Competition · AMC preparation · step 4 of 4
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?
Find the total points needed
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
Subtract 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 SubproblemsPush the fourth score to 100
Change 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.
To make the third test score as low as it can be, the fourth test score has to be pushed as high as the scoring rules allow.
▸ Why?
The third and fourth scores must together reach a fixed floor of 192 points, so they trade off against each other: every point the fourth score supplies is a point the third score no longer has to, and the third can only sink to its lowest when the fourth is at its largest.
▸ Why?
Those 192 needed points split with no gap and no overlap into the fourth score's share and the third score's share, so the two shares always add back to the same fixed 192 — lift one share and the other must drop by exactly as much.
Solve for the third score
With 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!
- Find the total points needed
- Subtract the two known scores
- Push the fourth score to 100
- Solve for the third score
A parent dashboard for the family lives at sensimlab.com.