A girls 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 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
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