AMC 10 · 2004 · #13
Easy mode Grade 4At a party, every man danced with exactly 3 women, and every woman danced with exactly 2 men. There were 12 men at the party. How many women were at the party?
Pick an answer.
AMC 10 2004 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: At a party, every man danced with exactly 3 women and every woman danced with exactly 2 men. There were 12 men. Find how many women were at the party.
Givens: There were 12 men.; Each man danced with exactly 3 women.; Each woman danced with exactly 2 men.
Unknowns: The number of women at the party.
Understand
Restated: At a party, every man danced with exactly 3 women and every woman danced with exactly 2 men. There were 12 men. Find how many women were at the party.
Givens: There were 12 men.; Each man danced with exactly 3 women.; Each woman danced with exactly 2 men.
Plan
Primary tool: #15 Organize Information in More Ways
Secondary: #4 Introduce a Variable
The key object to count is a man-woman dance-pair. That same collection of pairs can be counted two ways: once by going through the men and once by going through the women. Both counts describe the exact same set, so they must be equal. Counting from the men's side gives a number right away; counting from the women's side gives an expression with the unknown, and setting the two equal pins down the number of women.
Execute — Answer: D
3.OA.A.1 Step 1 Count the dance-pairs from the men
- Each man danced with exactly 3 women, so each man accounts for 3 man-woman dance-pairs.
- With 12 men, multiply to get the total number of dance-pairs at the party.
💡 Twelve equal groups of 3 pairs each is just 12 threes added up.
4.OA.A.3 Step 2 Count the same pairs from the women
- Let W be the number of women.
- Each woman danced with exactly 2 men, so each woman accounts for 2 dance-pairs, giving 2 times W pairs in total.
- This counts the exact same dances as before, just grouped by woman instead of by man, so it must equal 36.
💡 Counting one fixed pile of things two different ways has to give the same total.
3.OA.C.7 Step 3 Solve for the number of women
- The 36 dance-pairs are split into groups of 2, one group per woman.
- Divide 36 by 2 to find how many women there are.
- This gives 18, which is choice (D).
💡 If 36 pairs come 2 to a woman, the number of women is how many 2s fit in 36.
3.OA.A.1 Each man danced with exactly 3 women, so each man accounts for 3 man-woman dance 4.OA.A.3 Let W be the number of women. Each woman danced with exactly 2 men, so each woma 3.OA.C.7 The 36 dance-pairs are split into groups of 2, one group per woman. Divide 36 by Review
Reasonableness: Check the two counts match: 18 women times 2 men each is 36 dances, and 12 men times 3 women each is also 36. Both sides agree, so 18 is consistent. It also makes sense that there are more women than men, since each man reached 3 partners while each woman only reached 2.
Alternative: Use the ratio directly. For every 2 men there are 3 women in this dance pattern, so women outnumber men by a factor of 3 to 2. Multiplying 12 men by 3/2 gives 18 women, matching the answer.
CCSS standards used (min grade 4)
3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Multiplying 12 men by 3 partners each to count the total dance-pairs.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Recognizing that the same dance-pairs counted from the women's side must equal 36, giving 2 times W = 36.)3.OA.C.7Fluently multiply and divide within 100 (Dividing 36 dance-pairs by 2 per woman to find the number of women.)
⭐ Count the same connections from both sides and set the two counts equal — that equation hands you the missing number.
⭐ Count the same connections from both sides and set the two counts equal — that equation hands you the missing number.
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