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Difference between revisions of "1955 AHSME Problems/Problem 3"

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<math>(A)</math>
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== Problem==
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If each number in a set of ten numbers is increased by <math>20</math>, the arithmetic mean (average) of the ten numbers:
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<math> \textbf{(A)}\ \text{remains the same}\qquad\textbf{(B)}\ \text{is increased by 20}\qquad\textbf{(C)}\ \text{is increased by 200}\\ \textbf{(D)}\ \text{is increased by 10}\qquad\textbf{(E)}\ \text{is increased by 2} </math>
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==Solution 1==
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Let the sum of the 10 numbers be x. The mean is then <math>\frac{x}{10}</math>.  Then, since you're adding 20 to each number, the new sum of the numbers is x+200, since there are 10 numbers. Then, the new mean is <math>\frac{x+200}{10}</math>, which simplifies to <math>\frac{x}{10}+20</math>, which is <math>\fbox{B}</math>.
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==See Also==
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{{AHSME 50p box|year=1955|num-b=2|num-a=4}}
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{{MAA Notice}}

Latest revision as of 12:18, 4 May 2020

Problem

If each number in a set of ten numbers is increased by $20$, the arithmetic mean (average) of the ten numbers:

$\textbf{(A)}\ \text{remains the same}\qquad\textbf{(B)}\ \text{is increased by 20}\qquad\textbf{(C)}\ \text{is increased by 200}\\ \textbf{(D)}\ \text{is increased by 10}\qquad\textbf{(E)}\ \text{is increased by 2}$

Solution 1

Let the sum of the 10 numbers be x. The mean is then $\frac{x}{10}$. Then, since you're adding 20 to each number, the new sum of the numbers is x+200, since there are 10 numbers. Then, the new mean is $\frac{x+200}{10}$, which simplifies to $\frac{x}{10}+20$, which is $\fbox{B}$.

See Also

1955 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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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