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Difference between revisions of "1959 AHSME Problems/Problem 13"

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== Problem ==
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The arithmetic mean (average) of a set of <math>50</math> numbers is <math>38</math>. If two numbers, namely, <math>45</math> and <math>55</math>, are discarded, the mean of the remaining set of numbers is: <math>\textbf{(A)}\ 36.5 \qquad\textbf{(B)}\ 37\qquad\textbf{(C)}\ 37.2\qquad\textbf{(D)}\ 37.5\qquad\textbf{(E)}\ 37.52</math>
 
The arithmetic mean (average) of a set of <math>50</math> numbers is <math>38</math>. If two numbers, namely, <math>45</math> and <math>55</math>, are discarded, the mean of the remaining set of numbers is: <math>\textbf{(A)}\ 36.5 \qquad\textbf{(B)}\ 37\qquad\textbf{(C)}\ 37.2\qquad\textbf{(D)}\ 37.5\qquad\textbf{(E)}\ 37.52</math>
  

Latest revision as of 12:59, 16 July 2024

Problem

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.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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