AMC 8 · 2004 · #22

Grade 7 probability

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Percent / Fraction Recovery

ratio-proportionprobability-basicfraction-arithmeticlinear-equations-one-var convert-to-algebraidentify-subproblems ↑ Prerequisites: ratio-proportionfraction-arithmetic

📏 Medium solution 💡 3 insights

Problem

A party has only single women plus married couples (each married man comes with his wife). If you pick a woman at random, the chance she is single is

\frac{2}{5}

. What fraction of all the people in the room are married men?

Pick an answer.

(A)

$\frac13$

(B)

$\frac38$

(C)

$\frac25$

(D)

$\frac{5}{12}$

(E)

$\frac35$

AMC 8 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

The problem hides every actual count and only gives a ratio, so Tool #4 (Introduce a Variable) is the right opening move: name the three groups and turn the words into equations. Two relations drop out immediately — "each married man has a wife here" gives M = W, and the probability $\frac{2}{5}$ gives $\frac{S}{S+W}$ = $\frac{2}{5}$ . With two equations and three unknowns we can express everything in one letter and read off the answer fraction. Tool #9 (Try a Simpler Case) is the natural backup: pick concrete head counts that match the $\frac{2}{5}$ probability and just count.

1STEP 1

Name the three groups: S single women, W wives, M married men — together they are everyone in the room.

total people = S + W + M

2STEP 2

Each married man brought his wife and each wife has a husband here, so they pair one-to-one: M = W.

M = W

3STEP 3

A random woman is single with probability $\frac{S}{S+W}$ = $\frac{2}{5}$ ; cross-multiply to get 3S = 2W.

\frac{S}{S + W}

\frac{2}{5}

→ 5S = 2(S+W) → 3S = 2W

4STEP 4

Put all in terms of M: from M = W and 3S = 2W, substitute to get S = $\frac{2}{3}$ M (and W = M).

W = M, S =

\frac{2}{3}

5STEP 5

Married-men fraction $\frac{M}{S+W+M}$ = $\frac{M}{\frac{8}{3}M}$ = $\frac{3}{8}$ ; the M cancels, so choice (B).

\frac{M}{\frac{2}{3}M + M + M}

\frac{M}{\frac{8}{3}M}

\frac{3}{8}

→ (B)

Answer

\frac{3}{8}

Plug in concrete numbers. Take M = 3 married men, so there are W = 3 wives and S = 2 single women — that gives 2 singles out of 5 women, matching the

\frac{2}{5}

probability. Total people = 2 + 3 + 3 = 8, and the married men are 3 of them, so the fraction is

\frac{3}{8}

. Matches (B). A quick sanity gut-check: roughly half the women are married, and married women come with husbands, so married men should be a noticeable chunk of the room —

\frac{3}{8}

(a bit less than half) fits.

💡Key takeaway

Married men match wives one-for-one, so the count of men equals the count of wives. Add that to the $\frac{2}{5}$ probability and the answer $\frac{3}{8}$ falls out — actual head counts never mattered.

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