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2021 AMC 10B Problem 4

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2021 AMC 10B Problem 4AMC 10 Step by Step
Name:                             Date: September 13, 2026
Linear Equations & Word Problems · difficulty 1 of 5 · about 2 min · position P1-10

At a math contest, 5757 students are wearing blue shirts, and another 7575 students are wearing yellow shirts. The 132132 students are assigned into 6666 pairs. In exactly 2323 of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?

  1. A)

    2323

  2. B)

    3232

  3. C)

    3737

  4. D)

    4141

  5. E)

    6464

Answer

B

Key insight

The 23 blue–blue pairs use 46 blue students, so the other 11 blues each sit with a yellow, leaving 64 yellows in 32 pairs.

Solution

Follow the blue shirts. The 2323 all-blue pairs contain 2⋅23=462 \cdot 23 = 46 blue students, leaving 57−46=1157 - 46 = 11 blue students. Each of these is paired with a yellow-shirted student, so there are 1111 mixed pairs.

Those mixed pairs use 1111 yellow students, leaving 75−11=6475 - 11 = 64 yellow students who must be paired with each other. That makes 64/2=3264 / 2 = 32 all-yellow pairs.

Check: 23+11+32=6623 + 11 + 32 = 66 pairs.

The answer is (B) 32\boxed{\textbf{(B)}\ 32}.

Why this works

Pairs come in three kinds (blue–blue, blue–yellow, yellow–yellow), and counting one color pins down everything: the number of mixed pairs equals the number of blue students not in blue–blue pairs. Count students, not pairs, whenever a pairing problem gives student totals.

Alternative approach

Let yy be the number of yellow–yellow pairs; then there are 66−23−y66 - 23 - y mixed pairs. Yellow students: 2y+(43−y)=752y + (43 - y) = 75, so y=32y = 32.

The trap

Splitting the remaining 66 − 23 = 43 pairs somehow, instead of tracking that each leftover blue student consumes one yellow partner.

Common mistakes

  • Splitting the remaining 66 − 23 = 43 pairs somehow, instead of tracking that each leftover blue student consumes one yellow partner.
  • Forgetting that a blue–blue pair uses two blue students and computing 57−23=3457 - 23 = 34 leftover blues.

Techniques

Set up the equation/formula and compute; no special trick needed