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A girl is twice as old as her sister. Four years hence, the product of their ages (in years) will be160. Find their present ages.
Given:
A girl is twice as old as her sister. Four years hence, the product of their ages (in years) will be160.
To do:
We have to find their present ages.
Solution:
Let the present age of the sister be $x$ years.
This implies, the present age of the girl $=2x$ years.
Age of the sister 4 years later$=x+4$ years
Age of the girl 4 years later $=2x+4$ years
According to the question,
$(x+4)(2x+4)=160$
$2x^2+4x+8x+16=160$
$2x^2+12x+16-160=0$
$2x^2+12x-144=0$
$2(x^2+6x-72)=0$
$x^2+6x-72=0$
Solving for $x$ by factorization method, we get,
$x^2+12x-6x-72=0$
$x(x+12)-6(x+12)=0$
$(x+12)(x-6)=0$
$x+12=0$ or $x-6=0$
$x=-12$ or $x=6$
Age cannot be negative. Therefore, the value of $x$ is $6$.
$2x=2(6)=12$
The present age of the girl is $12$ years and the present age of the sister is $6$ years.