AMC 8 · 2006 · #12
Grade 6 rate-ratioPick an answer.
AMC 8 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Convert each percent to a count of correct answers by multiplying it (as a decimal) by that test's number of questions.
"Percent of a quantity as a rate per 100" is the Grade 6 way to read "70% of 10" as 7.
6.RP.A.3Solve An Easier Related ProblemAdd the three correct counts to get 50 right out of the 60 total questions.
Once the percents are gone, this is just whole-number addition — the Grade 5 fluency standard.
5.NBT.B.5Make A Systematic ListWrite the score as total correct over total questions and simplify: = .
and are the same ratio written with different sized parts — the Grade 6 ratio idea.
6.RP.A.1Make A Systematic ListDivide 5 by 6 to get about 83.3%, then pick the closest choice.
83.33% rounds to 83%, which is choice (D). Choice (C) 80% is the trap from naively averaging 70, 80, 90.
6.RP.A.3Solve An Easier Related ProblemWhen 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.