AMC 8 · 2025 · #17

Grade 5 rate-ratioarithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

fraction-arithmeticratio-proportionratefraction-multiplication identify-subproblemsdimensional-analysis ↑ Prerequisites: fraction-arithmeticmulti-digit-arithmetic

📏 Medium solution 💡 3 insights 📊 Diagram

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Problem

In the land of Markovia, there are three cities: $A$ , $B$ , and $C$ . There are 100 people who live in $A$ , 120 who live in $B$ , and 160 who live in $C$ . Everyone works in one of the three cities, and a person may work in the same city where they live. In the figure below, an arrow pointing from one city to another is labeled with the fraction of people living in the first city who work in the second city. (For example, $\frac{1}{4}$ of the people who live in $A$ work in $B$ .) How many people work in $A$ ?

Pick an answer.

(A)

(B)

(C)

(D)

115

(E)

160

AMC 8 2025 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.

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Toolkit + CCSS Solution

Understand

Restated: In Markovia, $100$ people live in city $A$, $120$ live in $B$, and $160$ live in $C$. Each labeled arrow on the diagram tells what fraction of a city's residents commute to work in another city (for example, $\tfrac{1}{4}$ of $A$'s residents work in $B$). Everyone works in exactly one of the three cities. Find the total number of workers in city $A$.

Givens: Population: $A = 100$, $B = 120$, $C = 160$; From $A$: $\tfrac{1}{4}$ work in $B$, $\tfrac{1}{5}$ work in $C$; From $B$: $\tfrac{1}{3}$ work in $A$, $\tfrac{1}{6}$ work in $C$; From $C$: $\tfrac{1}{8}$ work in $A$, $\tfrac{1}{10}$ work in $B$; Every resident works in exactly one of $A$, $B$, or $C$ (the remaining fraction stays home); Answer choices: (A) $55$, (B) $60$, (C) $85$, (D) $115$, (E) $160$

Unknowns: The total number of people whose workplace is city $A$

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram

The question "how many people work in $A$" mixes three independent populations ($A$, $B$, $C$ residents) into one total. Tool #7 (Identify Subproblems) splits that into three clean pieces — residents of $A$ working in $A$, residents of $B$ working in $A$, residents of $C$ working in $A$ — each a single fraction-times-population calculation that can be solved on its own and then added. Tool #1 (Draw a Diagram) is the natural companion: the arrow diagram already given is exactly the picture we need, and we just have to read the three arrows pointing **into** $A$ (plus deduce the self-loop $A \to A$ from what is *not* labeled leaving $A$).

Execute — Answer: D

#7 Identify Subproblems 5.NF.A.1 Step 1

Subproblem 1: count residents of $A$ who also work in $A$.
The diagram shows $\tfrac{1}{4}$ of $A$ leaves for $B$ and $\tfrac{1}{5}$ leaves for $C$.
Add those outgoing fractions using a common denominator of $20$ to see how much of $A$ commutes out.

$$\frac{1}{4} + \frac{1}{5} = \frac{5}{20} + \frac{4}{20} = \frac{9}{20}$$

💡 Adding two fractions with unlike denominators ($4$ and $5$) by rewriting both over $20$ is the Grade 5 unlike-denominator skill.

#1 Draw a Diagram 5.NF.A.1 Step 2

Since every resident of $A$ works in exactly one of the three cities, the remaining fraction of $A$ stays to work in $A$ itself.
Subtract the commuting fraction from $1$.

$$1 - \frac{9}{20} = \frac{11}{20}$$

💡 Reading the diagram tells us nothing leaves $A$ except the labeled arrows, so what's left of the whole ($1$) is the self-loop $A \to A$.

#7 Identify Subproblems 4.NF.B.4 Step 3

Apply that fraction to $A$'s population of $100$ to get the count of residents who live and work in $A$.

$$100 \times \frac{11}{20} = \frac{100}{20} \times 11 = 5 \times 11 = 55$$

💡 Multiplying a fraction by a whole number ($100 \times \tfrac{11}{20}$) is a Grade 4 fraction-of-a-set calculation.

