The present age of a man is twice that of his son. Eight years hence, their ages will be in the ratio 7 : 4. Find their present ages.

Let the age of son be x years.

The present age of a man is twice that of his son.

Man's age = 2x

After eight years, their ages will be in the ratio 7 : 4.

$⇒ \dfrac{\text{Man's age after 8 years}}{\text{Son's age after 8 years}} = \dfrac{7}{4}\\[1em] ⇒ \dfrac{\text{Man's present age + 8 years}}{\text{Son's present age + 8 years}} = \dfrac{7}{4}\\[1em] ⇒ \dfrac{2x + 8}{x + 8} = \dfrac{7}{4}\\[1em]$

By cross multiplying, we get

$⇒ 4(2x + 8) = 7(x + 8)\\[1em] ⇒ 8x + 32 = 7x + 56\\[1em] ⇒ 8x - 7x = 56 - 32\\[1em] ⇒ x = 24\\[1em]$

Present son's age = 24 years

Present man's age = 2x years

= 2 $\times$ 24 years

= 48 years

Hence, son's age is 24 years and man's age is 48 years.