1963 AHSME Problems/Problem 11

Problem

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

$\textbf{(A)}\ 38.5 \qquad \textbf{(B)}\ 37.5 \qquad \textbf{(C)}\ 37 \qquad \textbf{(D)}\ 36.5 \qquad \textbf{(E)}\ 36$

Solution

If the arithmetic mean of a set of $50$ numbers is $38$ , then the sum of the $50$ numbers equals $1900$ . Since $45$ and $55$ are being removed, subtract $100$ to get the sum of the remaining $48$ numbers, which is $1800$ . Therefore, the new mean is $37.5$ , which is answer choice $\boxed{\textbf{(B)}}$ .

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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1963 AHSME Problems/Problem 11

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