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 age.

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Answer 1

$\large \underline{ \underline{ \sf \: Solution : \: \: \: }}$

Let ,

The age of girl's sister = y

So , the girl's age = 2y

By the given condition ,

$\large \fbox{\fbox{\sf Girl's \: age × Girl's \: sister \: age = 160 }}$

$\sf \implies</p><p>2y × y = 160 \\ \\ \sf \implies</p><p>2 {y}^{2} = 160 \\ \\ \sf \implies</p><p> {y}^{2} = 80 \\ \\ \sf \implies </p><p>y = \sqrt{80} </p><p>$

$\large \underline{ \underline{ \sf \: Verification : \: \: \: }}$

Put the value of y = √80 in 2(y)² , we get

$\sf \implies 2 {( \sqrt{80} )}^{2} \\ \\ \sf \implies</p><p>2 × 80 \\ \\ \sf \implies </p><p>160$

Hence , the present age of girl's sister and girl are √80 and 2√80

Answer 2

$2y \times y = 160 \\ \\ 2y {}^{2} = 160 \\ \\ y {}^{2} = 80 \\ \\ y = \sqrt{80} \\ \\ put \: \sqrt{80} in2y {}^{2} \\ \\ 2( \sqrt{80) {}^{2} } \\ 2 \times 80 = 160 \\ \\ = 2 \sqrt{80} \\ \\ \\ so \: girls \: sister \: age \: is \: \sqrt{80} \: and \: girl \: age \: is \: \\ 2 \sqrt{80}$