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.

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Updated on: 10-Oct-2022

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