Question

five years ago a women's age was the sqaure of her son's age.ten years hence her age will be the twice of her son's age.find the present age of the son

Answer 1

Answer:

Let Son's age = x

Let Women's age = y

y - 5 = (x - 5)² -------------1

y + 10 = 2(x+10)----------2

Equation 1

y - 5 =x² - 10 x + 25

y = x² - 10x + 30

Equation 2

y = 2x +10

Substituting The Value of y in equation 1

2x + 10 = x² - 10x + 30

x² - 10x - 2x + 30 - 10 = 0

x² - 12x + 20 = 0

x² - 2x - 10x + 20 = 0

x(x - 2) -10(x - 2) = 0

(x - 10)(x - 2) = 0

x = 10 or x = 2[ x cant be 2 as five years ago it would be negative age]

Therefore, sons present age = 10

So sons age five years ago = 5 yrs

Substituting The Value of x in equation 2

y = 2x + 10

y = 2 * 10 + 10

y = 20 + 10

y = 30 yrs

Therefore, Women Present Age = 30 years

Hope it helps

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Answer 2

GiveN-

Five years ago a women's age was the sqaure of her son's age.Ten years hence her age will be the twice of her son's age.

To FinD:-

The present age of the son.

SolutioN:-

• Let the age of the son 5 years ago be x.
• Let the age of the mother 5 years ago be x².

After 10 years,

• Son's Age = (x + 15) years
• Mother's Age = (x² + 15) years

Ten years hence mother's age will be the twice of her son's age.

According to the question,

$\large\implies{\sf{x^2+15=2(x+15)}}$

$\large\implies{\sf{x^2+15=2x+30}}$

$\large\implies{\sf{x^2-2x+15-30=0}}$

$\large\implies{\sf{x^2-2x-15=0}}$

Splitting the middle term,

$\large\implies{\sf{x^2-5x+3x-15=0}}$

$\large\implies{\sf{x(x-5)+3(x-5)=0}}$

$\large\implies{\sf{(x-5)(x+3)=0}}$

• $\large{\sf{(x-5)=0}}$
• $\large\boxed{\bf{x=5.}}$

• $\large{\sf{(x+3)=0}}$
• $\large\boxed{\bf{x=-3.}}$

As age cannot be in negative so x is 5.

Now,

Present age of the son = 5 + 5 = 10.

So, present age of the son is 10 years.