Question

Four years ago, a mother 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 age of 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.
According to the question,
$x-4=4(y-4)\Rightarrow x-4=4y-16\Rightarrow x-4y=-\mathrm{12.....}\left(i\right)$
and $x+6=2\frac{1}{2}(y+6)\Rightarrow x+6=\frac{5}{2}y+15\Rightarrow x-\frac{5}{2}y=\mathrm{9.....}\left(ii\right)$
solving (i) and (ii), we get
$y=14$ and $x=44$
Hence, the present age of the mother is $44$ years and the present age of the daughter is $14$ years.