AMC 10 · 2021 · #4
Grade 4 arithmeticPick an answer.
AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #7 (Subproblems) chops the count by pair type: blue-blue → blue students used; mixed → blue students left over = yellow students used; yellow-yellow → yellow students left, halved. Tool #1 (Diagram) helps draw three labelled bins (BB, BY, YY) and watch each student get placed. Tool #3 (Eliminate) gives a quick parity safety net: total pairs = 23 + (mixed) + (yellow-yellow) = 66, so mixed + yellow-yellow = 43, ruling out (A) 23 at a glance.
23 blue-blue pairs × 2 students each lock up 46 blue students.
2 students per pair × number of pairs — Grade 3 multiplication within 100.
3.OA.C.7Identify SubproblemsLeftover blue students: 57 - 46 = 11, each forced to pair with a yellow.
Take away the ones already used — Grade 2 subtraction word problem.
2.OA.A.1Identify SubproblemsOne blue + one yellow per mixed pair, so 11 pairs use 11 yellow students.
1 yellow per mixed pair — Grade 3 multiplication/division word problem.
3.OA.A.3Draw A DiagramYellow left for yellow-yellow pairs: 75 - 11 = 64 students.
Subtract the mixed users from the yellow total — Grade 4 multi-digit subtraction.
4.NBT.B.4Identify SubproblemsHalve them: 64 ÷ 2 = 32 yellow-yellow pairs — choice (B).
2 students → 1 pair, so divide by 2 — Grade 3 division within 100.
3.OA.C.7Identify SubproblemsThis AMC 10 problem only needs Grade 4 multi-digit subtraction — track the 46 blue in BB pairs, the 11 leftover blue in mixed pairs, and the 64 leftover yellow making 32 YY pairs!