AMC 10 · 2021 · #4
Easy mode Grade 4Problem
Picture a room of 132 students at a math contest. 57 of them wear blue shirts. The other 75 wear yellow shirts.
The students all stand together in 66 pairs. So every student has one partner.
Look at the pairs. Exactly 23 pairs are made of two blue shirts together.
How many pairs are made of two yellow shirts together?
Pick an answer.
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.
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
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.
💡 $2$ students per pair $\times$ number of pairs — Grade 3 multiplication within $100$.
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.
💡 Take away the ones already used — Grade 2 subtraction word problem.
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.
💡 $1$ yellow per mixed pair — Grade 3 multiplication/division word problem.
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.
💡 Subtract the mixed users from the yellow total — Grade 4 multi-digit subtraction.
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).
💡 $2$ students $\to$ $1$ pair, so divide by $2$ — Grade 3 division within $100$.
3.OA.C.7 Count blue students inside blue-blue pairs. Each pair has $2$ students, so $23$ 2.OA.A.1 Find blue students in mixed pairs. The remaining blue students must be paired wi 3.OA.A.3 Each mixed pair has exactly one blue and one yellow student. So $11$ mixed pairs 4.NBT.B.4 Find yellow students left for yellow-yellow pairs. Of $75$ yellow students, $11$ 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.1Solve 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.3Solve multiplication and division word problems within 100 (Matching $11$ mixed pairs to $11$ yellow students in those pairs.)3.OA.C.7Fluently multiply and divide within 100 ($23 \times 2 = 46$ and $64 \div 2 = 32$.)4.NBT.B.4Fluently 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!
⭐ 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!
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