AMC 8 · 2018 · #10

Grade 5 arithmetic

Archetype Arithmetic Expression Decomposition

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

📏 Short solution 💡 2 insights

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?

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

Pick an answer.

(A)

$\frac{3}{7}$

(B)

$\frac{7}{12}$

(C)

$\frac{12}{7}$

(D)

$\frac{7}{4}$

(E)

$\frac{7}{3}$

View mode:

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

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

#7 Identify Subproblems 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 $\frac{1}{4}$.

$$\left\{\frac{1}{1},\; \frac{1}{2},\; \frac{1}{4}\right\} = \left\{1,\; \tfrac{1}{2},\; \tfrac{1}{4}\right\}$$

💡 A reciprocal $\frac{1}{x}$ is just a unit fraction — the Grade 3 idea of "one part out of $x$ equal parts."

#7 Identify Subproblems 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.

$$1 + \tfrac{1}{2} + \tfrac{1}{4} = \tfrac{4}{4} + \tfrac{2}{4} + \tfrac{1}{4} = \tfrac{7}{4}$$

💡 Adding fractions with unlike denominators by finding a common denominator is the core Grade 5 fraction skill.

#7 Identify Subproblems 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.

$$\text{average} = \dfrac{\tfrac{7}{4}}{3} = \tfrac{7}{4} \times \tfrac{1}{3} = \tfrac{7}{12}$$

💡 Dividing a fraction by a whole number is the Grade 5 "divide by multiplying by the reciprocal" move.

#7 Identify Subproblems 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.

$$\text{harmonic mean} = \dfrac{1}{\tfrac{7}{12}} = \tfrac{12}{7}$$

💡 Flipping a fraction to get its reciprocal is the same Grade 3 unit-fraction idea, just applied to $\frac{7}{12}$.

#3 Eliminate Possibilities 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).

$$\tfrac{12}{7} \;\Rightarrow\; \textbf{(C)}$$

💡 Comparing two fractions like $\frac{12}{7}$ and $\frac{7}{3}$ uses Grade 4 fraction-comparison reasoning.

[1] #7 3.NF.A.1 Subproblem 1 — Replace each number in $\{1, 2, 4\}$ with its reciprocal. The rec

[2] #7 5.NF.A.1 Subproblem 2a — Add the three reciprocals. Use the common denominator $4$ to rew

[3] #7 5.NF.B.7 Subproblem 2b — Divide that sum by $3$ (the count of numbers) to get the average

[4] #7 3.NF.A.1 Subproblem 3 — Take the reciprocal of the average, as the definition directs. Fl

[5] #3 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.1 Understand 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.2 Compare 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.1 Add 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.7 Apply 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 5)

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