AMC 8 · 2012 · #7
Grade 6 arithmeticPick an answer.
AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The goal (average = 95) is the end state, and we want to reverse-engineer the third score from it — that is Tool #11 (Work Backwards). Turn the average into a required total (380), peel away the two known scores, and we get the required sum T₃ + T₄. Then comes the trick: "smallest T₃" is hard to chase directly, but its mirror "largest T₄" is easy — that focus flip is Tool #16 (Change Focus). Tool #7 (Identify Subproblems) keeps the work clean: (a) what total do four tests need? (b) what's left after tests 1 and 2? (c) how small can T₃ be while T₄ stays legal?
Work backwards: an average of 95 on four tests means the four scores must total at least 95 × 4 = 380.
An average is just "total divided by count", so multiplying back gives the total — a Grade 6 statistics move.
6.SP.B.5Work BackwardsSubtract the two known scores — tests three and four must together supply 380 - 188 = 192 points.
Splitting four scores into "known" and "unknown" pieces is the subproblems move; the leftover subtraction is Grade 4 multi-digit arithmetic.
4.NBT.B.4Identify SubproblemsChange focus: to shrink the third score, push the fourth as high as the rules allow — the maximum test score is 100.
Instead of minimizing T₃ directly, minimize it by maximizing the other score — flipping the focus is Tool #16. The score-cap T₄ ≤ 100 is a simple inequality constraint.
6.EE.B.8Count The ComplementWith the fourth test at 100, the third must satisfy T₃ + 100 ≥ 192, so T₃ ≥ 92 — answer (B).
Subtracting 100 from both sides finishes the "work backwards" chain and lands on the minimum allowed third-test score.
6.EE.B.8Work BackwardsThis AMC 8 problem just needs Grade 6 "average = total ÷ count" and a simple inequality — push the other score to its max to find the smallest possible one!