Question

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 will be the sum of their ages in four years time?

Answer 1

Answer:

Let the present age of son be X

3 years ago:

Son's age = X - 3

so, father's age = 3 (x - 3) = 3x - 9

5 years hence:

Son's age = x + 5

So, father's age = 2 (x + 5) = 2x + 10

Since, the diff. in their ages will remain constant,

hence,

(3x - 9) - X = (2X + 10) - (X + 5)

3x - 9 - x = 2X + 10 - x - 5

2x - 9 = x + 5

2x - x = 5 + 9

x = 14 years

therefore, present age of son is 14 years.

so, present age of father = age of father 3 years back + 3 years.

= 3x - 9 + 3

= 3x - 6

= 3 × 14 - 6

= 42 - 6

= 36 years old.

Now,

4 years hence,

son's age = 14 + 4 = 18

and, father's age = 36 + 4 = 40

Sum of their ages 4 years from now = 40 + 18 = 58

I hope you understood:)