AMC 10 · 2021 · #2

Grade 4 counting
pair-countingparitymulti-digit-arithmeticset-partition identify-subproblemscomplementary-counting ↑ Prerequisites: multi-digit-arithmetic
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Problem
Fifty-seven students wear blue and seventy-five wear yellow. All one hundred thirty-two are split into sixty-six pairs. Exactly twenty-three pairs are blue with blue. Find how many pairs are yellow with yellow.

Pick an answer.

(A)
23
(B)
32
(C)
37
(D)
41
(E)
64
How to solve
Strategy Identify Subproblems

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.

1STEP 1

Count blue inside blue pairs

Count the students inside the blue pairs.

23 × 2 = 46 blue students in blue-blue pairs
2STEP 2

Find the leftover blue students

Every leftover blue is in a mixed pair.

57 - 46 = 11 blue students in mixed pairs
3STEP 3

Count yellow in mixed pairs

Each mixed pair holds one yellow.

11 mixed pairs → 11 yellow students in mixed pairs
4STEP 4

Count the leftover yellow

The rest are all in yellow pairs.

75 - 11 = 64 yellow students in yellow-yellow pairs
5STEP 5

Divide by two

Halving gives 32 pairs.

64 ÷ 2 = 32 → (B)
Answer
32
Total pair tally: 23 blue-blue + 11 mixed + 32 yellow-yellow = 66 ✓. Total student tally: 46 + 11 + 11 + 64 = 132 ✓. Both totals match the givens, so the count is consistent.
💡Key takeaway

This AMC 12 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!