Question

Six years before, the age of mother was
numerically equal to the square of son's age.
Three years hence, her age will be thrice the
age of her son then. Find the present ages of
the mother and son. ​

Answer 1

Age of mother x = 42 years

Age of son  y = 12 years

Step-by-step explanation:

Let the present age of mother = X

present age of the sun = y

Now from the given data six years before, the age of mother was

numerically equal to the square of son's age.

$x - 6 = ( y - 6 )^{2}$ -------- (1)

⇒ x - 6 = $y^{2}$ - 12 y + 36 ----- (2)

Three years hence, mother's age will be thrice the  age of her son.

⇒ x + 3 = 3 ( y + 3)

⇒ x + 3 = 3 y + 9

⇒ x = 3 y + 6 ------- (3)

Put the value of x from equation (3) to equation (2) we get

⇒ 3 y + 6 = $y^{2}$ - 12 y + 36

⇒ $y^{2}$ - 15 y + 36 = 0

By solving above equation we get y = 12, 3

Put y = 12 in equation (3), we get

⇒ x = 3 × 12 + 6

⇒ x = 42 years

And y = 12 years

Thus age of mother x = 42 years

Age of son  y = 12 years