Competition · AMC preparation · step 4 of 4
AMC 8 · 2010 · #9
Grade 6 rate-ratioPick an answer.
AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 (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.
Count Test 1 correct
Subproblem 1: take 80% of the 25 problems to get 20 correct on Test 1.
Finding a percent of a whole-number quantity is Grade 6 percent reasoning: 80% of 25 means 80/100 × 25.
6.RP.A.3Identify SubproblemsCount Test 2 correct
Subproblem 2: take 90% of the 40 problems to get 36 correct on Test 2.
Multiplying a decimal to hundredths by a whole number is exactly the Grade 5 decimal-arithmetic standard.
5.NBT.B.7Identify SubproblemsCount Test 3 correct
Subproblem 3: take 70% of the 10 problems to get 7 correct on Test 3.
Same decimal multiplication as the previous step — each test contributes its own whole-number count.
5.NBT.B.7Identify SubproblemsAdd correct and total
All three are now the same unit, so add them: 63 correct out of 75 total problems.
Adding counts is only legal because we converted percents to the common unit first — Tool #8's unit-check move.
Ryan's three test results combine into 63 correct answers out of a single 75-problem pool, and it is these pooled totals — not the average of 80%, 90%, and 70% — that the overall score is built from.
▸ Why?
The correct answers from the three tests are three separate, non-overlapping groups of right answers, so adding 20, 36, and 7 gives the whole count of right answers, 63.
▸ Why?
Split into pieces that neither overlap nor leave gaps, the parts add back to the whole they came from.
▸ Why?
The three problem counts are likewise non-overlapping parts of the single 75-problem pool, so adding 25, 40, and 10 gives the whole, 75.
▸ Why?
The pool is exactly the three tests laid side by side with nothing shared and nothing left out, so their sizes sum to the pool's size.
▸ Why?
Each test's correct count must first be found from its percent, because a percent by itself is not yet a count of problems: 80% of 25 is 20, 90% of 40 is 36, 70% of 10 is 7.
▸ Why?
A percent is just a count per hundred: 80% means 80 out of every hundred, so 80% of 25 is 80/100 of 25, which lands on the whole number 20.
▸ Why?
Averaging 80%, 90%, and 70% would treat the three tests as equally important, but the 40-problem test pours four times as many problems into the pool as the 10-problem test, so it must weigh more.
▸ Why?
Every problem counts once toward the whole pool, so a test with more problems is a larger part of that whole and pulls the overall fraction toward its own score.
Turn the ratio into a percent
Turn the combined count back into a percent: 63 out of 75 is 84%.
Expressing a part out of a whole as a percent is Grade 6 ratio reasoning — the inverse of step 1.
6.RP.A.3Analyze The UnitsWhen 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.
- Count Test 1 correct
- Count Test 2 correct
- Count Test 3 correct
- Add correct and total
- Turn the ratio into a percent
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