Question

Mary told her daughter, Seven years ago, I was seven times as old as you were then. Also three years from now, I shall be three times as old as you will be, Find the present age of Mary and her daughter.

Answer 1

Let the age of Mary be $x$ and the age of her daughter be $y$.

Seven years ago, Mary was $\left(x-7\right)$ years old and her daughter was $\left(y-7\right)$ years old.

Then, according to the question,

$x-7=7\left(y-7\right)$

$x-7=7y-49$

$x-7y=-42$ ....................(1)

Three years from now, Mary will be $\left(x+3\right)$ years old and her daughter will be $\left(y+3\right)$ years old.

According to question,

$x+3=3\left(y+3\right)$

$x+3=3y+9$

$x-3y=6$ ........................(2)

From equation (1) and (2),

Subtract equation $\left(1\right)$ from $\left(2\right)$,

$4y=48$ $\to y=12$

Put the value of $y$ in equation $\left(1\right)$

$x=7\times 12-42$ $\to y=42$

$x=42$ and $y=12$

Therefore, the present age of Mary is 42 and that of her daughter is 12.