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.

Answer 1

SOLUTION :  

Let the age her sister be x years.

Girl's age be = 2x years

After four years :  

Her sister's age = (x + 4) years

Girl's age = (2x + 4) years

A.T.Q

Given 4 years later the product of their age will be 160

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

2x² + 4x + 8x + 16 = 160

2x² + 12x + 16 - 160 = 0

2x² + 12x - 144 = 0

2(x² + 6x - 72) = 0

x² + 6x - 72 = 0

x² + 12x - 6x - 72 = 0

[By middle term splitting]

x(x + 12) - 6(x + 12) = 0

(x - 6) (x + 12) = 0

(x - 6) or  (x + 12) = 0

x = 6  or x = - 12  

Since, age can't be negative,x ≠ - 12 , Therefore, x = 6

Her sister's age = x = 6 years  

Girl's age = 2x = 2 × 6 = 12 years  

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

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Answer 2

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Here given that,

A girls is twice as old as her sister. Four years hence, the product of their ages will be 160.


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 age be (x/2+4) years

So, according to the questions,

(x+4)(x/2+4)=160
(x + 4) (x + 8) = 160×2
x2 + 8x + 4x + 32 = 320
x2 + 12x – 288 = 0
x2 – 12x + 24x – 288 = 0
x(x – 12) + 24(x – 12) = 0
(x – 12) (x + 24) = 0
x = 12  or x = – 24
Since, the can never be negative,
Therefore, the present age of the girl is = 12 years.
And her sister’s age will be,
x/2=12/2 = 6 years.

Hope its help you