Question

The age of a father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.

Answer 1

Persons	Present age	8 years hence	Future age
Father	$2x^2$	+ 8 years	$2x^2+8$
Son	x	+8 years	x+8

Now follow below $\downarrow$

Let the present age of the son be x years.
Present age of father = 2x years [according to the given condition ]
Eight years hence,
Son’s age = (x + 8) years
Father’s age = (2x+ 8) years
It is given that eight years hence, the age of the father will be 4 years more than three times the age of the son.

2x(x – 4) + 5(x – 4) = 0
(x – 4) (2x + 5) = 0
x = 4, -5/2
But, the age cannot be negative, so, x = 4.
Present age of son = 4 years
Present age of father = 2(4) years = 32 years