Question

If the average of $m$ numbers is $n^2$ and that of $n$ number is
$m^2$, then the average of $(m+n)$ numbers is

• A. $m-n$
• B. $mn$
• C. $(m+n)$
• D. $\frac{m}{n}$

Answer 1

The correct option is B $mn$

Average is the sum of the data values divided by the total number of data values.

Given that average of $m$ numbers is $n^2$

$n^2=\frac{\text{sum of}m\text{numbers}}{m}$

Therefore, sum of $m$ numbers $=m{n}^2$

Given that average of $n$ numbers is $m^2$

$m^2=\frac{\text{sum of}n\text{numbers}}{n}$

Therefore, sum of $n$ numbers $=n{m}^2$

Therefore, average of $(m+n)$ numbers is $\frac{m{n}^2+n{m}^2}{m+n}=\frac{mn(m+n)}{m+n}=mn$