The arithmetic mean (average) of a set of $50$ numbers is $38$ . If two numbers, namely, $45$ and $55$ , are discarded, the mean of the remaining set of numbers is :

36.5

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37

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37.2

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37.5

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37.52

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Solution

The correct option is D $37.5$
$A r i t h m e t i c s u m = \frac{s u m o f n u m b e r s}{n u m b e r o f n u m b e r s}$
$∴ S = 50 \times 38 = 1900; ∴ x = \frac{1900 - 45 - 55}{48} = 37.5$

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The arithmetic mean (average) of a set of 50 numbers is 38. If two numbers, namely, 45 and 55, are discarded, the mean of the remaining set of numbers is :

The correct option is D 37.5Arithmeticsum=sumofnumbersnumberofnumbers∴S=50×38=1900;∴x=1900−45−5548=37.5

The arithmetic mean (average) of a set of $50$ numbers is $38$ . If two numbers, namely, $45$ and $55$ , are discarded, the mean of the remaining set of numbers is :

The correct option is D $37.5$
$A r i t h m e t i c s u m = \frac{s u m o f n u m b e r s}{n u m b e r o f n u m b e r s}$
$∴ S = 50 \times 38 = 1900; ∴ x = \frac{1900 - 45 - 55}{48} = 37.5$