A.M. of a 50 set of numbers is 38 . If two numbers of set, namely 55 and 45 are discarded, A.M. of r

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8. Sequences and Series

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The $A.M.$ of a $50$ set of numbers is $38$. If two numbers of the set, namely $55$ and $45$ are discarded, the $A.M.$ of the remaining set of numbers is

A

$38.5$

B

$37.5$

C

$36.5$

D

$36$

Solution

(b) Given, $\frac{{\Sigma {x_i}}}{{50}} = 38,\,\,\,\therefore \Sigma {x_i} = 1900$

New value of $\Sigma {x_i} = 1900 – 55 – 45$$ = 1800$, $n = 48$

New mean $ = \frac{{1800}}{{48}}$$ = 37.5$.

Standard 11

Mathematics

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