1959 AHSME Problems/Problem 13

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: $\textbf{(A)}\ 36.5 \qquad\textbf{(B)}\ 37\qquad\textbf{(C)}\ 37.2\qquad\textbf{(D)}\ 37.5\qquad\textbf{(E)}\ 37.52$

Solution

Since the arithmetic mean of the $50$ numbers is $38$ , their sum must be $50*38 = 1900$ . After $45$ and $55$ are discarded, the sum decreases by $45 + 55 = 100$ , so it must become $1900 - 100 = 1800$ . But this means that the new mean of the remaining $50 - 2 = 48$ numbers must be $\frac{1800}{48} = 37.5$ . Thusly, our answer is $\boxed{\textbf{(D)}}$ , and we are done.

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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1959 AHSME Problems/Problem 13

Solution

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