2021 AMC 10B Problem 4

P1-10~2 minPrint

Linear Equations & Word Problems Basic Counting Set up the equation/formula and compute; no special trick needed

At a math contest, $57$ students are wearing blue shirts, and another $75$ students are wearing yellow shirts. The $132$ students are assigned into $66$ pairs. In exactly $23$ of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?

Show solution without answering

Key insight

The 23 blue–blue pairs use 46 blue students. The other 11 blue students each sit with a yellow, which leaves 64 yellows to pair with each other: 32 pairs.

Solution

There are only three kinds of pairs: blue–blue, blue–yellow, and yellow–yellow. Count the students, following one color at a time.

Blue students in blue–blue pairs. Each of the $23$ pairs has $2$ blue students, so that is $2 \times 23 = 46$ blue students.
Blue students left over. $57 - 46 = 11$ . They are not paired with another blue, so each one is paired with a yellow. That makes $11$ blue–yellow pairs.
Yellow students left over. Those $11$ mixed pairs use $11$ yellow students, leaving $75 - 11 = 64$ yellow students.
These $64$ must be paired with each other, which gives $64 \div 2 = 32$ yellow–yellow pairs.
Check the total: $23 + 11 + 32 = 66$ pairs. Correct.

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

Why this works

Once you know how many blue students are used up in blue–blue pairs, every other blue student must be in a mixed pair, and that fixes everything else. In pairing problems, count people rather than pairs whenever the problem gives you how many people there are of each kind.

Alternative approach

Let $y$ be the number of yellow–yellow pairs. There are $66 - 23 = 43$ pairs that are not blue–blue, so $43 - y$ of them are mixed. Count yellow students: each yellow–yellow pair has $2$ and each mixed pair has $1$ , so
$2y + (43 - y) = 75, \qquad y + 43 = 75, \qquad y = 32.$

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

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

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

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

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Algebra, Chapter 7: Proportion (§7.4)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques