Question

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

Answer 1

Answer:

Let the present age of her sister be x years.

Then, girl's present age = 2x years

Product of their ages 4 years hence = (x + 4)(2x + 4)

$\sf \therefore \: (x + 4)(2x + 4)=160$

$\sf \longrightarrow \: 2 {x}^{2} + 12x - 144 = 0$

$\sf \longrightarrow 2({x}^{2} + 6x - 72) = 0$

$\sf \longrightarrow {x}^{2} + 6x - 72 = 0$

$\sf \longrightarrow {x}^{2} + 12x - 6x - 72 = 0$

$\sf \longrightarrow (x + 12)(x - 6) = 0$

$\sf \longrightarrow x + 12 = 0 \: \: or \: \: x - 6 = 0$

$\sf \longrightarrow x = - 12 \: \: or \: \: x = 6$

$\sf \longrightarrow x = 6$

ㅤㅤㅤㅤㅤㅤㅤ[The age cannot be negative]

Hence, sister's present age is 6 years and girl's present age is 12 years.