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 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,

\frac{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.

Pick an answer.

(A)

(B)

(C)

(D)

115

(E)

160

AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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 → A from what is not labeled leaving A).

1STEP 1

Subproblem 1 — A's own workers: add A's two outgoing fractions, $\frac{1}{4}$ (to B) and $\frac{1}{5}$ (to C), over 20 to get $\frac{9}{20}$ .

\frac{1}{4}

\frac{1}{5}

\frac{5}{20}

\frac{4}{20}

\frac{9}{20}

2STEP 2

Everyone works somewhere, so the fraction staying in A is 1 minus $\frac{9}{20}$ , which leaves $\frac{11}{20}$ .

1 -

\frac{9}{20}

\frac{11}{20}

3STEP 3

Multiply A's population of 100 by $\frac{11}{20}$ to count residents who live and work in A: 55.

100 ×

\frac{11}{20}

\frac{100}{20}

× 11 = 5 × 11 = 55

4STEP 4

Subproblem 2 — the arrow B → A is $\frac{1}{3}$ , so $\frac{1}{3}$ of B's 120 residents work in A: 40.

120 ×

\frac{1}{3}

= 40

5STEP 5

Subproblem 3 — the arrow C → A is $\frac{1}{8}$ , so $\frac{1}{8}$ of C's 160 residents work in A: 20.

160 ×

\frac{1}{8}

= 20

6STEP 6

Add the three disjoint groups — 55 from A, 40 from B, 20 from C — to get 115 workers in A.

55 + 40 + 20 = 115 → (D)

Answer

115

Total population is 100 + 120 + 160 = 380. If everyone simply worked in their own city, A would have 100 workers; the diagram says only

\frac{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.

💡Key takeaway

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

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