#### Question

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.

#### Solution

Let the present age of girl be x years then, age of her sister x/2 years

Then, 4 years later, age of girl (x + 4) years and her sister’s age be `(x/2+4)Years`

Then according to question,

`(x+4)(x/2+4)=160`

(x + 4)(x + 8) = 160 x 2

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

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

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

x^{2} - 12x + 24x - 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

`x/2=12/2=6`

Hence, the present age of girl be 12 years and her sister’s age be 6 years.

Is there an error in this question or solution?

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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. Concept: Solutions of Quadratic Equations by Factorization.

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