Question

Three years ago, the father’s age was the square of his son’s age. 21 years hence, the father’s age will be twice the age of his son’s age. Find their present ages.

Answer 1

Let the present age of the son be x years

and the present age of the father be y years.

Three years ago, father’s age was the square of the son’s age:

y − 3 = (x − 3)²

y = (x − 3)² + 3     ...(1)

21 years hence, father’s age will be twice the son’s age:

y + 21 = 2(x + 21)

y = 2x + 42 − 21

y = 2x + 21     ...(2)

Equate (1) and (2)

(x − 3)² + 3 = 2x + 21

x² − 6x + 9 + 3 = 2x + 21

x² − 6x + 12 = 2x + 21

x² − 8x − 9 = 0

(x − 9) (x + 1) = 0

x = 9 or x = −1

From (2):

y = 2x + 21

= 2(9) + 21

= 39

Son’s present age = 9​ years

Father’s present age = 39​ years