Question

$A$'s age is twice as $B$'s. $4$ years ago, $A$ was three times as old as $B$. Find their ages.

• A. $12$ years and $8$ years
• B. $16$ years and $8$ years
• C. $17$ years and $6$ years
• D. $19$ years and $6$ years

Answer 1

The correct option is B $16$ years and $8$ years

Let the present age of $B^{\prime }s$ be $x$ years.
$\therefore A^{\prime }s$ age will be $2x$ years.
$4$ years ago, we have
$B^{\prime }s$ age is $(x-4)$ years.
$A^{\prime }s$ age is $(2x-4)$ years.
According to question, we have
$(2x-4)=3(x-4)$
$\Rightarrow 2x-4=3x-12$
$\Rightarrow x=8$
$\therefore B^{\prime }s$ age is $8$ years and $A^{\prime }s$ age is $2\times 8=16$ years.