AMC 10 · 2004 · #13
Grade 4 arithmeticPick an answer.
AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Count the dance-pairs from the men
Each of the 12 men danced with 3 women, so counting from the men gives 36 dance-pairs in all.
Twelve equal groups of 3 pairs each is just 12 threes added up.
3.OA.A.1Organize Information In More WaysCount the same pairs from the women
Let W be the number of women; each danced with exactly 2 men, so counting those same pairs by woman gives 2 × W, equal to 36.
Counting one fixed pile of things two different ways has to give the same total.
Counting one fixed pile of things two different ways has to give the same total.
▸ Why?
Each dance pair is counted once from each side, so the two tallies count the same objects.
▸ Why?
Each side's tally is a fixed number per person repeated once per person, which multiplying records.
Solve for the number of women
Split the 36 pairs into groups of 2, one per woman: 36 ÷ 2 gives 18 women, choice (D).
If 36 pairs come 2 to a woman, the number of women is how many 2s fit in 36.
3.OA.C.7Introduce A VariableCount the same connections from both sides and set the two counts equal — that equation hands you the missing number.
- Count the dance-pairs from the men
- Count the same pairs from the women
- Solve for the number of women