If the average of m numbers is n^{2} and that | vedclass

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(B) The sum of $m$ numbers is given by $m 	imes n^{2} = mn^{2}$.The sum of $n$ numbers is given by $n 	imes m^{2} = nm^{2}$.The total sum of $(m + n

Vedclass · Accepted Answer

$mn$

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(B) The sum of $m$ numbers is given by $m 	imes n^{2} = mn^{2}$.The sum of $n$ numbers is given by $n 	imes m^{2} = nm^{2}$.The total sum of $(m + n)$ numbers is $mn^{2} + nm^{2}$.The average of $(m + n)$ numbers is calculated as:$	ext{Average} = \frac{	ext{Total Sum}}{	ext{Total Count}} = \frac{mn^{2} + nm^{2}}{m + n}$.Factoring out $mn$ from the numerator:$	ext{Average} = \frac{mn(n + m)}{m + n}$.Since $(n + m) = (m + n)$,they cancel out:$	ext{Average} = mn$.

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

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