Question:
5 years ago a man was 7 times as old as his son. After 5 years he will be thrice as old as his son. Find their present ages.
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