Current age of ‘A’ is 20% more than that of ‘B’.

Solution

ATQ, Let the present age of ‘B’ = ‘5x’ years Then, present age of ‘A’ = 5x × 1.20 = ‘6x’ years 5 years ago, age of ‘A’ = (6x – 5) years 5 years ago, age of ‘C’ = (6x – 5) × (6/5) = {(36x – 30)/5} years So, present age of ‘C’ = {(36x – 30)/5} + 5 = {(36x – 5)/5} years Sum of present ages of ‘B’ and ‘C’ = 5x + {(36x – 5)/5} = {(25x + 36x – 5)/5} = {(61x – 5)/5} years So, {(61x – 5)/5} = 62 Or, 61x – 5 = 310 Or, x = (310 + 5) ÷ 61 Or, x = 5 So, sum of present ages of ‘A’ and ‘B’ = 6x + 5x = 11x = 11 × 5 = 55 years