Difference between revisions of "2018 AMC 8 Problems/Problem 10"
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| − | ==Problem | + | ==Problem== |
| − | The | + | 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> | ||
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| + | == Solution == | ||
| + | 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>. | ||
| + | |||
| + | == Video Solution (CRITICAL THINKING!!!)== | ||
| + | https://youtu.be/v-ZooJaqrWA | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/TkZvMa30Juo?t=2935 | ||
| + | |||
| + | ~ pi_is_3.141592653589793238462643383279502884197 | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/PA9X-2xLuxY | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2018|num-b=9|num-a=11}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 18:35, 25 December 2023
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!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/TkZvMa30Juo?t=2935
~ pi_is_3.141592653589793238462643383279502884197
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
