Question

$A$ and $B$ are friends and $A$ is older than $B$ by $2$ years. $A$'s father $D$ is twice as old as $A$ and $B$ is twice as old as his sister $C$. The ages of $D$ and $C$ differ by $40$ years. Find the ages of $A$ and $B$.

• A. Ages of $A$ and $B$ are $36$ and $20$ years respectively.
• B. Ages of $A$ and $B$ are $16$ and $14$ years respectively.
• C. Ages of $A$ and $B$ are $26$ and $24$ years respectively.
• D. Ages of $A$ and $B$ are $42$ and $24$ years respectively.

Answer 1

Answer: (C)

Ages of $A$ and $B$ are $26$ and $24$ years respectively.

It is given that $A$ and $B$ are friends and $A$ is older than $B$ by $2$ years:

$A-B=\mathrm{2............}\left(1\right)$

Now as $D$ is twice of $A$, therefore, $D^{\prime }s$ age is $2A$ years

and

$B$ is twice of his sister $C$, therefore,$C^{\prime }s$ age is $\frac{B}{2}$ years

So $D$ is older than $C$.

$2A-\left(\frac{B}{2}\right)=40\Rightarrow 4A-B=\mathrm{80.........}\left(2\right)$

Subtract equation 1 from equation 2:

$3A=78$

i.e. $A=26$

Substitute the value of $A$ in equation 1:

$26-B=2$

i.e. $B=24$

Hence, the ages of $A$ and $B$ are $26$ and $24$ respectively.