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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. - Mathematics

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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

x2 + 8x + 4x + 32 = 320

x2 + 12x + 32 - 320 = 0

x2 + 12x - 288 = 0

x2 - 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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APPEARS IN

 RD Sharma Solution for Class 10 Maths (2018 (Latest))
Chapter 4: Quadratic Equations
Ex. 4.9 | Q: 6 | Page no. 61
 RD Sharma Solution for Class 10 Maths (2018 (Latest))
Chapter 4: Quadratic Equations
Ex. 4.9 | Q: 6 | Page no. 61
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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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