The A.M. of a set of $50$ 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 :

36

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36.5

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37.5

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38.5

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Solution

The correct option is C $37.5$
Mean of remaining set $= \frac{50 \times 38 - (45 + 55)}{50 - 2} = \frac{1800}{48} = 37.5$

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The A.M. of a set of 50 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 :

The correct option is C 37.5Mean of remaining set =50×38−(45+55)50−2=180048=37.5

The A.M. of a set of $50$ 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 :

The correct option is C $37.5$
Mean of remaining set $= \frac{50 \times 38 - (45 + 55)}{50 - 2} = \frac{1800}{48} = 37.5$