2018 AMC 8 Problems/Problem 10: Difference between revisions
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==Solution== | ==Solution== | ||
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> | 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> | ||
==See Also== | |||
{{AMC8 box|year=2018|num-b=9|num-a=11}} | {{AMC8 box|year=2018|num-b=9|num-a=11}} | ||
{{MAA Notice}} | |||
Revision as of 18:53, 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 
See Also
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
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