AMC 8 · 2010 · #9
Easy mode Grade 6Problem
Ryan took three tests.
On a test with 25 problems, he got 80% correct. On a test with 40 problems, he got 90% correct. On a test with 10 problems, he got 70% correct.
Now think of all three tests as one big set of problems. What percent of the problems in the whole set did Ryan get correct?
Pick an answer.
AMC 8 2010 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: Ryan took three separate tests. He got $80\%$ correct on a $25$-problem test, $90\%$ correct on a $40$-problem test, and $70\%$ correct on a $10$-problem test. Treating all $75$ problems as one big pool, what percent did he get correct overall?
Givens: Test 1: $80\%$ correct out of $25$ problems; Test 2: $90\%$ correct out of $40$ problems; Test 3: $70\%$ correct out of $10$ problems; Answer choices: (A) $64$, (B) $75$, (C) $80$, (D) $84$, (E) $86$
Unknowns: The overall percent correct across all three tests combined
Understand
Restated: Ryan took three separate tests. He got $80\%$ correct on a $25$-problem test, $90\%$ correct on a $40$-problem test, and $70\%$ correct on a $10$-problem test. Treating all $75$ problems as one big pool, what percent did he get correct overall?
Givens: Test 1: $80\%$ correct out of $25$ problems; Test 2: $90\%$ correct out of $40$ problems; Test 3: $70\%$ correct out of $10$ problems; Answer choices: (A) $64$, (B) $75$, (C) $80$, (D) $84$, (E) $86$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #8 Analyze the Units
The three tests are independent subproblems (Tool #7): for each one, convert its percent into an actual count of correct problems. Once we have three counts in the same unit ("problems correct"), we can add them and compare to the total. Tool #8 (Analyze the Units) reminds us we cannot average the three percents directly — "percent" is a ratio, not a count, so $\tfrac{80 + 90 + 70}{3}$ has no meaning here. We must move to the common unit of "problems" first, then return to percent at the end.
Execute — Answer: D
6.RP.A.3 Step 1 Subproblem 1: convert $80\%$ of $25$ into a count of problems correct on Test 1.
💡 Finding a percent of a whole-number quantity is Grade 6 percent reasoning: $80\%$ of $25$ means $\tfrac{80}{100} \times 25$.
5.NBT.B.7 Step 2 Subproblem 2: convert $90\%$ of $40$ into a count of problems correct on Test 2.
💡 Multiplying a decimal to hundredths by a whole number is exactly the Grade 5 decimal-arithmetic standard.
5.NBT.B.7 Step 3 Subproblem 3: convert $70\%$ of $10$ into a count of problems correct on Test 3.
💡 Same decimal multiplication as the previous step — each test contributes its own whole-number count.
4.NBT.B.4 Step 4 Now that all three counts share the unit "problems correct," add them to get the total correct, and add the three test sizes to get the total number of problems.
💡 Adding counts is only legal because we converted percents to the common unit first — Tool #8's unit-check move.
6.RP.A.3 Step 5 Convert the combined count back to a percent: $\tfrac{\text{correct}}{\text{total}} \times 100\%$.
💡 Expressing a part out of a whole as a percent is Grade 6 ratio reasoning — the inverse of step 1.
6.RP.A.3 Subproblem 1: convert $80\%$ of $25$ into a count of problems correct on Test 1. 5.NBT.B.7 Subproblem 2: convert $90\%$ of $40$ into a count of problems correct on Test 2. 5.NBT.B.7 Subproblem 3: convert $70\%$ of $10$ into a count of problems correct on Test 3. 4.NBT.B.4 Now that all three counts share the unit "problems correct," add them to get the 6.RP.A.3 Convert the combined count back to a percent: $\tfrac{\text{correct}}{\text{tota Review
Reasonableness: The naive (wrong) answer of averaging the three percents gives $\tfrac{80 + 90 + 70}{3} = 80\%$ — which is choice (C), a classic trap. The correct $84\%$ is higher than $80\%$ because the largest test ($40$ problems) is the one with the highest score ($90\%$), so it pulls the weighted average up. Our answer sits between the lowest score ($70\%$) and the highest ($90\%$), and leans toward $90\%$ as expected.
Alternative: Tool #6 (Guess and Check) on the choices: the total is $75$ problems, so each percent choice corresponds to a count $\tfrac{p}{100} \times 75$. (A) $64\% \to 48$, (B) $75\% \to 56.25$, (C) $80\% \to 60$, (D) $84\% \to 63$, (E) $86\% \to 64.5$. Only (D) gives the integer $63$ we computed.
CCSS standards used (min grade 6)
4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the three correct counts ($20 + 36 + 7 = 63$) and the three test sizes ($25 + 40 + 10 = 75$).)5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Multiplying decimals such as $0.90 \times 40 = 36$ and $0.70 \times 10 = 7$ to convert each test's percent into a count.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems, including finding a percent of a quantity (Going from percent to count ($80\%$ of $25 = 20$) and back from count to percent ($\tfrac{63}{75} = 84\%$) — the percent reasoning that frames the whole problem.)
⭐ When tests have different sizes, you can't just average the percents — turn each percent into a count of correct problems first, add them up, then convert back. That's Grade 6 percent reasoning at work.
⭐ When tests have different sizes, you can't just average the percents — turn each percent into a count of correct problems first, add them up, then convert back. That's Grade 6 percent reasoning at work.
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