Question

Paul is x years old and his father’s age is twice the square of Paul’s age. Ten years hence, the father’s age will be four times Paul’s age. Find their present ages.

Answer 1

Age of Paul = x years
Father's age = 2x²
10 years hence,
Age of Paul = x + 10
and father's age = 2x² + 10
According to the conditions,
2x² + 10 = 4(x + 10)
⇒ 2x² + 10 = 4x + 40
⇒ 2x² + 10 - 4x - 40 = 0
⇒ 2x² - 4x - 30 = 0
⇒ x² - 2x - 15 = 0  ...(Dividing by 2)
⇒ x² - 5x + 3x - 15 = 0
⇒ x(x - 5) + 3(x - 5) = 0
⇒ (x - 5)(x + 3) = 0
Either x - 5 = 0,
then x = 5
or
x + 3 = 0,
then x = -3,
but it is  not possible as it is in negative.
∴ Age of Paul = 5 years.
and his father's age
= 2x²
= 2(5)²
= 2 x 25
= 50 years.