Question

The ratio of present ages of $A$ and $B$ is $4:5.$ $18$ years ago, this ratio was $11:16.$ Find the sum of their present ages.

• A. $90$ years
• B. $105$ years
• C. $110$ years
• D. $80$ years

Answer 1

The correct option is A $90$ years

Let present ages of $A$ and $B$ be $4x$ and $5x$ respectively.
$18$ years ago,
$\frac{4x-18}{5x-18}=\frac{11}{16}$
$\Rightarrow 64x-288=55x-198$
$\Rightarrow 64x-55x=288-198$
$\Rightarrow 9x=90$
$\Rightarrow x=10$
$\therefore$ Sum of their present ages $=4\times 10+5\times 10=90$ years