Difference between revisions of "2018 AMC 8 Problems/Problem 10"
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== Solution == | == Solution == | ||
− | The sum of the reciprocals is <math>\frac | + | The sum of the reciprocals is <math>\frac{1}{1} + \frac{1}{2} + \frac{1}{4}= \frac{7}{4}</math>. Their average is <math>\frac{7}{12}</math>. Taking the reciprocal of this gives <math>\boxed{\textbf{(C) }\frac{12}{7}}</math> |
==See Also== | ==See Also== |
Revision as of 17:04, 5 January 2022
Problem
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
. Taking the reciprocal of this gives
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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