#7 Identify Subproblems 4.NF.B.4 Step 4

Subproblem 2: count residents of $B$ who work in $A$.
The arrow $B \to A$ is labeled $\tfrac{1}{3}$, so apply that fraction to $B$'s population of $120$.

$$120 \times \frac{1}{3} = 40$$

💡 Taking $\tfrac{1}{3}$ of $120$ is the classic "fraction of a quantity" move from Grade 4.

#7 Identify Subproblems 4.NF.B.4 Step 5

Subproblem 3: count residents of $C$ who work in $A$.
The arrow $C \to A$ is labeled $\tfrac{1}{8}$, so apply that fraction to $C$'s population of $160$.

$$160 \times \frac{1}{8} = 20$$

💡 Same fraction-of-a-set move: $\tfrac{1}{8}$ of $160$ is just $160 \div 8$.

#7 Identify Subproblems 4.OA.A.3 Step 6

Combine the three subproblem answers.
Every worker in $A$ is counted exactly once because the three groups (lives in $A$, lives in $B$, lives in $C$) are disjoint by residence.

$$55 + 40 + 20 = 115 \;\Rightarrow\; \textbf{(D)}$$

💡 Summing the partial counts from three subproblems is the final step of a Grade 4 multi-step word problem.

[1] #7 5.NF.A.1 Subproblem 1: count residents of $A$ who also work in $A$. The diagram shows $\t

[2] #1 5.NF.A.1 Since every resident of $A$ works in exactly one of the three cities, the remain

[3] #7 4.NF.B.4 Apply that fraction to $A$'s population of $100$ to get the count of residents w

[4] #7 4.NF.B.4 Subproblem 2: count residents of $B$ who work in $A$. The arrow $B \to A$ is lab

[5] #7 4.NF.B.4 Subproblem 3: count residents of $C$ who work in $A$. The arrow $C \to A$ is lab

[6] #7 4.OA.A.3 Combine the three subproblem answers. Every worker in $A$ is counted exactly onc

Review

Reasonableness: Total population is $100 + 120 + 160 = 380$. If everyone simply worked in their own city, $A$ would have $100$ workers; the diagram says only $\tfrac{11}{20}$ ($55$) of $A$'s residents stay, but it also pulls in $40$ from $B$ and $20$ from $C$ — a net gain of $15$. So $A$ ends up with $115$, slightly above its own population, which matches choice (D). A quick sanity check on conservation: total workers $= 380$, and the answer $115$ for $A$ leaves $265$ workers to be split between $B$ and $C$, which is plausible given $B$ and $C$ together house $280$ residents.

Alternative: Tool #3 (Eliminate Possibilities) gives a fast sanity filter. Just from arrows pointing into $A$, we already collect $120 \times \tfrac{1}{3} = 40$ from $B$ and $160 \times \tfrac{1}{8} = 20$ from $C$ — that's $60$ before counting any $A$-resident. So the answer must be at least $60$, eliminating (A) $55$ and (B) $60$ (the latter would require zero $A$-residents to stay, which contradicts the labeled outgoing fractions summing to only $\tfrac{9}{20} < 1$). Only (D) $115$ is consistent with adding a positive $A$-self count.

CCSS standards used (min grade 5)

5.NF.A.1 Add and subtract fractions with unlike denominators (Computing $\tfrac{1}{4} + \tfrac{1}{5} = \tfrac{9}{20}$ and then $1 - \tfrac{9}{20} = \tfrac{11}{20}$ to find the fraction of $A$'s residents who stay in $A$.)
4.NF.B.4 Multiply a fraction by a whole number (Computing $100 \times \tfrac{11}{20} = 55$, $120 \times \tfrac{1}{3} = 40$, and $160 \times \tfrac{1}{8} = 20$ — the three "fraction of a population" counts.)
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Adding the three subproblem totals $55 + 40 + 20 = 115$ to combine workers from all three resident groups.)

⭐ This AMC 8 problem only needs Grade 5 fraction addition (and Grade 4 "fraction of a number") that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 5)

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