Difference between revisions of "2018 AMC 8 Problems/Problem 10"
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<math>\textbf{(A) }\frac{3}{7}\qquad\textbf{(B) }\frac{7}{12}\qquad\textbf{(C) }\frac{12}{7}\qquad\textbf{(D) }\frac{7}{4}\qquad \textbf{(E) }\frac{7}{3}</math> | <math>\textbf{(A) }\frac{3}{7}\qquad\textbf{(B) }\frac{7}{12}\qquad\textbf{(C) }\frac{12}{7}\qquad\textbf{(D) }\frac{7}{4}\qquad \textbf{(E) }\frac{7}{3}</math> | ||
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| + | ==Solution== | ||
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| + | The sum of the reciprocals is <math>\frac 11 + \frac 12 + \frac 14= \frac 74</math>. Their average is <math>\frac 7{12}</math>. Thus our answer is <math>\boxed{\textbf{(C) }\frac{12}{7}}</math> | ||
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{{AMC8 box|year=2018|num-b=9|num-a=11}} | {{AMC8 box|year=2018|num-b=9|num-a=11}} | ||
Revision as of 13:11, 21 November 2018
Problem 10
The harmonic mean of a set of non-zero numbers is the reciprocal of the average of the reciprocals of the numbers. What is the harmonic mean of 1, 2, and 4?
Solution
The sum of the reciprocals is 
. Their average is 
. Thus our answer is 
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
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| All AJHSME/AMC 8 Problems and Solutions | ||
