Difference between revisions of "2018 AMC 8 Problems/Problem 10"

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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>
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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>
  
 
==See Also==
 
==See Also==

Revision as of 16:42, 22 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?

$\textbf{(A) }\frac{3}{7}\qquad\textbf{(B) }\frac{7}{12}\qquad\textbf{(C) }\frac{12}{7}\qquad\textbf{(D) }\frac{7}{4}\qquad \textbf{(E) }\frac{7}{3}$

Solution

The sum of the reciprocals is $\frac 11 + \frac 12 + \frac 14= \frac 74$. Their average is $\frac 7{12}$. Taking the reciprocal of this gives $\boxed{\textbf{(C) }\frac{12}{7}}$

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
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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