A girls is twice as old as her sister. Four years hence,

Solution:

Let the present age of girl be $x$ years then, age of her sister $\left(\frac{x}{2}\right)$ years

Then, 4 years later, age of girl $=(x+4)$ years and her sister's age be $\left(\frac{x}{2}+4\right)$ years

Then according to question,

$(x+4)\left(\frac{x}{2}+4\right)=160$

$(x+4)(x+8)=160 \times 2$

$x^{2}+8 x+4 x+32=320$

$x^{2}+12 x+32-320=0$

$x^{2}+12 x-288=0$

$x^{2}+12 x-288=0$

$x^{2}-12 x+24 x-288=0$

$x(x-12)+24(x-12)=0$

$(x-12)(x+24)=0$

So, either 

$(x-12)=0$

$x=12$

Or

$(x+24)=0$

$x=-24$

But the age never be negative

Therefore, when $x=12$ then

$\frac{x}{2}=\frac{12}{2}$

$=6$

Hence, the present age of girl be $=12$ years and her sister's age be 6 years