AMC 8 · 2019 · #19
Grade 4 logiccountingPick an answer.
AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The cleanest move is to split the 10 games each top team plays into two groups: (i) games against the three bottom teams (where the top team is free to win every time) and (ii) games against the other two top teams (where every point earned by one top team is a point denied to another). That is exactly Tool #7 (Identify Subproblems). After we compute the best score using this split, Tool #3 (Eliminate Possibilities) lets us check the leftover answer choices — especially 26 and 30 — to confirm 24 is the maximum that is actually achievable.
Call the top three A, B, C, the rest D, E, F. Each plays 10 games: 6 outside its group, 4 inside — so handle the two kinds separately.
Breaking the 10 games into two cleaner groups is Tool #7 — each piece is now a small, easy counting problem.
4.OA.A.3Identify SubproblemsSubproblem 1: each leader beats D, E, F in both meetings — 6 wins — for 18 points from outside games, with no clash between leaders.
Multiplying 6 × 3 to count points from 6 wins is a Grade 3 multiplication word-problem move.
3.OA.A.3Identify SubproblemsSubproblem 2: the 6 inside games pay 3 points only when decisive, so 18 at most, split three ways to 6 points each — if reachable.
Comparing decisive games (3 pts handed out) versus draws (2 pts) shows draws waste points, so for the maximum we look for decisive games only.
4.OA.A.3Identify SubproblemsMake it reachable: in each pair (A–B, A–C, B–C) split the two games one win apiece, so every leader goes 2–2 inside for 6 points.
Solving a two-step word problem (count wins, then turn wins into points) is a Grade 3 four-operations skill.
3.OA.D.8Identify SubproblemsAdd the two parts: 24 points per leader — achievable and equal, so the answer is (C).
Adding two two-digit numbers (18+6) is the Grade 2 fluency step that closes the argument.
2.NBT.B.5Identify SubproblemsBigger choices fail: 30 needs all wins (impossible among leaders); 26 needs 24 inside points but only 18 exist — so 24 is the max.
Multiple-choice elimination using a clean upper bound is Tool #3 — it pins the answer at 24.
4.OA.A.3Eliminate PossibilitiesThis AMC 8 problem only needs Grade 4 multi-step word-problem reasoning you already know — split the games into two groups, multiply, and add!