Question

Four years ago, a mother was four times as old as her daughter. Six years later, the mother will be two and a half times as old as her daughter at that time. Find the present ages of the mother and her daughter.

Answer 1

Let the present age of the mother be x years.

and the present age of the daughter be y year.

x - 4 = 4(y - 4)

⇒ x - 4 = 4y - 16

⇒ x - 4y = -12                ...(1)

And,

x + 6 = 2`1/2`(y + 6)

x + 6 = `5/2`(y + 6)

2(x + 6) = 5(y + 6)

2x + 12 = 5y + 30

2x - 5y = 30 - 12

2x - 5y = 18             ...(2)

Multiply 2 in equation (1)

2x - 8y = -24         ...(3)

subtract equation (2) from (3)

2x - 8y = -24
2x - 5y = 18

-3y = -42

y = `42/3`

y = 14 year

From equation (1)

x - 4(14) = -12

x - 56 = -12

x = 56 - 12

x = 44 year