Question

Five years ago a woman's age was the square of her son's age. Ten years hence her age will be twice of her son's age. find i) the age of the son five years ago ii) the present age of the women.

Answer 1

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 yrs