The average of present ages of 'Emily', 'Frank', and 'George' is 25 years. Four years ago, the average of ages of 'Frank', 'George', and 'Harry' was 18 years. If the sum of the present ages of 'Emily' and 'Harry' is 58 years, then what was the age of 'Emily' three years ago from now?

Solution

ATQ, Sum of the present ages of 'Emily', 'Frank', and 'George' = 25 x 3 = 75 years. Sum of the present ages of 'Frank', 'George', and 'Harry' = (18 x 3) + (4 x 3) = 66 years. Difference between the present ages of 'Emily' and 'Harry' = 75 - 66 = 9 years. Let the present ages of 'Emily' and 'Harry' be 'e' years and 'h' years, respectively. e - h = 9 -----------(i) Sum of present ages 'E' and 'H' = 58 years. e + h = 58 ---------(ii) Adding equations (i) and (ii), we get: 2e = 67 Or, 'e' = 67 / 2 = 33.5 So, the age of 'Emily' three years ago from now = e - 3 = 33.5 - 3 = 30.5 years.