
2(v + 3) Equation1 ⇒ =3 Equation2 (v − 2) ⇒ 2v + 6 = 3v − 6 ⇒ v = 12
Substituting this value of v in Equation 1 ⇒ 2 × 12 × 15 = 9x 2 × 12 × 15 2 × 4 × 15 => x = = = 2 × 4 × 5 = 40 9 3 Hence distance = 40 km
7. A and B walk around a circular track. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. If they start at 8 a.m. from the same point in opposite directions, how many times shall they cross each other before 9.30 a.m.? A. 5
B. 6
C. 7
D. 8
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Relative speed = Speed of A + Speed of B (∴ they walk in opposite directions) = 2 + 3 = 5 rounds per hour
=> They cross each other 5 times in 1 hour and 2 times in 1/2 hour
Time duration from 8 am to 9.30 am = 1.5 hour
Hence they cross each other 7 times before 9.30 am
8. Two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction. What time will they take to be 8.5 km apart? A. 17 hr
B. 14 hr
C. 12 hr
D. 19 hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Relative speed = 5.5 - 5 = .5 kmph (because they walk in the same direction) distance = 8.5 km
time =
distance 8.5 = = 17 hr speed .5
9. In covering a distance of 30 km, Arun takes 2 hours more than Anil. If Arun doubles his speed, then he would take 1 hour less than Anil. What is Arun's speed? A. 8 kmph
B. 5 kmph
C. 4 kmph
D. 7 kmph
Hide Answer | Notebook | Discuss Here is the answer and explanation
Answer : Option B Explanation : Let the speed of Arun = x kmph and the speed of Anil = y kmph
distance = 30 km
We know that
distance = time speed
Hence 30 30 − = 2...........(Equation1) x y 30 30 − = 1...........(Equation2) y 2x Equation1 + Equation2 ⇒ ⇒
30 =3 2x
⇒
15 =3 x
⇒
5 =1 x
30 30 − =3 x 2x
⇒x=5 Hence Arun's speed = 5 kmph
10. A car travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. What is the average speed for the first 320 km of the tour? A. 70.24 km/hr
B. 74. 24 km/hr
C. 71.11 km/hr
D. 72.21 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : ------------------------------------------Solution 1 (Quick) --------------------------------------------
If a car covers a certain distance at x kmph and an equal distance at y kmph, 2xy the average speed of the whole journey = kmph. x+y By using the same formula, we can find out the average speed quickly 2 × 64 × 80 2 × 64 × 80 2 × 32 × 40 average speed = = = 64 + 80 144 36 =
2 × 32 × 10 64 × 10 = = 71.11 kmph 9 9
-------------------------------------------Solution 2 (Fundamentals)--------------------------------------------Car travels first 160 km at 64 km/hr
Time taken to travel first 160 km =
distance 160 = speed 64
Car travels next160 km at 80 km/hr
Time taken to travel next 160 km =
distance 160 = speed 80
Total distance traveled = 160 + 160 = 2 × 160 Total time taken = Average speed =
=
160 160 + 64 80 Total distance traveled Total time taken
=
2 × 160 160 160 + 64 80
2 2 × 64 × 80 2 × 64 × 80 2 × 8 × 80 640 = = = = 1 1 80 + 64 144 18 9 + 64 80
= 71.11 km/hr
11. A Man travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly on bicycle at 9 km/hr. What is the distance travelled on foot? A. 12 km
B. 14 km
C. 16 km
D. 18 km
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : ----------------------------------------------------------------------Solution 1 ----------------------------------------------------------------------Let the time in which he travelled on foot = x hr Then the time in which he travelled on bicycle = (9 - x) hr
distance = speed x time
⇒ 4x + 9(9 − x) = 61 ⇒ 4x + 81 − 9x = 61 ⇒ 5x = 20 ⇒x=4
⇒ distance travelled on foot = 4x = 4 × 4 = 16 km -----------------------------------------------------------------------Solution 2-----------------------------------------------------------------------let the distance he Then the distance he travelled on bicycle = (61-x) km
Time =
⇒
Distance Speed
x (61 − x) + =9 4 9
⇒ 9x + 4 × 61 − 4x = 36 × 9 ⇒ 5x + 244 = 324 ⇒ 5x = 324 − 244 = 80 ⇒x=
80 = 16 km 5
12. Walking 6/7th of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance? A. 1 hr 42 min
B. 1 hr
C. 2 hr Hide Answer | Notebook
D. 1 hr 12 min | Discuss
Here is the answer and explanation Answer : Option D Explanation : New speed = 6/7 of usual speed
Speed and time are inversely proportional. Hence new time = 7/6 of usual time
Hence, 7/6 of usual time - usual time = 12 minutes => 1/6 of usual time = 12 minutes => usual time = 12 x 6 = 72 minutes = 1 hour 12 minutes
13. A man goes to his office from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming, what is the distance between his house and office? A. 3 km
B. 4 km
C. 5 km
D. 6 km
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
If a car covers a certain distance at x kmph and an equal distance at y kmph, 2xy the average speed of the whole journey = kmph x+y Hence, average speed =
2×3×2 12 = km/hr 2+3 5
Total time taken = 5 hours 12 ⇒ Distance travelled = × 5 = 12 km 5 ⇒ Distance between his house and office =
