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### Question 1

Each year the ages of three friends are in an AP(arithmetic progression). The age of middle friend today is 48 years. What would be the sum of their ages 10 years from now?

**A**

174 years.

**B**

175 years.

**C**

173 years.

**D**

176 years.

**Soln.**

**Ans: a**

Let the present ages of the three friends be a - d, a and a + d. As of today a = 48. Ten years later their ages would be a + 10 - d, a + 10, a + 10 + d. Adding these we get 3a + 30 which equals 3 × 48 + 30 = 174 years.

### Question 2

Each year the ages of three friends are in an AP(arithmetic progression). The age of middle friend today is 18 years. What would be the sum of their ages 10 years from now?

**A**

84 years.

**B**

85 years.

**C**

83 years.

**D**

86 years.

**Soln.**

**Ans: a**

Let the present ages of the three friends be a - d, a and a + d. As of today a = 18. Ten years later their ages would be a + 10 - d, a + 10, a + 10 + d. Adding these we get 3a + 30 which equals 3 × 18 + 30 = 84 years.

### Question 3

The ages of two friends are 4 and 25 years respectively. They are looking for a special third friend whose age is in-between their ages. What should be the age of the third friend if the ages of all three have to be in a GP(geometric progression)?

### Question 4

The age of father tortoise is 12 times the age of his son. After 42 years his age will be 5 times the age of his son. What would be the ratio of their ages 240 years from today?

**A**

2.

**B**

5.

**C**

3.

**D**

4.

**Soln.**

**Ans: a**

If the age of the son today is s, the age of the father is 12s. After 42 years, we have 12s + 42 = 5(s + 42). Solving for s, we get s = 24 years. The ratio of their ages after 240 years = ${12s + 240}/{s + 240}$. Substituting s and simplifying we get the ratio as 2.

### Question 5

The sum of the ages of 14 calves of a whale born at a gap of 8 years is 924. What is the age of the youngest calf?

**A**

14 years.

**B**

15 years.

**C**

13 years.

**D**

16 years.

**Soln.**

**Ans: a**

The ages of the calves are in an AP with d = 8, n = 14, and sum S = 924. We have to find the first term a. We know S = ${n/2} × (2a + (n-1)d)$ Putting the values 924 = ${14/2} × (2a + (14-1)×8)$. Solving, get a = 14 years.

This Blog Post/Article "Problems on Ages Quiz Set 015" by Parveen (Hoven) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.

Updated on 2020-02-07. Published on: 2016-04-23