Question

Five years ago, a woman’s age was the square of her son’s age. Ten years hence, her age will be twice that of her son’s age. Find:

1. the age of the son five years ago.
2. the present age of the woman.

Answer 1

Let the present age of the son be x years
and the present age of the woman be y years.

Five years ago, woman’s age was the square of her son’s age:

y − 5 = (x − 5)²

y = (x − 5)² + 5     ...(1)

Ten years hence, woman’s age will be twice the son’s age:

y + 10 = 2(x + 10)

y = 2x + 20 − 10

y = 2x + 10     ...(2)

(x − 5)² + 5 = 2x + 10

x² − 10x + 30 = 2x + 10

x² − 12x + 20 = 0

(x − 10) (x − 2) = 0

x = 10 or x = 2

If x = 2, son’s age 5 years ago would be negative → reject.

x = 10

From (2):

y = 2x + 10

= 2(10) + 10

= 30

(a) Son’s age five years ago:

10 − 5

= 5 years​

(b) Woman’s present age:

30 Years