12 = 6 km 2
14. A man rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. What is his average speed for the entire trip approximately? A. 11.2 kmph
B. 10 kmph
C. 10.2 kmph
D. 10.8 kmph
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Total distance travelled = 10 + 12 = 22 km distance 10 = hr speed 12 distance 12 Time taken to travel 12 km at an average speed of 10 km/hr = = hr speed 10 10 12 Total time taken = + hr 12 10 Time taken to travel 10 km at an average speed of 12 km/hr =
distance 22 22 × 120 = = time 10 12 (10 × 10) + (12 × 12) ( ) + 12 10 22 × 120 11 × 120 11 × 60 660 = = = ≈ 10.8 kmph 244 122 61 61
Average speed =
15. An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in A. 660 km/hr
B. 680 km/hr
1
2 hours, it must travel at a speed of: 3
C. 700 km/hr
D. 720 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : -------------------------------------------------------------------------------Solution 1 (Recommended) --------------------------------------------------------------------------------
Speed and time are inversely proportional 1 ⇒ Speed ∝ (when distance is constant) Time
Here distance is constant and Speed and time are inversely proportional 1 Speed ∝ Time ⇒
Speed1 Time2 = Speed2 Time1
240 ⇒ = Speed2
(1
2 ) 3 5
5 ( ) 3 240 ⇒ = Speed2 5 ⇒
240 1 = Speed2 3
⇒ Speed2 = 240 × 3 = 720 km/hr ---------------------------------------------------------------------------------Solution 2-------------------------------------------------------------------------------
New time = 1
2 5 hr = hr 3 3
Hence, new speed =
Distance 240 × 5 = = 240 × 3 = 720 km/hr Time 5 3
16. A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. What is the speed of the car? A. 80 kmph
B. 102 kmph
C. 120 kmph
D. 140 kmph
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation :
Let speed of the car = x kmph Then speed of the train =
(100 + 50) 150 3 x= x = x kmph 100 100 2
Time taken by the car to travel from A to B = Time taken by the train to travel from A to B =
75 hours x 75 3 ( x) 2
+
12.5 hours 60
Since Both start from A at the same time and reach point B at the same time 75 = x
75 3 ( x) 2
+
12.5 60
25 12.5 = x 60 x=
25 × 60 = 2 × 60 = 120 12.5
17. In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. What is the duration of the flight ?
1 hour 2
A.
2 hour
B.
1
C.
1 hour 2
D.
1 hour
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Let the duration of the flight = x hours Given that distance = 600 km
Speed =
Distance 600 = ................(Equation1) Time x
Duration of the flight due to the slow down = (x + New speed =
30 1 ) hours = (x + ) hours 60 2
600 ................(Equation2) 1 (x + ) 2
From Equations 1 and 2, Reduction in Speed =
600 − x
600 1 (x + ) 2
Given that Reduction in average speed =200 km/hr ⇒
600 − x
600 = 200 1 (x + ) 2
⇒
3 − x
⇒
3 6 − =1 x 2x + 1
⇒
3(2x + 1) − 6x x(2x + 1)
⇒
6x + 3 − 6x x(2x + 1)
⇒
3 =1 x(2x + 1)
3 =1 1 (x + ) 2
=1
=1
⇒ 2x 2 + x − 3 = 0................(Equation3) From here, you can get the answer using Trial and error method. If you try with the values given as the choices, you can see the value of x = 1 satisfies the equation 3. Hence answer is 1 hour Or, we can solve the equation 3 to get the answer ⇒ 2x 2 + x − 3 = 0 ⇒ (2x + 3)(x − 1) = 0 ⇒ x = 1(Removing the -ve value for x) Hence answer is 1 hour
18. If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. What is the actual distance travelled by him? A. 80 km
B. 70 km
C. 60 km
D. 50 km
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Assume that the person would have covered x km if travelled at 10 km/hr
⇒ Speed =
Distance x = .....(Equation1) Time 10
Give that the person would have covered (x + 20) km if travelled at 14 km/hr
⇒ Speed =
(x + 20) Distance = .....(Equation2) Time 14
From Equations 1 and 2,
(x + 20) x = 10 14 14x = 10x + 200 4x = 200 x=
200 = 50 4
19. The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, What is the the speed of the first train? A. 85 km/hr
B. 87.5 km/hr
C. 90 km/hr
D. 92.5 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : -------------------------------------------------------------------------------Solution 1 (Recommended) --------------------------------------------------------------------------------
Speed and time are inversely proportional (when distance is constant) 1 ⇒ Speed ∝ (when distance is constant) Time Here distance is constant and hence speed and time are inversely proportional 1 Speed ∝ Time ⇒
Speed1 Time2 = Speed2 Time1
⇒
7 4 = 8 Time1
⇒ Time1 =
4×8 hr 7
⇒ Speed of the first train =
=
Distance 400 = Time1 4×8 ( ) 7
100 × 7 = 12.5 × 7 = 87.5 km/hr 8
---------------------------------------------------------------------------------Solution 2------------------------------------------------------------------------------speed of the trains are 7x and 8x respectively. Given that second train runs 400 km in 4 hours.
