The sum of present age of a father and his son is 80 years. When the father's age is equal to the present age of the son the sum of their ages was 40 years find their present age.

Solution:

Let x and y be the present ages of father and son respectively.

Given,

Sum of present ages of father and son is 80 years.

$\implies x + y = 80$ - (i)

When the father's age is equal to the present age of the son, their sum was 40 years

$\implies x - (x-y) + y - (x-y) = 40$

$\implies x - x + y + y - x + y = 40$

$\implies 3y - x = 40$ - (ii)

Adding equation (i) and equation (ii), we get,

$or, x + y +3y - x = 80+40$

$or, 4y = 120$

$\therefore y = 30$

Put value of y in equation (i), we get,

$x + y = 80$

$or, x + 30=80$

$\therefore x = 50$

Hence, the required present ages of the father and the son are 50 years and 30 years, respectively.

Brief Explanation:

When father was 50 and son was 30, the difference between their ages was 20 years which is given by (x-y).

When the father was 30 (equal to present age of the son), the son was (30-20) 10 years old. Which means we need to subtract (x-y) from x and y, respectively.