AMC 8 · 2006 · #12
Grade 6 rate-ratioProblem
Antonette gets 70% on a 10-problem test, 80% on a 20-problem test and 90% on a 30-problem test. If the three tests are combined into one 60-problem test, which percent is closest to her overall score?
Pick an answer.
AMC 8 2006 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: Antonette scores $70\%$ on a $10$-question test, $80\%$ on a $20$-question test, and $90\%$ on a $30$-question test. If all three tests are merged into one $60$-question test, which choice is closest to her overall percent correct?
Givens: Test 1: $70\%$ correct out of $10$ questions; Test 2: $80\%$ correct out of $20$ questions; Test 3: $90\%$ correct out of $30$ questions; Combined test has $10 + 20 + 30 = 60$ questions; Answer choices: (A) $40$, (B) $77$, (C) $80$, (D) $83$, (E) $87$
Unknowns: The percent closest to Antonette's overall score on the $60$-question combined test
Understand
Restated: Antonette scores $70\%$ on a $10$-question test, $80\%$ on a $20$-question test, and $90\%$ on a $30$-question test. If all three tests are merged into one $60$-question test, which choice is closest to her overall percent correct?
Givens: Test 1: $70\%$ correct out of $10$ questions; Test 2: $80\%$ correct out of $20$ questions; Test 3: $90\%$ correct out of $30$ questions; Combined test has $10 + 20 + 30 = 60$ questions; Answer choices: (A) $40$, (B) $77$, (C) $80$, (D) $83$, (E) $87$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #9 Solve an Easier Problem
The trap is averaging $70$, $80$, $90$ to get $80$ (choice C). That ignores test size. Tool #9 (Solve an Easier Problem) reframes the question: instead of mixing percents, count actual questions answered correctly on each test — those are concrete numbers we can add. Tool #2 (Make a Systematic List) then walks through the three tests in order, recording correct count and total count for each, so the combined total is just two sums to divide. No algebra needed.
Execute — Answer: D
6.RP.A.3 Step 1 - Convert each percent into a count of correct answers.
- For each test, multiply the percent (as a decimal) by the number of questions.
- This swaps the messy percent-mixing problem for a clean count-and-add problem.
💡 "Percent of a quantity as a rate per 100" is the Grade 6 way to read "$70\%$ of $10$" as $7$.
5.NBT.B.5 Step 2 - List the totals in a small table.
- Sum the correct counts from the three tests to get total correct on the combined test, and sum the question counts to get the combined test length.
💡 Once the percents are gone, this is just whole-number addition — the Grade 5 fluency standard.
6.RP.A.1 Step 3 - Write the overall score as a ratio and simplify.
- The combined percent correct is total correct over total questions, expressed as a percentage.
💡 $\dfrac{50}{60}$ and $\dfrac{5}{6}$ are the same ratio written with different sized parts — the Grade 6 ratio idea.
6.RP.A.3 Step 4 - Convert $\dfrac{5}{6}$ to a percent and pick the closest choice.
- Divide $5$ by $6$ (or multiply numerator and denominator until the denominator is $100$).
💡 $83.33\%$ rounds to $83\%$, which is choice (D). Choice (C) $80\%$ is the trap from naively averaging $70, 80, 90$.
6.RP.A.3 Convert each percent into a count of correct answers. For each test, multiply th 5.NBT.B.5 List the totals in a small table. Sum the correct counts from the three tests to 6.RP.A.1 Write the overall score as a ratio and simplify. The combined percent correct is 6.RP.A.3 Convert $\dfrac{5}{6}$ to a percent and pick the closest choice. Divide $5$ by $ Review
Reasonableness: The bigger tests (the $20$-question and $30$-question ones) score higher than $70\%$, and the $30$-question test — the heaviest — scores the highest at $90\%$. So the overall average should pull toward $90\%$, landing above the naive mean of $80\%$. Our answer $83.3\%$ sits between $80\%$ and $90\%$ and is closer to $80\%$, which makes sense because the $30$-question test is only half of the $60$-question total. Choice (D) $83$ fits; choice (E) $87$ would mean the $90\%$ test dominated more than it really does.
Alternative: Tool #5 (Look for a Pattern): the three tests have sizes in the ratio $1 : 2 : 3$, so each percent should be weighted by that ratio. Overall percent $= \dfrac{1 \cdot 70 + 2 \cdot 80 + 3 \cdot 90}{1 + 2 + 3} = \dfrac{70 + 160 + 270}{6} = \dfrac{500}{6} \approx 83.3\%$. Same answer (D), reached by weighted average instead of raw counts.
CCSS standards used (min grade 6)
6.RP.A.3Find a percent of a quantity as a rate per 100; solve problems involving finding the whole given a part and the percent (Reading "$70\%$ of $10$", "$80\%$ of $20$", "$90\%$ of $30$" as the counts $7$, $16$, $27$, and converting the final fraction $\dfrac{5}{6}$ back into the percent $83.\overline{3}\%$.)5.NBT.B.5Fluently multiply multi-digit whole numbers using the standard algorithm (Adding $7 + 16 + 27 = 50$ and $10 + 20 + 30 = 60$ — basic whole-number arithmetic once the percents are gone.)6.RP.A.1Understand the concept of a ratio and use ratio language to describe a ratio relationship (Writing the combined score as the ratio $\dfrac{50}{60}$ and simplifying it to $\dfrac{5}{6}$ before converting to a percent.)
⭐ When tests have different sizes, you can't just average the percents. Turn each percent into a count of correct answers, add them up, and divide by the total number of questions — then the right percent falls out.
⭐ When tests have different sizes, you can't just average the percents. Turn each percent into a count of correct answers, add them up, and divide by the total number of questions — then the right percent falls out.
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