⇒ Speed of the 2nd train =
Distance 400 = = 100 km/hr Time 4
⇒ 8x = 100 100 ⇒x= = 12.5 8 ⇒ Speed of the first train = 7x = 7 × 12.5 = 87.5 km/hr
20. It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. What is the ratio of the speed of the train to that of the car? A. 3 : 4
B. 2 : 3
C. 1 : 2
D. 1 : 3
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let speed of the train = x kmph and speed of the car = y kmph
Time needed for traveling 600 km if 120 km by train and the rest by car = 8 hr
⇒
120 (600 − 120) + =8 x y
⇒
120 480 + = 8.............(Equation 1) x y
⇒
15 60 + = 1.............(Equation 1) x y
Time needed for traveling 600 km if 200 km by train and the rest by car = 8 hr 20 min
⇒
200 (600 − 200) 20 1 25 + =8 =8 = x y 60 3 3
⇒
200 400 25 + = x y 3
⇒
8 16 1 + = x y 3
⇒
24 48 + = 1.............(Equation 2) x y
Solving Equation1 and Equation2 Here Equation1 = Equation2 = 1 ⇒
15 60 24 48 + = + x y x y
⇒
12 9 = y x
⇒
4 3 = y x
⇒
x 3 = y 4
⇒ x :y= 3 :4
21. Arun is traveling on his cycle and has calculated to reach point A at 2 pm if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 pm? A. 8 kmph
B. 10 kmph
C. 12 kmph
D. 14 kmph
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let the distance be x km
if travels at 10 kmph, Arun will reach point A at 2 pm if travels at 15 kmph, Arun will reach point 12 noon => Time taken when traveling at 10 km = Time taken when traveling at 15 km + 2 hours
⇒
x x = +2 10 15
⇒
x x − =2 10 15
⇒ 3x − 2x = 2 × 30 ⇒ x = 60 ⇒ Distance = 60 km Examine the any statement say, if travels at 10 kmph, Arun will reach point A at 2 pm
⇒ Time taken =
Distance 60 = = 6 hours Speed 10
⇒ He must have started 6 hours back 2 pm, ie, at 8 am ⇒ Now he wants to reach at 1 pm. ie; time to be taken = 5 hours ⇒ Speed needed =
Distance 60 = = 12 kmph Time 5
22. A car travels at an average of 50 miles per hour for
2
hours?
1 1 hours and then travels at a speed of 70 miles per hour for 1 hours. How far did the car travel in the entire 4 2 2
A. 210 miles
B. 230 miles
C. 250 miles
D. 260 miles
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation :
Speed1 = 50 miles/hour 1 5 Time1 = 2 hour = hour 2 2 Distance1 = Speed1 × Time1 = 50 ×
Speed2 = 70 miles/hour 1 3 Time2 = 1 hour = hour 2 2 Distance2 = Speed2 × Time2 = 70 ×
5 = 25 × 5 = 125 miles 2
3 = 35 × 3 = 105 miles 2
Total Distance = Distance1 + Distance2 = 125 + 105 = 230 miles
23. The speed of a bus increases by 2 km after every one hour. If the distance travelling in the first one hour was 35 km. what was the total distance travelled in 12 hours? A. 422 km
B. 552 km
C. 502 km
D. 492 km
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Given that distance travelled in 1st hour = 35 km and speed of the bus increases by 2 km after every one hour
Hence distance travelled in 2nd hour = 37 km Hence distance travelled in 3rd hour = 39 km ...
Total Distance Travelled = [35 + 37 + 39 + ... (12 terms)]
This is an Arithmetic Progression(AP) with first term, a=35, number of terms,n = 12 and common difference, d=2.
The sequence a , (a + d), (a + 2d), (a + 3d), (a + 4d), . . . is called an Arithmetic Progression(AP) where a is the first term and d is the common difference of the AP Sum of the first n terms of an Arithmetic Progression(AP), n S n = [2a + (n − 1) d] 2 where n = number of terms Hence, [35 + 37 + 39 + ... (12 terms)] 12 = S 12 = [2 × 35 + (12 − 1) 2] 2 = 6 [70 + 22] = 6 × 92 = 552 Hence the total distance travelled = 552 km
24. Sound is said to travel in air at about 1100 feet per second. A man hears the axe striking the tree, 11/5 seconds after he sees it strike the tree. How far is the man from the wood chopper? A. 1800 ft
B. 2810 ft
C. 3020 ft
D. 2420 ft
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Speed of the sound = 1100 ft/s Time = 11/5 second
Distance = Speed × Time = 1100 ×
11 = 220 × 11 = 2420 ft 5
25. An athlete runs 200 metres race in 24 seconds. What is his speed? A. 20 km/hr
B. 25 km/hr
C. 27.5 km/hr
D. 30 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Speed = =
Distance 200 200 18 = m/s = × km/hr Time 24 24 5
40 × 3 km/hr = 10 × 3 km/hr = 30 km/hr 4
26. A train is moving at the speed of 80 km/hr. What is its speed in metres per second?
2 m/s 9 1 C. 21 m/sec 9 A.
22
B.
22 m/s
D.
21 m/s
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Speed = 80 km/hr = 80 ×
5 5 200 2 m/s = 40 × m/s = m/s = 22 m/s 18 9 9 9
27. The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travel towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 Km/hr. At what time will they meet? A. 10.30 a.m. C. 12 noon
B. 10 a.m. D. 11 a.m.
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Assume that they meet x hours after 8 a.m.
