Question

The average of $5$ distinct positive integers is $33$. If the average of the three largest numbers within this set is $39$ then the difference of the maximum and minimum possible values of the median of the $5$ numbers is

• A. $12$
• B. $14$
• C. $8$
• D. $9$

Answer 1

The correct option is A $12$

Let the $5$ numbers be $a,b,c,d,e$ in ascending order
and
$\frac{a+b+c+d+e}{5}=33$
$\Rightarrow a+b+c+d+e=165\dots \left(1\right)$

and $\frac{c+d+e}{3}=39$
$\Rightarrow c+d+e=117\dots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$
$\Rightarrow a+b=48\dots \left(3\right)$

median of $a,b,c,d,e$ is $c$ So,
we need to find maximum and minimum possible value of $c$

For minimum value of $c$. $a$ and $b$ have to be minimum.
$a+b=48$
as $a$ and $b$ are distinct So,
$a=23,b=25$
then $c=26$

For maximum value of $c$. $d$ and $e$ have to be maximum.
As $c+d+e=117$
as $c,d$ and $e$ are distinct So,
$e=40,d=39$
then $c=38$.

Now difference $38-26=12$