AMC 8 · 2023 · #5

Grade 4 rate-ratio

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Direct Ratio Scale-Up

ratio-proportionfraction-arithmetic ratio-proportionpattern-recognition ↑ Prerequisites: ratio-proportionfraction-arithmetic

📏 Short solution 💡 2 insights

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Problem

A lake has 250 trout and many other fish. A biologist nets a sample of 180 fish and finds 30 trout. If the trout-to-total ratio is the same in the sample and in the whole lake, how many fish live in the lake?

Pick an answer.

(A)

1250

(B)

1500

(C)

1750

(D)

1800

(E)

2000

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

Read the sample as a repeating pattern: out of every 180 fish, 30 are trout — that means in every group of 6 fish, exactly 1 is a trout (Tool #5). The whole lake follows the same pattern, so we just count how many groups of 6 fit around the 250 trout we know about. We don't need algebra: scaling a sample to a population is a smaller, friendlier version of the same multiplication problem (Tool #9). Since this is multiple-choice, Tool #3 lets us double-check the answer against the listed options at the end.

1STEP 1

Simplify the sample: 30 trout out of 180 fish reduces to 1 trout in every 6 fish.

(30 trout)/(180 fish) = (1 trout)/(6 fish)

2STEP 2

The lake shares that ratio, so its 250 trout mean the lake is built from 250 groups of 6 fish.

250 trout → 250 groups of 6 fish

3STEP 3

Count them all: 250 groups × 6 fish each gives 1500 fish in the lake.

250 × 6 = 1500 fish

4STEP 4

Cross-check the choices: 1500 is (B); the others like $\frac{250}{2000}$ = $\frac{1}{8}$ break the $\frac{1}{6}$ ratio.

1500 → (B)

Answer

1500

Does 1500 make sense? In the sample, 1 out of every 6 fish was a trout, so the total should be about 6 times the trout count. 6 × 250 = 1500, which fits. And

\frac{250}{1500}

\frac{1}{6}

matches

\frac{30}{180}

\frac{1}{6}

, so the ratio condition is satisfied. The answer (B) 1500 is consistent.

💡Key takeaway

This AMC 8 problem only needs Grade 4 equivalent fractions and multiplicative comparison you already know!

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