AMC 8 · 2018 · #10
Grade 5 arithmeticProblem
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?
Pick an answer.
AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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).
Givens: The set of numbers is $\{1, 2, 4\}$; Harmonic mean $=$ reciprocal of (average of the reciprocals); Answer choices: (A) $\frac{3}{7}$, (B) $\frac{7}{12}$, (C) $\frac{12}{7}$, (D) $\frac{7}{4}$, (E) $\frac{7}{3}$
Unknowns: The harmonic mean of $1$, $2$, and $4$, expressed as one of the listed fractions
Understand
Restated: 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).
Givens: The set of numbers is $\{1, 2, 4\}$; Harmonic mean $=$ reciprocal of (average of the reciprocals); Answer choices: (A) $\frac{3}{7}$, (B) $\frac{7}{12}$, (C) $\frac{12}{7}$, (D) $\frac{7}{4}$, (E) $\frac{7}{3}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities
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 $\ge 1$ 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.
Execute — Answer: C
3.NF.A.1 Step 1 - Subproblem 1 — Replace each number in $\{1, 2, 4\}$ with its reciprocal.
- The reciprocal of $x$ is $\frac{1}{x}$, so the reciprocals are $1$, $\frac{1}{2}$, and $\frac{1}{4}$.
💡 A reciprocal $\frac{1}{x}$ is just a unit fraction — the Grade 3 idea of "one part out of $x$ equal parts."
5.NF.A.1 Step 2 - Subproblem 2a — Add the three reciprocals.
- Use the common denominator $4$ to rewrite $1 = \frac{4}{4}$ and $\frac{1}{2} = \frac{2}{4}$, then add numerators.
💡 Adding fractions with unlike denominators by finding a common denominator is the core Grade 5 fraction skill.
5.NF.B.7 Step 3 - Subproblem 2b — Divide that sum by $3$ (the count of numbers) to get the average.
- Dividing a fraction by a whole number means multiplying the denominator by that whole number.
💡 Dividing a fraction by a whole number is the Grade 5 "divide by multiplying by the reciprocal" move.
3.NF.A.1 Step 4 - Subproblem 3 — Take the reciprocal of the average, as the definition directs.
- Flipping $\frac{7}{12}$ swaps numerator and denominator.
💡 Flipping a fraction to get its reciprocal is the same Grade 3 unit-fraction idea, just applied to $\frac{7}{12}$.
4.NF.A.2 Step 5 - Cross-check against the answer list.
- $\frac{12}{7} \approx 1.71$, which sits between $1$ and the arithmetic mean $\frac{1+2+4}{3} = \frac{7}{3} \approx 2.33$ — a known sanity check for the harmonic mean.
- It matches choice (C) exactly, and we can rule out (A) $\frac{3}{7}$ and (B) $\frac{7}{12}$ (both less than $1$) and (E) $\frac{7}{3}$ (the arithmetic mean, a common distractor).
💡 Comparing two fractions like $\frac{12}{7}$ and $\frac{7}{3}$ uses Grade 4 fraction-comparison reasoning.
3.NF.A.1 Subproblem 1 — Replace each number in $\{1, 2, 4\}$ with its reciprocal. The rec 5.NF.A.1 Subproblem 2a — Add the three reciprocals. Use the common denominator $4$ to rew 5.NF.B.7 Subproblem 2b — Divide that sum by $3$ (the count of numbers) to get the average 3.NF.A.1 Subproblem 3 — Take the reciprocal of the average, as the definition directs. Fl 4.NF.A.2 Cross-check against the answer list. $\frac{12}{7} \approx 1.71$, which sits bet Review
Reasonableness: 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 $\frac{7}{3} \approx 2.33$, so a valid harmonic mean must satisfy $1 \le H \le \tfrac{7}{3}$. Our answer $\frac{12}{7} \approx 1.71$ 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.
Alternative: Use Tool #6 (Guess and Check) on the choices via the equivalent definition: $H$ is the harmonic mean iff $\frac{1}{H} = \frac{1}{3}\left(\frac{1}{1} + \frac{1}{2} + \frac{1}{4}\right) = \frac{7}{12}$. So we just need the choice whose reciprocal is $\frac{7}{12}$. Flipping each option: (A) $\frac{7}{3}$, (B) $\frac{12}{7}$, (C) $\frac{7}{12}$ ✓, (D) $\frac{4}{7}$, (E) $\frac{3}{7}$. Only (C) matches.
CCSS standards used (min grade 5)
3.NF.A.1Understand a fraction as quantity formed by parts of a whole (Forming each reciprocal $\frac{1}{x}$ as the Grade 3 unit-fraction idea, and flipping $\frac{7}{12}$ to $\frac{12}{7}$ as the same idea applied to the final answer.)4.NF.A.2Compare two fractions with different numerators and different denominators (Comparing $\frac{12}{7}$, $\frac{7}{3}$, and $\frac{7}{4}$ when checking the answer against the listed choices.)5.NF.A.1Add and subtract fractions with unlike denominators (Adding $1 + \frac{1}{2} + \frac{1}{4} = \frac{7}{4}$ by rewriting each term over the common denominator $4$.)5.NF.B.7Apply and extend understanding of division to divide unit fractions by whole numbers (Dividing the sum $\frac{7}{4}$ by $3$ to obtain the average $\frac{7}{12}$, the second piece of the harmonic-mean definition.)
⭐ This AMC 8 problem only needs Grade 5 fraction arithmetic — adding fractions and dividing a fraction by a whole number — that you already know!
⭐ 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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