AMC 8 · 2003 · #7
Grade 6 arithmeticPick an answer.
AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The question only asks for the difference of two averages, not either average by itself, which is why Blake's 78 never has to be used. Tool #7 (Identify Subproblems) splits the work into two clean pieces: (1) add the four per-test score differences to get the total difference, then (2) divide that total by the number of tests to get the difference of the averages. Tool #11 (Work Backwards) here just means reading the mean formula in reverse — average = total ÷ count, so difference of averages = difference of totals ÷ count.
Subproblem 1: read the four per-test gaps (Jenny minus Blake) straight off the problem: +10, -10, +20, +20.
The Grade 6 "signed quantities for opposite directions" move: + means Jenny scored higher, - means lower.
6.NS.C.5Identify SubproblemsAdd the four signed gaps: +10 and -10 cancel, leaving 20 + 20 = 40 as the total point gap.
Adding integers with opposite signs is Grade 6 — +10 and -10 cancel to 0, then 20 + 20 = 40.
6.NS.C.6Identify SubproblemsSubproblem 2: same test count, so the average gap is the total gap over 4 — Jenny's average minus Blake's is .
Grade 6 mean formula run in reverse — if you take the same number of tests, the gap between averages is just the gap between totals shrunk by that count.
6.SP.B.5Work BackwardsDivide to finish: = 10, choice (A).
Notice Blake's 78 was never used: the difference of averages only depends on the per-test gaps.
6.SP.B.5Identify SubproblemsWhen a question asks for the difference of two averages over the same number of tests, you don't need either average. Add the per-test gaps and divide by the count — here, 40 ÷ 4 = 10.