5 years ago a man was 7 times as old as his son.

Solution:

Let the present age of the son be x years and that of the father be $f$ years.

5 years back, the father was 7 times as old as his son.

$\therefore(f-5)=7(x-5)$

$f=7 x-35+5$

$f=7 x-30$    ....(i) 

After 5 years, ages of the father and son will be $(f+5)$ and $(x+5)$, respectively.

After 5 years, the father will be three times older than his son.

$\therefore(f+5)=3(x+5)$

$7 x-30+5=3 x+15$          [inserting the value of $\mathrm{f}$ from equation (1)]

$7 x-25=3 x+15$

$7 x-3 x=25+15$

$4 x=40$

$x=\frac{40}{4}=10$

Therefore, the present age of the son is 10 years.

Father's present age $=(7 x-30)=7(10)-30$

$=70-30=40$ years