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== | ||
| − | 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}, which simplifies to \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>. |
Revision as of 11:47, 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 
.