Then, train1,starting from A , travelling towards B, travels x hours till the trains meet
⇒ Distance travelled by train1 in x hours = Speed × Time = 60x Then, train2, starting from B , travelling towards A, travels (x-1) hours till the trains meet
⇒ Distance travelled by train2 in (x-1) hours = Speed × Time = 75(x-1) Total distance travelled = Distance travelled by train1 + Distance travelled by train2 => 330 = 60x + 75(x-1) => 12x + 15(x-1) = 66 => 12x + 15x - 15 = 66 => 27x = 66 + 15 = 81 => 3x = 9 => x = 3 Hence the trains meet 3 hours after 8 a.m., i.e. at 11 a.m.
28. A man walking at the rate of 5 km/hr crosses a bridge in 15 minutes. What is the length of the bridge (in metres)? A. 1250
B. 1280
C. 1320
D. 1340
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Speed = 5 km/hr Time = 15 minutes =
1 hour 4
Length of the bridge = Distance Travelled by the man = Speed × Time = 5 × = 5×
1 km 4
1 × 1000 metre = 1250 metre 4
29. A train travelled at an average speed of 100 km/hr, stopping for 3 minutes after every 75 km. How long did it take to reach its destination 600 km from the starting point? A. 6 hrs 21 min
B. 7 hrs 14 min
C. 7 hrs 22 min
D. 6 hrs
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation :
Time taken to travel 600 km =
Distance 600 = = 6 hour Speed 100
Now we need to find out the number of stops in the 600 km travel. Given that train stops after every 75 km.
600 =8 75 It means train stops 7 times before 600 km and 1 time just after 600 km. Hence we need to take only 7 stops into consideration for the 600 km travel. Hence, total stopping time in the 600 km travel = 7 x 3 = 21 minutes
Total time needed to reach the destination = 6 hours + 21 minutes = 6 hrs 21 min
30. A person travels from A to B at a speed of 40 km/hr and returns by increasing his speed by 50%. What is his average speed for both the trips? A. 60 km/hr
B. 56 km/hr
C. 52 km/hr
D. 48 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : ------------------------------------------Solution 1 (Quick) --------------------------------------------
If a car covers a certain distance at x kmph and an equal distance at y kmph, 2xy the average speed of the whole journey = kmph. x+y By using the same formula, we can find out the average speed quickly Speed with which he travels from A to B = x = 40 km/hr (100 + 50) Speed with which he travels from B to A = x = 40 × = 60 km/hr 100 2 × 40 × 60 average speed = = 48 km/hr 40 + 60 -------------------------------------------Solution 2 (Fundamentals)--------------------------------------------Assume that distance from A to B = x km Speed with which he travels from A to B = x = 40 km/hr
distance x = hr speed 40 (100 + 50) Speed with which he travels from B to A = 40 × = 60 km/hr 100 distance x Time to travel from B to A = = speed 60 Time to travel from A to B =
Total distance traveled = x + x = 2x x x + 40 60
Total time taken = Average speed =
=
Total distance traveled Total time taken
2x = x x + 44 60
2 2 × 2400 = = 2 × 24 = 48 km/hr 1 1 40 + 60 + 60 40
31. A man in a train notices that he can count 21 telephone posts in one minute. If they are known to be 50 metres apart, at what speed is the train travelling? A. 61 km/hr
B. 56 km/hr
C. 63 km/hr
D. 60 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : The man in the train notices that he can count 21 telephone posts in one minute. Number of gaps between 21 posts are 20 and Two posts are 50 metres apart. It means 20 x 50 meters are covered in 1 minute.
Distance = 20 × 50 meter = 1 hour 60
Time = 1 minute = Speed =
20 × 50 km = 1 km 1000
Distance = Time
1 = 60 km/hr 1 ( ) 60
32. A truck covers a distance of 550 metres in 1 minute whereas a train covers a distance of 33 kms in 45 minutes. What is the ratio of their speed? A. 2 : 1
B. 1 : 2
C. 4 : 3
D. 3 : 4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation :
Speed of the truck =
Distance 550 = = 550 meters/minute Time 1
Speed of the train =
Distance 33 33000 = km/minute = meters/minute Time 45 45
Speed of the truck Speed of the train
=
=
550 33000 ( ) 45
=
550 × 45 55 × 45 = 33000 3300
11 × 45 11 × 9 9 3 = = = 660 132 12 4
Hence, Speed of the truck : Speed of the train = 3 : 4
33. A person has to cover a distance of 6 km in 45 minutes. If he covers one-half of the distance in two-thirds of the total time; to cover the remaining distance in the remaining time, what should be his speed in km/hr? A. 14 km/hr
B. 12 km/hr
C. 10 km/hr
D. 8 km/hr
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : The person needs to cover 6 km in 45 minutes
Given that he covers one-half of the distance in two-thirds of the total time => He covers half of 6 km in two-thirds of 45 minutes => He covers 3 km in 30 minutes
Hence, now he need to cover the remaining 3 km in the remaining 15 minutes Distance = 3 km Time = 15 minutes = 1/4 hour
Required Speed =
Distance 3 = Time 1 ( ) 4
= 12 km/hr
I mportan t Formu las - Time an d W ork 1.
