AMC 8 · 2018 · #10

Grade 5 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

fraction-arithmeticmean-median-mode-rangeformula-substitution identify-subproblems ↑ Prerequisites: fraction-arithmetic

📏 Short solution 💡 2 insights

Problem

The problem defines the harmonic mean of a set of non-zero numbers as the reciprocal of the average of the reciprocals. Apply that definition to the set {1, 2, 4} and pick the matching fraction from (A)-(E).

Pick an answer.

(A)

$\frac{3}{7}$

(B)

$\frac{7}{12}$

(C)

$\frac{12}{7}$

(D)

$\frac{7}{4}$

(E)

$\frac{7}{3}$

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The definition of harmonic mean is a chain of three little operations packed into one sentence: (i) replace each number with its reciprocal, (ii) average those reciprocals, (iii) take the reciprocal of that average. Tool #7 (Identify Subproblems) is the natural fit — we explicitly split the chain into three small, easy subproblems and do each one cleanly. Tool #3 (Eliminate Possibilities) is a strong sanity check at the end: the answer must be greater than 1 (since the harmonic mean of numbers ≥ 1 is at least 1) and less than the arithmetic mean $\frac{7}{3}$ , which already rules out (A) and (B). That funnels us toward (C), (D), or (E) before we even finish computing.

1STEP 1

Subproblem 1 — Replace each number in {1, 2, 4} with its reciprocal: 1, $\frac{1}{2}$ , and $\frac{1}{4}$ .

{

\frac{1}{1}

\frac{1}{2}

\frac{1}{4}

} = {1,

\frac{1}{2}

\frac{1}{4}

}

2STEP 2

Subproblem 2a — Over the common denominator 4, add the reciprocals: $\frac{4}{4}$ + $\frac{2}{4}$ + $\frac{1}{4}$ = $\frac{7}{4}$ .

1 +

\frac{1}{2}

\frac{1}{4}

\frac{4}{4}

\frac{2}{4}

\frac{1}{4}

\frac{7}{4}

3STEP 3

Subproblem 2b — Divide that sum by 3 (how many numbers) for the average: $\frac{7}{4}$ × $\frac{1}{3}$ = $\frac{7}{12}$ .

average = 7/4/3 =

\frac{7}{4}

\frac{1}{3}

\frac{7}{12}

4STEP 4

Subproblem 3 — As the definition directs, flip the average $\frac{7}{12}$ to its reciprocal: $\frac{12}{7}$ .

harmonic mean = 1/7/12 =

\frac{12}{7}

5STEP 5

Cross-check: $\frac{12}{7}$ ≈ 1.71 sits between 1 and the arithmetic mean $\frac{7}{3}$ , so the answer is (C).

\frac{12}{7}

→ (C)

Answer

\frac{12}{7}

The harmonic mean of positive numbers always lies between the smallest input and the arithmetic mean. Here the smallest input is 1 and the arithmetic mean is

\frac{7}{3}

≈ 2.33, so a valid harmonic mean must satisfy 1 ≤ H ≤

\frac{7}{3}

. Our answer

\frac{12}{7}

≈ 1.71 fits cleanly in that window. Choices (A) and (B) are below 1 and (E) equals the arithmetic mean, so they are immediately impossible; (D)

\frac{7}{4}

= 1.75 is plausible by size but doesn't survive the actual computation. Only (C) survives both the sanity bounds and the exact calculation.

💡Key takeaway

This AMC 8 problem only needs Grade 5 fraction arithmetic — adding fractions and dividing a fraction by a whole number — that you already know!

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