Question

A boy is twice as old as her sister. Four year hence, the product of their ages (in years) will be 160. Find their present ages.

Answer 1

Let the age of sister be x years.

It is given in question that the boy is twice as old as her sister.

⇒ Boy's age = 2x

Four years hence, the product of their ages = 160

⇒ (Sister's age + 4) x (Boy's age + 4) = 160

⇒ (x + 4) × (2x + 4) = 160

⇒ x × (2x + 4) + 4 × (2x + 4) = 160

⇒ 2x² + 4x + 8x + 16 = 160

⇒ 2x² + 12x + 16 − 160 = 0

⇒ 2x² + 12x − 144 = 0

⇒ x² + 6x − 72 = 0

⇒ x² + 12x − 6x − 72 = 0

⇒ x(x + 12) − 6(x + 12) = 0

⇒ (x + 12)(x − 6) = 0

⇒ (x + 12) = 0 or (x − 6) = 0

⇒ x = −12 or x = 6

Since age cannot be negative,

∴ Present age of sister = 6 years

Boy's age = 2 x 6 = 12 years

Thus, the present age of boy = 12 years and sister = 6 years.