Question

In a class of $5$ students, average weight of the $4$ lightest students is $40$ kg, average weight of the $4$ heaviest students is $45$ kg. Then the difference between the maximum and minimum possible average weight is

• A. $2$
• B. $3$
• C. $1.5$
• D. $2.5$

Answer 1

The correct option is B $3$

Let the $5$ students in ascending order of their weight be $a,b,c,d,e$
Given : $\frac{a+b+c+d}{4}=40$
$\Rightarrow a+b+c+d=160\cdots \left(1\right)$

Now,
$\frac{b+c+d+e}{4}=45$
$\Rightarrow b+c+d+e=180\cdots \left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$, we get
$e=20+a\cdots \left(3\right)$

For minimum possible average we need minimum value of $a$ and $e$
Minimum value of $e$ is $45$ and $b=c=d=45$
Minimum value of $a$ is $45-20=25$
Now least average
$=\frac{25+45+45+45+45}{5}=41$

For maximum possible average we need maximum possible value of $a$ and $e$
Maximum value of $a$ is $40$ and $b=c=d=40$
Maximum value of $e$ is $40+20=60$
Now maximum average
$=\frac{40+40+40+40+60}{5}=44$

Now, the required difference
$=44-41=3$