Three years ago, a father was three times as old as his son. In five years time, the father will be twice as old as his son. What is the sum of their ages in four years time?

Answer 1

Let's denote the present age of the father as F and the present age of the son as S.

According to the first statement, three years ago, the father was three times as old as his son:
F - 3 = 3(S - 3)
F - 3 = 3S - 9
F = 3S - 6

According to the second statement, in five years time, the father will be twice as old as his son:
F + 5 = 2(S + 5)
F + 5 = 2S + 10
F = 2S + 5

Now, we can equate both equations to find the ages of the father and son. So:
3S - 6 = 2S + 5
3S - 2S = 5 + 6
S = 11

Substituting the value of S back into the second equation:
F = 2(11) + 5
F = 22 + 5
F = 27

In four years time, the son will be 11 + 4 = 15 years old, and the father will be 27 + 4 = 31 years old. The sum of their ages in four years time will be 15 + 31 = 46.

Therefore, the sum of their ages in four years time is 46.