Problems on Ages Quiz Set 018

The ages of the calves are in an AP with d = 4, n = 10, and sum S = 280. We have to find the first term a. We know S = ${n/2} × (2a + (n-1)d)$ Putting the values 280 = ${10/2} × (2a + (10-1)×4)$. Solving, get a = 10 years.

Let the ages of three friends be 2r, 23r and 3r. The youngest of these is 2r. We have been given their sum 3 years back. So (2 + 23 + 3)r - (3 × 3) = 159. Solving, we get r = 6. The youngest is 2 × 6 = 12 years.

The ratio of ages of A and B is given as ${29/10}$, which is same as: $2{9/10}$. So we can write the present ages of A and B, respectively, as 29r and 10r years. 4 years later the age of A is $29r + 4 = 381$ which gives r = 13. The age of B, therefore, is 10r = 10 × 13 = 130 years.

The ratio of ages of A and B is given as ${9/4}$, which is same as: $2{1/4}$. So we can write the present ages of A and B, respectively, as 9r and 4r years. 3 years later the age of A is $9r + 3 = 102$ which gives r = 11. The age of B, therefore, is 4r = 4 × 11 = 44 years.

Let the ages after 5 years be a - d, a and a + d. The sum is given to us. So (a - d) + a + (a + d) = 3a = 99. We get a = 99/3 = 33. So, the age of the middle friend today is a - 5 = 28 years.