If A can do a piece of work in n days, work done by A in 1 day = 1/n
2.
If A does 1/n work in a day, A can finish the work in n days
3.
If M1 men can do W1 work in D1 days working H1 hours per day and M2 men can do W2 work in D2 days working H2 hours per day (where all men work at the same rate), then M1 D1 H1 / W1 = M2 D2 H2 / W2
4.
If A can do a piece of work in p days and B can do the same in q days, A and B together can finish it in pq / (p+q) days
5.
If A is thrice as good as B in work, then Ratio of work done by A and B = 3 : 1 Ratio of time taken to finish a work by A and B = 1 : 3
1. P is able to do a piece of work in 15 days and Q can do the same work in 20 days. If they can work together for 4 days, what is the fraction of work left? A. 8/15
B. 7/15
C. 11/15
D. 2/11
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Amount of work P can do in 1 day = 1/15 Amount of work Q can do in 1 day = 1/20 Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60 Amount of work P and Q can together do in 4 days = 4 × (7/60) = 7/15 Fraction of work left = 1 – 7/15= 8/15
2. P can lay railway track between two stations in 16 days. Q can do the same job in 12 days. With the help of R, they completes the job in 4 days. How much days does it take for R alone to complete the work? A. 9(3/5) days
B. 9(1/5) days
C. 9(2/5) days
D. 10 days
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Amount of work P can do in 1 day = 1/16 Amount of work Q can do in 1 day = 1/12 Amount of work P, Q and R can together do in 1 day = 1/4 Amount of work R can do in 1 day = 1/4 - (1/16 + 1/12) = 3/16 – 1/12 = 5/48 => Hence R can do the job on 48/5 days = 9 (3/5) days
3. P, Q and R can do a work in 20, 30 and 60 days respectively. How many days does it need to complete the work if P does the work and he is assisted by Q and R on every third day? A. 10 days
B. 14 days
C. 15 days
D. 9 days
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Amount of work P can do in 1 day = 1/20 Amount of work Q can do in 1 day = 1/30 Amount of work R can do in 1 day = 1/60 P is working alone and every third day Q and R is helping him Work completed in every three days = 2 × (1/20) + (1/20 + 1/30 + 1/60) = 1/5 So work completed in 15 days = 5 × 1/5 = 1 Ie, the work will be done in 15 days
4. A is thrice as good as B in work. A is able to finish a job in 60 days less than B. They can finish the work in - days if they work together. A. 18 days
B. 22 ½ days
C. 24 days
D. 26 days
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : If A completes a work in 1 day, B completes the same work in 3 days Hence, if the difference is 2 days, B can complete the work in 3 days => if the difference is 60 days, B can complete the work in 90 days => Amount of work B can do in 1 day= 1/90 Amount of work A can do in 1 day = 3 × (1/90) = 1/30 Amount of work A and B can together do in 1 day = 1/90 + 1/30 = 4/90 = 2/45 => A and B together can do the work in 45/2 days = 22 ½ days
5. A can do a particular work in 6 days . B can do the same work in 8 days. A and B signed to do it for Rs. 3200. They completed the work in 3 days with the help of C. How much is to be paid to C?
A. Rs. 380
B. Rs. 600
C. Rs. 420
D. Rs. 400
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Amount of work A can do in 1 day = 1/6 Amount of work B can do in 1 day = 1/8 Amount of work A + B can do in 1 day = 1/6 + 1/8 = 7/24 Amount of work A + B + C can do = 1/3 Amount of work C can do in 1 day = 1/3 - 7/24 = 1/24 work A can do in 1 day: work B can do in 1 day: work C can do in 1 day = 1/6 : 1/8 : 1/24 = 4 : 3 : 1 Amount to be paid to C = 3200 × (1/8) = 400
6. 6 men and 8 women can complete a work in 10 days. 26 men and 48 women can finish the same work in 2 days. 15 men and 20 women can do the same work in - days. A. 4 days
B. 6 days
C. 2 days
D. 8 days
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b Work done by 6 men and 8 women in 1 day = 1/10 => 6m + 8b = 1/10 => 60m + 80b = 1 --- (1) Work done by 26 men and 48 women in 1 day = 1/2 => 26m + 48b = ½ => 52m + 96b = 1--- (2)
Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200 Work done by 15 men and 20 women in 1 day = 15/100 + 20/200 =1/4 => Time taken by 15 men and 20 women in doing the work = 4 days
7. A can do a piece of work in 4 hours . A and C together can do it in just 2 hours, while B and C together need 3 hours to finish the same work. B alone can complete the work in --- days. A. 12 hours
B. 6 hours
C. 8 hours
D. 10 hours
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Work done by A in 1 hour = 1/4 Work done by B and C in 1 hour = 1/3 Work done by A and C in 1 hour = 1/2 Work done by A,B and C in 1 hour = 1/4+1/3 = 7/12 Work done by B in 1 hour = 7/12 – 1/2 = 1/12 => B alone can complete the work in 12 hour
8. P can do a work in the same time in which Q and R together can do it. If P and Q work together, the work can be completed in 10 days. R alone needs 50 days to complete the same work. then Q alone can do it in A. 30 days
B. 25 days
C. 20 days
D. 15 days
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Work done by P and Q in 1 day = 1/10 Work done by R in 1 day = 1/50 Work done by P, Q and R in 1 day = 1/10 + 1/50 = 6/50
