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.
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 × 25.
6.RP.A.3Identify SubproblemsSubproblem 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 SubproblemsSubproblem 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 SubproblemsAll 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.
4.NBT.B.4Analyze The UnitsTurn 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.