BPSC 66th Prelims Prelims 27 Dec 2020 Solved Paper

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Question : 20 of 150

Marks: +1, -0

If the average of

m

numbers is

n^{2}

and that of

n

numbers is

m^{2}

, then the average of mun numbers is

$n \/ m$
$m \/ n$
${mn}$
$m-n$
None of the above/More than one of the above

Solution:

Given that average of

m

numbers

=n^{2}

{n}^{2}=

sum of

{m}

numbers

\/ {m}

sum of

{m}

numbers

= {mn}^{2}

Given that average of

n

numbers

= {m}^{2}

{m}^{2}=

sum of

n

numbers

\/ n

sum of

n

numbers

= {nm}^{2}

Average of

( {m}+ {n})

numbers is

=( {mn}^{2}+ {nm}^{2}) \/( {m}+ {n})

=\{mn(m+n)\} \/(m+n)

= {mn}

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