Question:
Five years ago a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.
Solution:
Five years ago:
Let the age of the son be $x$ years.
Therefore, the age of the father will be $7 \mathrm{x}$ years.
$\therefore$ Present age of the son $=(\mathrm{x}+5)$ years
Present age of the father $=(7 \mathrm{x}+5)$ years
After five years:
Age of the son $=(\mathrm{x}+5+5)=(\mathrm{x}+10)$ years
Age of the father $=(7 \mathrm{x}+5+5)=(7 \mathrm{x}+10)$ years
According to the question,
$7 \mathrm{x}+10=3(\mathrm{x}+10)$
or $7 \mathrm{x}-3 \mathrm{x}=30-10$
or $4 \mathrm{x}=20$
or $\mathrm{x}=5$
$\therefore$ Present age of the son $=(5+5)=10$ years.
Present age of the father $=(7 \times 5+5)=40$ years.