But Work done by P in 1 day = Work done by Q and R in 1 day . Hence the above equation can be written as Work done by P in 1 day × 2 = 6/50 => Work done by P in 1 day = 3/50 => Work done by Q and R in 1 day = 3/50 Hence work done by Q in 1 day = 3/50 – 1/50 = 2/50 = 1/25 So Q alone can do the work in 25 days
9. A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work? A. 37 ½ days
B. 22 days
C. 31 days
D. 22 days
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Work done by A in 20 days = 80/100 = 8/10 = 4/5 Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 --- (1) Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B) Work done by A and B in 1 day = 1/15 ---(2) Work done by B in 1 day = 1/15 – 1/25 = 2/75 => B can complete the work in 75/2 days = 37 ½ days
10. Machine P can print one lakh books in 8 hours. Machine Q can print the same number of books in 10 hours while machine R can print the same in 12 hours. All the machines started printing at 9 A.M. Machine P is stopped at 11 A.M. and the remaining two machines complete work. Approximately at what time will the printing of one lakh books be completed? A. 3 pm
B. 2 pm
C. 1:00 pm
D. 11 am
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C
Explanation : Work done by P in 1 hour = 1/8 Work done by Q in 1 hour = 1/10 Work done by R in 1 hour = 1/12 Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120 Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60
From 9 am to 11 am, all the machines were operating. Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60
Pending work = 1- 37/60 = 23/60 Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11 which is approximately equal to 2
Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm
11. P can finish a work in 18 days. Q can finish the same work in 15 days. Q worked for 10 days and left the job. how many days does P alone need to finish the remaining work? A. 8
B. 5
C. 4
D. 6
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Work done by P in 1 day = 1/18 Work done by Q in 1 day = 1/15
Work done by Q in 10 days = 10/15 = 2/3
Remaining work = 1 – 2/3 = 1/3 Number of days in which P can finish the remaining work = (1/3) / (1/18) = 6
12. 3 men and 7 women can complete a work in 10 days . But 4 men and 6 women need 8 days to complete the same work . In how many days will 10 women complete the same work? A. 50
B. 40
C. 30
D. 20
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Work done by 4 men and 6 women in 1 day = 1/8 Work done by 3 men and 7 women in 1 day = 1/10 Let 1 man does m work in 1 day and 1 woman does w work in 1 day. The above equations can be written as 4m + 6w = 1/8 ---(1) 3m + 7w = 1/10 ---(2) Solving equation (1) and (2) , we get m=11/400 and w=1/400 Amount of work 10 women can do in a day = 10 × (1/400) = 1/40 Ie, 10 women can complete the work in 40 days
13. A and B can finish a work 30 days if they work together. They worked together for 20 days and then B left. A finished the remaining work in another 20 days. In how many days A alone can finish the work? A. 60
B. 50
C. 40
D. 30
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Amount of work done by A and B in 1 day = 1/30 Amount of work done by A and B in 20 days = 20 × (1/30) = 20/30 = 2/3
Remaining work – 1 – 2/3 = 1/3 A completes 1/3 work in 20 days Amount of work A can do in 1 day = (1/3)/20 = 1/60 => A can complete the work in 60 days
14. A can complete a work in 12 days with a working of 8 hours per day. B can complete the same work in 8 days when working 10 hours a day. If A and B work together, working 8 hours a day, the work can be completed in --- days. A. 5 5⁄11
B. 4 5⁄11
C. 6 4⁄11
D. 6 5⁄11
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : A can complete the work in 12 days working 8 hours a day => Number of hours A can complete the work = 12×8 = 96 hours => Work done by A in 1 hour = 1/96
B can complete the work in 8 days working 10 hours a day => Number of hours B can complete the work = 8×10 = 80 hours => Work done by B in 1 hour = 1/80
Work done by A and B in 1 hour = 1/96 + 1/80 = 11/480 => A and B can complete the work in 480/11 hours A and B works 8 hours a day
Hence total days to complete the work with A and B working together = (480/11)/ (8) = 60/11 days = 5 5⁄11 days
15. P is 30% more efficient than Q. P can complete a work in 23 days. If P and Q work together, how much time will it take to complete the same work? A. 9
B. 11
C. 13
D. 15
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Work done by P in 1 day = 1/23 Let work done by Q in 1 day = q q × (130/100) = 1/23 => q = 100/(23×130) = 10/(23×13) Work done by P and Q in 1 day = 1/23 + 10/(23×13) = 23/(23×13)= 1/13 => P and Q together can do the work in 13 days
16. P, Q and R can complete a work in 24, 6 and 12 days respectively. The work will be completed in --days if all of them are working together. A. 2
B. 3 3⁄7
C. 4 ¼
D. 5
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Work done by P in 1 day = 1/24 Work done by Q in 1 day = 1/6 Work done by R in 1 day = 1/12 Work done by P,Q and R in 1 day = 1/24 + 1/6 + 1/12 = 7/24 => Working together, they will complete the work in 24/7 days = 3 3⁄7 days
17. 10 men can complete a work in 7 days. But 10 women need 14 days to complete the same work. How many days will 5 men and 10 women need to complete the work?
