Question

Ten years hence, a man's age will be twice the age of his son. Ten years ago, the man was four times as old as his son. Find their present ages.

• A. Man = 50 years Son = 25 years
• B. Man = 60 years Son = 20 years
• C. Man = 50 years Son = 20 years
• D. Man = 50 years Son = 10 years

Answer 1

The correct option is C Man = 50 years Son = 20 years

Let a man's present age be 'x' years
Let the present age of his son be 'y' years

$\begin{array}{cccc}& \text{PAST}& \text{PRESENT}& \text{FUTURE}\\ \text{MAN}& x-10& x& x+10\\ \text{SON}& y-10& y& y+10\end{array}$
$x+10=2(y+10)$
$\Rightarrow x+10=2y+20\Rightarrow x+10-2y-20=0$
$\therefore x-2y-10=0$..........(1)
$x-10=4(y-10)$
$\Rightarrow x-10=4y-40\Rightarrow x-10-4y+40=0$
$\therefore x-4y+30=0$..........(2)
$x-2y-10=0$..........(1)
$x-4y+30=0$..........(2)
Subtracting (2) from (1), we get:
$x-2y-10=0$
$x-4y+30=0$
______________
$2y-40=0\Rightarrow 2y=40$
$\therefore y=20$
Substituting $y=20$ in (1), we get:
$x-2y-10=0\Rightarrow x-2\left(20\right)-10=0$
$\Rightarrow x-40-10=0\Rightarrow x-50=0$
$\therefore x=50$
Thus, man's present age = x yrs = 50 yrs
his son's present age = y yrs = 20 yrs