Question

$A$ is elder to $B$ by $2$ years. $A$'s father $F$ is twice as old as $A$ and $B$ is twice as old as his sister $S$. If the ages of the father and sister differ by $40$ years, find the age of $A$.

Answer 1

Let $a$, $b$, $f$ and $s$ be the ages of $A$, $B$, $F$ and $S$ respectively.

$A$ is elder to $B$ by 2 years. So, $a=b+2$. $⟹a-b=2$

$F$ is twice as old as $A$. So, $f=2a$.

$B$ is twice as old as $S$. So, $b=2s$.

Finally, the ages of $F$ and $S$ differ by 40 years. So, $f-s=40$.

Substituting $f=2a$ and $s=\frac{b}{2}$ into $f-s=40$, we get $2a-\frac{b}{2}=40$.

$⟹4a-b=80$.

Subtracting the equation $a-b=2$ from $4a-b=80$, we get

$3a=78$.

$⟹a=26$

So, the age of $A$ is 26 years.