Problem 22 · 2004 AMC 8
Medium
Fractions, Decimals & Percents
set-up-variables
At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is 2/5. What fraction of the people in the room are married men?
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Answer: B — 3/8.
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Hint 1 of 2
The fraction 2/5 is begging you to pick a friendly number of women — choose 5, so 2 are single and 3 are married, with no ugly fractions. The hidden link: every married woman comes with exactly one husband, so married men = married women.
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Hint 2 of 2
The technique is plug in a convenient total (pick the denominator) plus spotting the pairing: married men and married women come in couples, so their counts are equal. Then it's just a head-count.
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Approach: pick 5 women, count heads
- Let there be 5 women (matches the /5). Then 2/5 × 5 = 2 are single and the other 3 are married.
- Key link: married couples pair up, so the 3 married women bring 3 married men — and the room has only single women and married couples, no one else.
- Total people = 5 women + 3 men = 8. Married men out of everyone = 3/8.
- Sanity check on the trap: 'single women = 2/5' is the fraction of women, not of people; the married-men fraction (3/8) uses the bigger people-total, so it should be smaller than a naive 3/5 — and it is. Picking a concrete total turns abstract-fraction problems into plain counting.
Another way — algebra with a variable:
- Let women = w. Single women = (2/5)w, married women = (3/5)w = married men.
- Total people = w + (3/5)w = (8/5)w.
- Married men / people = (3/5)w ÷ (8/5)w = 3/8 — the w cancels, confirming the choice of 5 didn't matter.
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