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2018 AMC 8 Problems/Problem 10: Difference between revisions

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==Problem 10==
==Problem==
The [i]harmonic mean[/i] 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 ==
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/zlmUJMwHpqk
~EzLx
== Video Solution==
[//youtu.be/TkZvMa30Juo?t=2935 OmegaLearn video]
==Video Solution==
https://youtu.be/PA9X-2xLuxY
~savannahsolver
==See Also==
{{AMC8 box|year=2018|num-b=9|num-a=11}}
{{MAA Notice}}
[[Category:Introductory Algebra Problems]]

Latest revision as of 12:39, 6 June 2025

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 (CRITICAL THINKING!!!)

https://youtu.be/zlmUJMwHpqk

~EzLx

Video Solution

OmegaLearn video

Video Solution

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

~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. Error creating thumbnail: File missing

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