Five years ago a man was seven times as old as his son.

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.