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

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==Problem 10==
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==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==
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== 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>
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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 by OmegaLearn ==
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https://youtu.be/TkZvMa30Juo?t=2935
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~ pi_is_3.14
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==Video Solution==
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https://youtu.be/PA9X-2xLuxY
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~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 03:46, 16 January 2023

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?

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

Solution

The sum of the reciprocals is $\frac{1}{1} + \frac{1}{2} + \frac{1}{4}= \frac{7}{4}$. Their average is $\frac{7}{12}$. Taking the reciprocal of this gives $\boxed{\textbf{(C) }\frac{12}{7}}$.

Video Solution by OmegaLearn

https://youtu.be/TkZvMa30Juo?t=2935

~ pi_is_3.14

Video Solution

https://youtu.be/PA9X-2xLuxY

~savannahsolver

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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