Difference between revisions of "1955 AHSME Problems/Problem 3"
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| − | <math>(A)</math> | + | == Problem== |
| − | {{ | + | 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> | ||
| + | ==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>. | ||
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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 
, the arithmetic mean (average) of the ten numbers:
Solution 1
Let the sum of the 10 numbers be x. The mean is then 
. 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 
, which simplifies to 
, which is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
