Question:
The difference between the ages of two cousins is 10 years. 15 years ago, if the elder one was twice as old as the younger one, find their present ages.
Solution:
Let the age of the younger cousin be $\mathrm{x}$.
Then, the age of the elder cousin will be $(\mathrm{x}+10)$.
15 years ago :
Age of the younger cousin $=(\mathrm{x}-15)$
Age of elder $\operatorname{cousin}=(\mathrm{x}+10-15)$
$=(x-5)$
$\therefore(x-5)=2(x-15)$
$\Rightarrow x-5=2 x-30$
$\Rightarrow x-2 x=-30+5$
$\Rightarrow-x=-25$
$\Rightarrow x=25$
Therefore, the present age of the younger cousin is 25 years.
Present age of elder cousin $=(\mathrm{x}+10)=(25+10)=35$ years