2018 AMC 8 Problem 10

Attempt Problem 10 of the 2018 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 8 solutions, or check the answer key.

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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 1,1, 2,2, and 4?4?
37 \dfrac{3}{7}
712 \dfrac{7}{12}
127 \dfrac{12}{7}
74 \dfrac{7}{4}
73 \dfrac{7}{3}
Answer: C
Concepts:harmonic meanfraction
Small hint:
Average the reciprocals of 1,2,1,2, and 44.
Big hint:
The harmonic mean is the reciprocal of that average.
Video solution:
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Written solution:
The reciprocals of 11, 22, and 44 are 11\dfrac11, 12\dfrac12, and 14\dfrac14, respectively. The average of these reciprocals is (1+12+14)3=(74)3=712.\begin{align*}\dfrac{\left(1 + \dfrac12 + \dfrac14\right)}{3} &= \dfrac{\left(\dfrac{7}{4}\right)}{3} \\&= \dfrac{7}{12}. \end{align*}
As the harmonic mean is the reciprocal of the average of the reciprocals of the numbers (which we just calculated to be 712\dfrac{7}{12}), we conclude that the harmonic mean is 127.\dfrac{12}{7}.
Thus, the correct answer is C.
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