AMC 8 · 2023 · #5
Grade 4 rate-ratioProblem
A lake contains 250 trout, along with a variety of other fish. When a marine biologist catches and releases a sample of 180 fish from the lake, 30 are identified as trout. Assume that the ratio of trout to the total number of fish is the same in both the sample and the lake. How many fish are there in the lake?
Pick an answer.
AMC 8 2023 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: 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?
Givens: The lake contains exactly $250$ trout (the rest are other kinds of fish); A sample of $180$ fish is netted from the lake; Of those $180$ sampled fish, $30$ are trout; The ratio of trout to total fish is the SAME in the sample and in the whole lake; Answer choices: (A) 1250, (B) 1500, (C) 1750, (D) 1800, (E) 2000
Unknowns: The total number of fish in the lake (trout + all other fish)
Understand
Restated: 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?
Givens: The lake contains exactly $250$ trout (the rest are other kinds of fish); A sample of $180$ fish is netted from the lake; Of those $180$ sampled fish, $30$ are trout; The ratio of trout to total fish is the SAME in the sample and in the whole lake; Answer choices: (A) 1250, (B) 1500, (C) 1750, (D) 1800, (E) 2000
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
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.
Execute — Answer: B
4.NF.A.1 Step 1 - Find the pattern hidden in the sample.
- Out of $180$ sampled fish, $30$ are trout, so we can group the sample into equal piles where each pile has the same trout-to-total mix.
- Dividing both numbers by $30$, the simplest pile is: $1$ trout among every $6$ fish.
- So the sample is $30$ copies of the pattern "1 trout in 6 fish".
💡 Recognizing that $\tfrac{30}{180}$ and $\tfrac{1}{6}$ are equivalent fractions is exactly the Grade 4 equivalent-fractions skill.
4.OA.A.2 Step 2 - Use that same pattern for the whole lake.
- Because the lake has the same trout-to-total mix, the lake is also built out of "1 trout, 6 fish" groups.
- The lake holds $250$ trout, so it must be made of $250$ such groups — one for each trout.
- This is the same kind of scaling we did in the sample, just with bigger counts (Tool #9 in action).
💡 "For each trout, there are $6$ fish total" is a Grade 4 multiplicative comparison: $\text{fish} = 6 \times \text{trout}$.
4.NBT.B.5 Step 3 - Count up the total.
- Each of the $250$ groups contains $6$ fish, so the lake holds $250 \times 6$ fish in all.
- Multiplying a multi-digit number by a single digit, $250 \times 6 = 1500$.
💡 Multiplying a 3-digit number by a 1-digit number is a Grade 4 fluency standard.
4.NF.A.1 Step 4 - Verify against the answer choices.
- Our total is $1500$, which matches choice (B).
- The other choices ($1250, 1750, 1800, 2000$) would each give a different trout-to-fish ratio — for example, $\tfrac{250}{2000} = \tfrac{1}{8}$, not $\tfrac{1}{6}$ — so they can be eliminated.
💡 Checking whether $\tfrac{250}{1500}$ reduces to $\tfrac{1}{6}$ is again the Grade 4 equivalent-fractions check.
4.NF.A.1 Find the pattern hidden in the sample. Out of $180$ sampled fish, $30$ are trout 4.OA.A.2 Use that same pattern for the whole lake. Because the lake has the same trout-to 4.NBT.B.5 Count up the total. Each of the $250$ groups contains $6$ fish, so the lake hold 4.NF.A.1 Verify against the answer choices. Our total is $1500$, which matches choice (B) Review
Reasonableness: 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. $6 \times 250 = 1500$, which fits. And $\tfrac{250}{1500} = \tfrac{1}{6}$ matches $\tfrac{30}{180} = \tfrac{1}{6}$, so the ratio condition is satisfied. The answer (B) 1500 is consistent.
Alternative: We could use Tool #13 (Convert to Algebra) and set up the proportion $\tfrac{30}{180} = \tfrac{250}{F}$, then cross-multiply to get $F = \tfrac{180 \times 250}{30} = 1500$. The pattern-and-scaling path above is friendlier for an elementary student because it avoids cross-multiplication.
CCSS standards used (min grade 4)
4.NF.A.1Explain why a fraction is equivalent to another fraction (Reducing the sample ratio $\tfrac{30}{180}$ to its simplest form $\tfrac{1}{6}$ and checking that $\tfrac{250}{1500} = \tfrac{1}{6}$.)4.OA.A.2Multiply or divide to solve word problems involving multiplicative comparison (Recognizing that "$1$ trout per $6$ fish" means the total number of fish is $6$ times the number of trout.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Computing the final product $250 \times 6 = 1500$.)
⭐ This AMC 8 problem only needs Grade 4 equivalent fractions and multiplicative comparison you already know!
⭐ This AMC 8 problem only needs Grade 4 equivalent fractions and multiplicative comparison you already know!
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