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Difference between revisions of "1955 AHSME Problems/Problem 3"
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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>
 
<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>
 
==Solution 1==
 
==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 13:18, 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
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All AHSME Problems and Solutions


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