AMC 10 · 2021 · #4

Grade 4 arithmetic

Archetype Overlap / Complement Counting └ Sub-archetype Two-Set Inclusion-Exclusion: Solve for Missing Region

pair-countingparitymulti-digit-arithmeticset-partition identify-subproblemscomplementary-counting ↑ Prerequisites: multi-digit-arithmetic

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Problem

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?

Pick an answer.

(A)

~23

(B)

~32

(C)

~37

(D)

~41

(E)

~64

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: $57$ students wear blue and $75$ wear yellow. All $132$ are split into $66$ pairs. Exactly $23$ pairs are blue-blue. How many pairs are yellow-yellow?

Givens: Blue students: $57$; Yellow students: $75$; Total students: $57 + 75 = 132$; Total pairs: $66$; Blue-blue pairs: $23$; Answer choices: (A) $23$, (B) $32$, (C) $37$, (D) $41$, (E) $64$

Unknowns: Number of yellow-yellow pairs

Understand

Restated: $57$ students wear blue and $75$ wear yellow. All $132$ are split into $66$ pairs. Exactly $23$ pairs are blue-blue. How many pairs are yellow-yellow?

Givens: Blue students: $57$; Yellow students: $75$; Total students: $57 + 75 = 132$; Total pairs: $66$; Blue-blue pairs: $23$; Answer choices: (A) $23$, (B) $32$, (C) $37$, (D) $41$, (E) $64$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities

Tool #7 (Subproblems) chops the count by pair type: blue-blue $\to$ blue students used; mixed $\to$ blue students left over $=$ yellow students used; yellow-yellow $\to$ 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 $= 23 + \text{(mixed)} + \text{(yellow-yellow)} = 66$, so mixed $+$ yellow-yellow $= 43$, ruling out (A) $23$ at a glance.

Execute — Answer: B

#7 Identify Subproblems 3.OA.C.7 Step 1

Count blue students inside blue-blue pairs.
Each pair has $2$ students, so $23$ pairs hold $23 \times 2 = 46$ blue students.

$$23 \times 2 = 46 \text{ blue students in blue-blue pairs}$$

💡 $2$ students per pair $\times$ number of pairs — Grade 3 multiplication within $100$.

#7 Identify Subproblems 2.OA.A.1 Step 2

Find blue students in mixed pairs.
The remaining blue students must be paired with yellow students.
So mixed pairs have $57 - 46 = 11$ blue students.

$$57 - 46 = 11 \text{ blue students in mixed pairs}$$

💡 Take away the ones already used — Grade 2 subtraction word problem.

#1 Draw a Diagram 3.OA.A.3 Step 3

Each mixed pair has exactly one blue and one yellow student.
So $11$ mixed pairs use $11$ yellow students.

$$11 \text{ mixed pairs} \;\Rightarrow\; 11 \text{ yellow students in mixed pairs}$$

💡 $1$ yellow per mixed pair — Grade 3 multiplication/division word problem.

#7 Identify Subproblems 4.NBT.B.4 Step 4

Find yellow students left for yellow-yellow pairs.
Of $75$ yellow students, $11$ are in mixed pairs, leaving $75 - 11 = 64$ to pair only with other yellow students.

$$75 - 11 = 64 \text{ yellow students in yellow-yellow pairs}$$

💡 Subtract the mixed users from the yellow total — Grade 4 multi-digit subtraction.

#7 Identify Subproblems 3.OA.C.7 Step 5

Divide the remaining yellow students into pairs of $2$.
$64 \div 2 = 32$ yellow-yellow pairs.
That is choice (B).

$$64 \div 2 = 32 \;\Rightarrow\; \textbf{(B)}$$

💡 $2$ students $\to$ $1$ pair, so divide by $2$ — Grade 3 division within $100$.

[1] #7 3.OA.C.7 Count blue students inside blue-blue pairs. Each pair has $2$ students, so $23$

[2] #7 2.OA.A.1 Find blue students in mixed pairs. The remaining blue students must be paired wi

[3] #1 3.OA.A.3 Each mixed pair has exactly one blue and one yellow student. So $11$ mixed pairs

[4] #7 4.NBT.B.4 Find yellow students left for yellow-yellow pairs. Of $75$ yellow students, $11$

[5] #7 3.OA.C.7 Divide the remaining yellow students into pairs of $2$. $64 \div 2 = 32$ yellow-

Review

Reasonableness: 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.

Alternative: Tool #16 (Change Focus / Complement) — instead of tracking yellow, ask "how many pairs are NOT yellow-yellow?" That's $23$ (blue-blue) plus $11$ (mixed) $= 34$, so yellow-yellow $= 66 - 34 = 32$. Same answer (B).

CCSS standards used (min grade 4)

2.OA.A.1 Solve one- and two-step word problems using addition and subtraction within 100 (Computing $57 - 46 = 11$ blue students left over for mixed pairs.)
3.OA.A.3 Solve multiplication and division word problems within 100 (Matching $11$ mixed pairs to $11$ yellow students in those pairs.)
3.OA.C.7 Fluently multiply and divide within 100 ($23 \times 2 = 46$ and $64 \div 2 = 32$.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers ($75 - 11 = 64$ to find yellow students in yellow-yellow pairs.)

⭐ This 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 4)

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