A. 5
B. 6
C. 7
D. 8
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Work done by 10 men in 1 day = 1/7 Work done by 1 man in 1 day = (1/7)/10 = 1/70
Work done by 10 women in 1 day = 1/14 Work done by 1 woman in 1 day = 1/140
Work done by 5 men and 10 women in 1 day = 5 × (1/70) + 10 × (1/140) = 5/70 + 10/140 = 1/7 => 5 men and 10 women can complete the work in 7 days
18. Kamal will complete work in 20 days. If Suresh is 25% more efficient than Kamal, he can complete the work in --- days. A. 14
B. 16
C. 18
D. 20
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Work done by Kamal in 1 day = 1/20 Work done by Suresh in 1 day = (1/20) × (125/100) = 5/80 = 1/16 => Suresh can complete the work in 16 days
19. Anil and Suresh are working on a special assignment. Anil needs 6 hours to type 32 pages on a computer and Suresh needs 5 hours to type 40 pages. If both of them work together on two different computers, how much time is needed to type an assignment of 110 pages?
A. 7 hour 15 minutes
B. 7 hour 30 minutes
C. 8 hour 15 minutes
D. 8 hour 30 minutes
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Pages typed by Anil in 1 hour = 32/6 = 16/3 Pages typed by Suresh in 1 hour = 50/5 = 8 Pages typed by Anil and Suresh in 1 hour = 16/3 + 8 = 40/3 Time taken to type 110 pages when Anil and Suresh work together = 110 × 3 /40 = 33/4 = 8 ¼ hours = 8 hour 15 minutes
20. P and Q can complete a work in 20 days and 12 days respectively. P alone started the work and Q joined him after 4 days till the completion of the work. How long did the work last? A. 5 days
B. 10 days
C. 14 days
D. 22 days
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Work done by P in 1 day = 1/20 Work done by Q in 1 day = 1/12
Work done by P in 4 days = 4 × (1/20) = 1/5 Remaining work = 1 – 1/5 = 4/5
Work done by P and Q in 1 day = 1/20 + 1/12 = 8/60 = 2/15 Number of days P and Q take to complete the remaining work = (4/5) / (2/15) = 6 Total days = 4 + 6 = 10
21. P takes twice as much time as Q or thrice as much time as R to finish a piece of work. They can finish the work in 2 days if work together. How much time will Q take to do the work alone? A. 4
B. 5
C. 6
D. 7
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let P takes x days to complete the work Then Q takes x/2 days and R takes x/3 days to finish the work Amount of work P does in 1 day = 1/x Amount of work Q does in 1 day = 2/x Amount of work R does in 1 day = 3/x Amount of work P,Q and R do in 1 day = 1/x + 2/x + 3/x = 1/x (1 + 2 + 3) = 6/x 6/x = 2 => x = 12 => Q takes 12/2 days = 6 days to complete the work
22. P and Q can complete a work in 15 days and 10 days respectively. They started the work together and then Q left after 2 days. P alone completed the remaining work. The work was finished in --- days. A. 12
B. 16
C. 20
D. 24
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Work done by P in 1 day = 1/15 Work done by Q in 1 day = 1/10
Work done by P and Q in 1 day = 1/15 + 1/10 = 1/6
Work done by P and Q in 2 days = 2 × (1/6) = 1/3 Remaining work = 1 – 1/3 = 2/3
Time taken by P to complete the remaining work 2/3 = (2/3) / (1/15) = 10 days Total time taken = 2 + 10 = 12 days
23. P and Q can do a work in 30 days. Q and R can do the same work in 24 days and R and P in 20 days. They started the work together, but Q and R left after 10 days. How many days more will P take to finish the work? A. 10
B. 15
C. 18
D. 22
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Let work done by P in 1 day = p, Work done by Q in 1 day = q, Work done by R in 1 day = r
p + q = 1/30 q + r = 1/24 r + p = 1/20
Adding all the above, 2p + 2q + 2r = 1/30 + 1/24+ 1/20 = 15/120 = 1/8 => p + q + r = 1/16 => Work done by P,Q and R in 1 day = 1/16
Work done by P, Q and R in 10 days = 10 × (1/16) = 10/16 = 5/8 Remaining work = 1 = 5/8 = 3/8
Work done by P in 1 day = Work done by P,Q and R in 1 day - Work done by Q and R in 1 day = 1/16 – 1/24 = 1/48
Number of days P needs to work to complete the remaining work = (3/8) / (1/48) = 18
24. P works twice as fast as Q. If Q alone can complete a work in 12 days, P and Q can finish the work in --- days A. 1
B. 2
C. 3
D. 4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Work done by Q in 1 day = 1/12 Work done by P in 1 day = 2 × (1/12) = 1/6
Work done by P and Q in 1 day = 1/12 + 1/6 = ¼ => P and Q can finish the work in 4 days
25. A work can be finished in 16 days by twenty women. The same work can be finished in fifteen days by sixteen men. The ratio between the capacity of a man and a woman is A. 1:3
B. 4:3
C. 2:3
D. 2:1
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Work done by 20 women in 1 day = 1/16 Work done by 1 woman in 1 day = 1/(16×20)
Work done by 16 men in 1 day = 1/15 Work done by 1 man in 1 day = 1/(15×16)
Ratio of the capacity of a man and woman =1/(15×16) : 1/(16×20) = 1/15 : 1/20 = 1/3 :1/4 = 4:3
26. P and Q need 8 days to complete a work. Q and R need 12 days to complete the same work. But P, Q and R together can finish it in 6 days. How many days will be needed if P and R together do it? A. 3
B. 8
C. 12
D. 4
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Let work done by P in 1 day = p work done by Q in 1 day =q Work done by R in 1 day = r p + q = 1/8 ---(1) q + r= 1/12 ---(2) p+ q+ r = 1/6 ---(3) (3) – (2) => p = 1/6 - 1/12 = 1/12 (3) – (1) => r = 1/6 – 1/8 = 1/24 p + r = 1/12 + 1/24 = 3/24 = 1/8 => P and R will finish the work in 8 days
27. P can do a work in 24 days. Q can do the same work in 9 days and R can do the same in 12 days. Q and R start the work and leave after 3 days. P finishes the remaining work in --- days.
