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 mother and herdaughter.​

Answer 1

let mother's age be y years and his daughter's age be x years

ATP

four years ago x-4=4(y-4)

divide by 4 x-1=y-4

x=y-4+1

x=y-3 ---------------------------(1)

six years later x+6=2.5 (y+6)

$x + 6 = 2.5y + 15$

$x = 2.5y + 15 - 6$

$x = 2.5y + 9 - - - - - - (2)$

by 1 and 2

$y - 3 = 2.5y + 9$

$y - 2.5y = 9 + 3$

$1.5y = 12$

$y = \frac{12}{1.5} = \frac{12 \times 10}{15}$

$y = 8$

put y in equation 1

x=y-3=8-3=5

•°•mother's age is 8 years and daughter's age is 5 years