Question

$18$ years ago, a man was three times as old as his son. Now the man is twice as old as his son. The sum of the present ages of the man and his son is

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

Answer 1

The correct option is D $108$ years

Let the son's age $18$ years ago be $x$ years.
Then, man's age $18$ years ago $=3x$ years.
$(3x+18)=2(x+18)$
$\Rightarrow 3x+18=2x+36$
$\Rightarrow x=18$
Sum of their present ages
$=(3x+18+x+18)$ years
$=(4x+36)$ years
$=(4\times 18+36)$ years
$=108$ years