BANK & INSURANCE (AGE BASED PROBLEMS) PART 2

Total Questions: 45

31. Seven years hence from now, the ratio of ages of mother and son will be 13:5, respectively. Five years ago from now, mother's age was five times the age of son. Find the sum of their present ages.

Correct Answer: (d) 58 years
Solution:

Let the age of son five years ago was 'x' years

So, the age of Mother five years ago was = '5x' years

The present age of mother and son will be (5x + 5) years and (x + 5) years, respectively.

According to the question,

((5x + 5 + 7)/(x + 5 + 7)) = (13/5)

25x + 60 = 13x + 156

12x = 96

x = 8

The present age of mother = 5x + 5 = 5 × 8 + 5 = 45 years

The present age of son = x + 5 = 8 + 5 = 13 years

The sum of their present ages = 45 + 13 = 58 years

32. Age of 'B', 2 years hence from now, will be 75% more than the age of 'A', 2 years ago from now. The sum of present ages of 'A' and 'B' is 66 years. How many years ago was 'B' twice as old as 'A'?

Correct Answer: (a) 12 years
Solution:

Let the present age of 'A' = 'x' years

Let the present age of 'B' = 'y' years

2 years ago, age of 'A' = (x − 2) years

2 years hence from now, age of 'B'
= (x − 2) × 1.75 = (1.75x − 3.5) years

So, present age of 'B' = 1.75x − 3.5 − 2 = (1.75x − 5.5) years

So, sum of present ages of 'A' and 'B'
= x + (1.75x − 5.5) = 2.75x − 5.5 = 66

So, x = (66 + 5.5) ÷ 2.75 = 26

So, present ages of 'A' and 'B' are 26 years and 40 years, respectively

Difference between ages of 'A' and 'B' = 14 years

Therefore, 'B' was twice as old as 'A', when 'A' was 14 years old

So, 'B' was twice as old as 'A' 12 years ago.

33. Vipin's age 2 years hence from now will be exactly half of his father's age 4 years ago from now. Present age of Vipin's father is 10% more than present age of Vipin's mother. Find Vipin's age, 7 years hence from now, if the present average age of the given three people is 34 years.

Correct Answer: (b) 25 years
Solution:

Let the present age Vipin's mother be 10x years

So, present age of Vipin's father = 10x + 10 × 0.1 = 11x years

Present age of Vipin = ((1/2) × (11x − 4) − 2) years

According to question:

((1/2) × (11x − 4) − 2) + 10x + 11x = (34 × 3)

11x − 4 − 4 + 42x = (102 × 2)

53x − 8 = 204

53x = 212

x = 4

So, present age of Vipin
= (1/2) × (11 × 4 − 4) − 2 = (20 − 2) years = 18 years

And, age of Vipin after 7 years from now

= 18 + 7 = 25 years

34. The present age of Ram is two years more than thrice the present age of Shyam. Ten years hence from now, the age of Ram will be eight years more than twice the age of Shyam at that time. Find the sum of the present ages of Ram and Shyam.

Correct Answer: (d) 66 years
Solution:

Let the present age of Shyam be 'x' years

The present age of Ram = (3x + 2) years

Ten years hence from now the age of Shyam = (x + 10) years

Ten years hence from now the age of Ram
= (3x + 2 + 10) = (3x + 12) years

Given,

3x + 12 = 2(x + 10) + 8

3x + 12 = 2x + 20 + 8

x = 16

The present age of Shyam = x = 16 years

The present age of Ram
= 3x + 2 = 3 × 16 + 2 = 50 years

The sum of the present ages of Ram and Shyam = 50 + 16 = 66 years

35. 5 years ago from now, ratio of ages of 'P' and 'Q' was 7:9, respectively. Present average age of 'P' and 'R' is 52 years. Find the present age of 'S' which is 36% more than the present age of 'Q'. If 4 years hence from now, the average age of 'Q' and 'R' will be 61 years.

