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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.


Solution

Let the present age of the mother be $$ x $$ years
and the present age of the daughter be $$ y $$ year.
According to the question,
$$ x-4=4(y-4)\Rightarrow x-4=4y-16 \Rightarrow x-4y=-12.....(i) $$
and $$ x+6=2\frac{1}{2}(y+6)\Rightarrow x+6 =\frac{5}{2}y+15 \Rightarrow x-\frac{5}{2}y=9.....(ii) $$
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.

Mathematics

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