# Problems on Ages Quiz Set 016

### 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

The ages of two friends are 4 and 9 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)?

A

6 years.

B

7 years.

C

5 years.

D

8 years.

Soln.
Ans: a

If the age of the third friend is x. Then for a GP, x = \$√(4 × 9)\$ = 6 years.

### Question 3

The square root of twice my present age is equal to the sum of the square roots of my ages 10 years back and 10 years hence. What is my present age?

A

10 years.

B

11 years.

C

9 years.

D

20 years.

Soln.
Ans: a

Let the present age be x. Then \$√(x - 10) + √(x + 10)\$ = \$√{2x}\$. Squaring both sides, \$(√(x - 10))^2 + (√(x + 10))^2 + 2√{x^2 - 10^2}\$ = \$2x\$ which gives \$x - 10 + x + 10 + 2√{x^2 - 10^2}\$ = \$2x\$. Cancelling, and simplifying, \$2√(x^2 - 100) = 0\$, which leads to x = 10 years.

### Question 4

The ratio of ages of P and Q today is \${29/40}\$. After 6 years, their ages will be in the ratio \${151/206}\$. What is the age of P today?

A

145 years.

B

146 years.

C

144 years.

D

147 years.

Soln.
Ans: a

Let the ages of P and Q be 29x and 40x. After 6 years the ratio would be \${29x + 6}/{40x + 6}\$ = \${151/206}\$. Solving, we get x = 5. So age of P = 29 × 5 = 145.

### Question 5

The ages of two friends are 3 and 27 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)?

A

9 years.

B

10 years.

C

8 years.

D

11 years.

Soln.
Ans: a

If the age of the third friend is x. Then for a GP, x = \$√(3 × 27)\$ = 9 years.