A. 7
B. 8
C. 9
D. 10
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option D Explanation : Work done by P in 1 day = 1/24 Work done by Q in 1 day = 1/9 Work done by R in 1 day = 1/12
Work done by Q and R in 1 day = 1/9 + 1/12 = 7/36 Work done by Q and R in 3 days = 3×7/36 = 7/12 Remaining work = 1 – 7/12 = 5/12 Number of days in which P can finish the remaining work = (5/12) / (1/24) = 10
28. If daily wages of a man is double to that of a woman, how many men should work for 25 days to earn Rs.14400? Given that wages for 40 women for 30 days are Rs.21600. A. 12
B. 14
C. 16
D. 18
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Wages of 1 woman for 1 day =
Wages of 1 man for 1 day =
21600 × 2 40 × 30
Wages of 1 man for 25 days =
Number of men =
21600 40 × 30
21600 × 2 × 25 40 × 30
14400 21600 × 2 × 25 ( ) 40 × 30
=
144 216 × 50 ( ) 40 × 30
=
144 = 16 9
29. P,Q and R together earn Rs.1620 in 9 days. P and R can earn Rs.600 in 5 days. Q and R in 7 days can earn Rs.910. How much amount does R can earn per day? A. Rs.40
B. Rs.70
C. Rs.90
D. Rs.100
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option B Explanation : Amount Earned by P,Q and R in 1 day = 1620/9 = 180 ---(1) Amount Earned by P and R in 1 day = 600/5 = 120 ---(2) Amount Earned by Q and R in 1 day = 910/7 = 130 ---(3)
(2)+(3)-(1) => Amount Earned by P , Q and 2R in 1 day - Amount Earned by P,Q and R in 1 day = 120+130-180 = 70
=>Amount Earned by R in 1 day = 70
30. Assume that 20 cows and 40 goats can be kept for 10 days for Rs.460. If the cost of keeping 5 goats is the same as the cost of keeping 1 cow, what will be the cost for keeping 50 cows and 30 goats for 12 days? A. Rs.1104
B. Rs.1000
C. Rs.934
D. Rs.1210
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option A Explanation : Assume that cost of keeping a cow for 1 day = c , cost of keeping a goat for 1 day = g
Cost of keeping 20 cows and 40 goats for 10 days = 460
Cost of keeping 20 cows and 40 goats for 1 day = 460/10 = 46 => 20c + 40g = 46 => 10c + 20g = 23 ---(1) Given that 5g = c
Hence equation (1) can be written as 10c + 4c = 23 => 14c =23 => c=23/14
cost of keeping 50 cows and 30 goats for 1 day = 50c + 30g = 50c + 6c (substituted 5g = c) = 56 c = 56×23/14 = 92
Cost of keeping 50 cows and 30 goats for 12 days = 12×92 = 1104
31. There is a group of persons each of whom can complete a piece of work in 16 days, when they are working individually. On the first day one person works, on the second day another person joins him, on the third day one more person joins them and this process continues till the work is completed. How many days are needed to complete the work? A. 3 1⁄4 days
B. 4 1⁄3 days
C. 5 1⁄6 days
D. 6 1⁄5 days
Hide Answer | Notebook | Discuss Here is the answer and explanation Answer : Option C Explanation : Work completed in 1st day = 1/16 Work completed in 2nd day = (1/16) + (1/16) = 2/16 Work completed in 3rd day = (1/16) + (1/16) + (1/16) = 3/16
…
An easy way to attack such problems is from the choices. You can see the choices are very close to each other. So just see one by one.
For instance, The first choice given in 3 1⁄4 The work done in 3 days = 1/16 + 2/16 + 3/16 = (1+2+3)/16 = 6/16 The work done in 4 days = (1+2+3+4)/16 = 10/16 The work done in 5 days = (1+2+3+4+5)/16 = 15/16, almost close, isn't it? The work done in 6 days = (1+2+3+4+5+6)/16 > 1 Hence the answer is less than 6, but greater than 5. Hence the answer is 5 1⁄6days.
(Just for your reference, work done in 5 days = 15/16. Pending work in 6th day = 1 – 15/16 = 1/16. In 6th day, 6 people are working and work done = 6/16. To complete the work 1/16, time required = (1/16) / (6/16) = 1/6 days. Hence total time required = 5 + 1/6 = 5 1⁄6 days )