Question

Six years later the man's age will be three times the age of his son and three years ago he was nine times as old as his son. Find their present ages.

• A. The present age of man is $55$ years and present age of son is $11$ years.
• B. The present age of man is $40$ years and present age of son is $8$ years.
• C. The present age of man is $50$ years and present age of son is $10$ years.
• D. The present age of man is $30$ years and present age of son is $6$ years.

Answer 1

The correct option is D The present age of man is $30$ years and present age of son is $6$ years.

Let man's present age be $x$ years and son's present age be $y$.
After six years age of man will be $(x+6)$ and age of boy will be $(y+6)$.
According to problem, we get
$x+6$ $=$ $3(y+6)$
$x-3y$ $=$ $12$ .........(1)
Three years ago, age of father will be $y-3$ and age of son will be $x-3$
$x-3$ $=$ $9(y-3)$
$x-9y$ $=$ $-24$ ......(2)
Hence we get two equations as
$x-3y$ $=$ $12$ .........(1)
$x-9y$ $=$ $-24$ ......(2)
Put $x$ $=$ $3y+12$ in equation (2)
$\Rightarrow (3y+12)-9y$ $=$ $-24$
$\Rightarrow -6y$ $=$ $-36$
$\Rightarrow y$ $=$ $6$
Putting value of $y$ in equation (1), we get

$x$ $=$ $6$
On solving equation (1) and (2), we get

$x$ $=$ $30$ and $y$ $=$ $6$
So, the present age of man $=$ $30$ years and present age of son $=$ $6$ years.