Difference between revisions of "2018 AMC 8 Problems/Problem 10"
m (→Solution) |
(→Solution) |
||
| Line 4: | Line 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 11 + \frac 12 + \frac 14= \frac 74</math>. Their average is <math>\frac 7{12}</math>. Taking the reciprocal of this gives <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>. Taking the reciprocal of this gives <math>\boxed{\textbf{(C) }\frac{12}{7}}</math> | ||
Revision as of 19:31, 23 December 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 
. 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
