Question

Three years ago, a father was eight times as old as his son. In three years time, if the father's age

exceeds twice his son's age by six years, then, the ratio of their present ages is /​

Answer 1

Answer:

Age of son is 9 years at present and father is 33 years old on present day.

Detailed solution is given below for better understanding.

Let the present age of son = s

and the present age of man = m

One year before:-

Age of son was (s-1) and age of man was (m-1)

As per first case: 4*(s-1)= (m-1) ……………………..……………………….….…………(i)

After 6 years:-

Age of son will be (s+6) and age of man will be (m+6)

Man will be 9 years elder than twice the son’s age

As per second case given: 2*(s+6)+9 = (m+6) ….………………………..…………(ii)

From eq (ii).

2s+12+9 = m+6

2s+21 = m+6

2s = m+6–21

2s = m- 15

4s = 2m-30 ………………………………………………………………………………….…………(iii)