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 her
daughter​

Answer 1

Answer:

let x be the present age of mother

let y be the present age of daughter

Answer 2

Answer:

The mother's age is 44 years and her daughter's age is 14 years.

Given:

• 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.

To find:

• The present age of mother and her daughter.

Solution:

Let the present ages of mother and her daughter be x and y years respectively.

According to the first condition.

=> x - 4 = 4 (y - 4)

=> x - 4 = 4y - 16

=> x - 4y = - 12...(1)

According to the second condition.

=> x + 6 = 2½ (y + 6)

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

=> x + 6 = 2.5 (y + 6)

=> x + 6 = 2.5y + 15

=> x - 2.5y = 9...(2)

Subtract equation (1) from equation (2), we get

⠀⠀x - 2.5y = 9

⠀-

⠀⠀x -⠀4y = - 12

_______________

⠀⠀=> 1.5y = 21

Multiply both sides by 2, we get

⠀⠀=> 3y = 42

⠀⠀=> y = 42/3

⠀⠀=> y = 14

Substitute y = 14 in equation (2), we get

⠀⠀=> x - 2.5 (14) = 9

⠀⠀=> x = 9 + 35

⠀⠀=> x = 44

Therefore, the mother's age is 44 years and her daughter's age is 14 years.