Question

Eighteen years ago, a father was three times as old as his son. Now the father is only twice as old as his son. Then the sum of the present ages of the son and the father is

• A. $54$
• B. $72$
• C. $105$
• D. $108$

Answer 1

The correct option is D $108$

Let the present age of father be $x$ and that of his Son Be $y$

Then $18$ years ago

Father's age $=x-18$

Son's age $=y-18$

But given that

$x-18=3(y-18)$

$\Rightarrow x-18=3y-54$

$\Rightarrow x-3y+36=0$ (1)

Also given that $x=2y$ (2)

From (1) & (2)

$x-3y=36=0$

$\Rightarrow 2y-3y+36=0$

$\Rightarrow y=36$

Putting $y=36$ in (2)

$x=2y$

$=2\times 36$

$=72$

Sum of their present ages

$=x+y$

$72+36$

$=108$