Question

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

Answer 1

Let the age of girls sister be of $x$ years

Given that :

Girl is twice as old as her sister.

$\Rightarrow$ Girl's age$2\times x$years$=2x$years.

Given that, after $4$ years, the product of their ages will be $160$.

$\Rightarrow$ Girls age after $4$ years$=2x+4$years

and sisters age after $4$ years$=x+4$years

Given: $(2x+4)(x+4)=160$

$\Rightarrow 2x^2+8x+4x+16-160=0$

$\Rightarrow 2x^2+12x-144=0$

$\Rightarrow 2(x^2+6x-72)=0$

$\Rightarrow x^2+6x-72=0$

$\Rightarrow x^2+12x-6x-72=0$

$\Rightarrow x(x+12)-6(x+12)=0$

$\Rightarrow (x+12)(x-6)=0$

$\Rightarrow x=-12$ or $x=6$

Since, age cannot be negative

So, $x=6$

$\therefore$ Age of girls sister is $x=6$ years

And age of girl is $2x=2\times 6=12$years.

Hence, the present ages of girl and her sister are $12$ years and $6$ years respectively.​