Difference between revisions of "2018 AMC 8 Problems/Problem 10"
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| − | ==Problem | + | ==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? | 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? | ||
<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> | ||
| − | ==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>. |
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| + | == Video Solution (CRITICAL THINKING!!!)== | ||
| + | https://youtu.be/zlmUJMwHpqk | ||
| + | |||
| + | ~EzLx | ||
| + | |||
| + | == Video Solution== | ||
| + | [//youtu.be/TkZvMa30Juo?t=2935 OmegaLearn video] | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/PA9X-2xLuxY | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
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| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 13:39, 6 June 2025
Contents
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 
.
Video Solution (CRITICAL THINKING!!!)
~EzLx
Video Solution
Video Solution
~savannahsolver
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
