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

(Solution 1)
(Solution 1)
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==Solution 1==
 
==Solution 1==
 
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>.
 
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|before=num-a=1|num-a=3}}
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{{MAA Notice}}

Revision as of 13:17, 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 (ProblemsAnswer KeyResources)
Preceded by
num-a=1
Followed by
Problem 3
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All AHSME Problems and Solutions


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