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

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) $\times$ (2x + 4) = 160

⇒ x $\times$ (2x + 4) + 4 $\times$ (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.