AMC 10 · 2004 · #13

Grade 4 arithmetic

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Direct Ratio Scale-Up

double-countingratio-proportion convert-to-algebra ↑ Prerequisites: multi-digit-arithmetic

📏 Short solution 💡 1 insight

Problem

At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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.

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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

#15 Organize Information in More Ways 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.

$$12 \times 3 = 36 \text{ dance-pairs}$$

💡 Twelve equal groups of 3 pairs each is just 12 threes added up.

#4 Introduce a Variable 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.

$$2 \times W = 36$$

💡 Counting one fixed pile of things two different ways has to give the same total.

#4 Introduce a Variable 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).

$$W = 36 \div 2 = 18$$

💡 If 36 pairs come 2 to a woman, the number of women is how many 2s fit in 36.

[1] #15 3.OA.A.1 Each man danced with exactly 3 women, so each man accounts for 3 man-woman dance

[2] #4 4.OA.A.3 Let W be the number of women. Each woman danced with exactly 2 men, so each woma

[3] #4 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.1 Interpret 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.3 Solve 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.7 Fluently 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.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Count the dance-pairs from the men

Count the same pairs from the women

Solve for the number of women

Review

CCSS standards used (min grade 4)

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