Correct Answer: (d) 68 years
Solution:

Let the age of 'P', 5 years ago from now was '7x' years

And, age of 'Q', 5 years ago from now = 7x × (9/7) = '9x' years

So, present age of 'P' = (7x + 5) years

And, present age of 'Q' = (9x + 5) years

Sum of present ages of 'P' and 'R' = (52 × 2) = 104 years

So, present age of 'R' = 104 − (7x + 5) = (99 − 7x) years

According to question:

9x + 5 + 99 − 7x = (61 × 2) − 8

2x + 104 = 122 − 8

2x = 114 − 104

2x = 10

x = 5

So, present age of 'Q' = (9 × 5 + 5) = 50 years

And, present age of 'S' = (50 × 1.36) = 68 years

36. Sum of present ages of 'A' and 'B' is 20 years and sum of present ages of 'A' and 'C' is 30 years. If 6 years hence from now, the ages of 'A' and 'C' will be in ratio 3:4 respectively, then find the sum of present ages of 'A', 'B' and 'C'.

Correct Answer: (b) 38 years
Solution:

Let the present ages of 'A', 'B' and 'C' be 'a' year, 'b' years and 'c' years respectively.

ATQ:

a + b = 20 ...(I)

a + c = 30 ...(II)

ATQ:

(a + 6)/(c + 6) = 3/4

4a + 24 = 3c + 18

(4a + 6)/3 = c ...(III)

On substituting the value of 'c' in equation (II), we have:

[a + {(4a + 6)/3}] = 30

3a + 4a + 6 = 90

7a = 84

So, a = 12

So, b = 20 − 12 = 8

And c = 30 − 12 = 18

So, sum of present ages of 'A', 'B' and 'C' = 12 + 8 + 18 = 38 years

37. The sum of present age of 'A' and present average age of 'B' and 'C' is 47 years, whereas sum of present age of 'B' and present average age of 'A' and 'C' is 49 years. If the difference between present ages of 'A' and 'B' is 'D' years, then find the value of 'D'.

Correct Answer: (b) 4
Solution:

Let the present ages of 'A', 'B' and 'C' be 'a', 'b' and 'c' years respectively.

ATQ:

a + [(b + c)/2] = 47

2a + b + c = 94 ...(I)

Also,

b + [(a + c)/2] = 49

2b + a + c = 98 ...(II)

On subtracting equation (I) from equation (II), we have:

b − a = 4

So, D = 4

38. Present ages of 'A' and 'B' are in the ratio 2:5, respectively. If B's age, 13 years hence from now will be six times of A's age, 6 years ago from now, then find the present age of 'A'.

Correct Answer: (a) 14 years
Solution:

Let the present ages of 'A' and 'B' be '2x' years and '5x' years, respectively.

ATQ:

(5x + 13) = 6 × (2x − 6)

5x + 13 = 12x − 36

7x = 49

x = 7

So, present age of 'A' = 2 × 7 = 14 years

39. The sum of present ages of 'A' and 'B' is 45 years. 2 years hence from now, age of 'B' will be 25% less than that of 'A'. If the present average age of 'B' and 'C' is 32 years, then find the present age of 'C'.

Correct Answer: (c) 45 years
Solution:

Let the present age of 'A' = 'y' years

Then, present age of 'B' = (45 − y) years

2 years hence from now, ages of 'A' and 'B' will be
(y + 2) years and (47 − y) years, respectively

According to the question,

(y + 2) × 0.75 = 47 − y

 0.75y + 1.5 = 47 − y

1.75y = 45.5

So, y = 45.5 ÷ 1.75 = 26

So, present age of 'B' = 45 − 26 = 19 years

And, sum of present ages of 'B' and 'C'

= 32 × 2 = 64 years

So, present age of 'C' = 64 − 19 = 45 years

40. After 10 years, A's age will be twice that of B's age. A's present age is 6 times that of C. If B's eighth birthday was celebrated 2 years ago, then what is C's present age?

Correct Answer: (b) 5
Solution:

Let C's present age be x year.

Then, A's present age = 6x.

Let B's present age be y.

Then, after 10 years,

6x + 10 = 2(y + 10)

⇒ 6x + 10 = 2y + 20

⇒ 6x − 2y = 10

⇒ y = 3x − 5

'B's eighth birthday was celebrated 2 years ago, so,

B's present age = 10

Also, B's present age = y = 3x − 5

⇒ 10 = 3x − 5

⇒ x = 15/3 = 5