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

At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is $\frac25$ . What fraction of 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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: 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 $\tfrac{2}{5}$. What fraction of all the people in the room are married men?

Givens: Everyone at the party is either a single woman, a married woman (wife), or her married-man husband; Every married man has exactly one wife also at the party, so $\#\text{married men} = \#\text{wives}$; $P(\text{a random woman is single}) = \dfrac{2}{5}$; Answer choices: (A) $\tfrac{1}{3}$, (B) $\tfrac{3}{8}$, (C) $\tfrac{2}{5}$, (D) $\tfrac{5}{12}$, (E) $\tfrac{3}{5}$

Unknowns: The fraction $\dfrac{\#\text{married men}}{\#\text{all people in the room}}$

Understand

Plan

Primary tool: #4 Introduce a Variable

Secondary: #9 Try a Simpler Case (WLOG)

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 $\tfrac{2}{5}$ gives $\tfrac{S}{S+W} = \tfrac{2}{5}$. 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 $\tfrac{2}{5}$ probability and just count.

Execute — Answer: B

#4 Introduce a Variable 6.EE.A.2 Step 1

Name the three groups.
Let $S$ = number of single women, $W$ = number of wives (married women), and $M$ = number of married men.
These three groups cover every person in the room.

$$\text{total people} = S + W + M$$

💡 Grade 6 expression-writing: give the unknown counts letters so we can connect them with equations.

#4 Introduce a Variable 6.EE.B.6 Step 2

Use the "married men come with their wives" clue.
Every married man at the party brought his wife, and every wife there is married to one of those men, so the men and the wives are in one-to-one pairs.

$$M = W$$

💡 A one-to-one pairing always gives an equal-count equation — the cleanest possible relation.

#4 Introduce a Variable 7.SP.C.5 Step 3

Turn the probability into an equation.
"A random woman is single" means we sample from all the women, so the denominator is $S + W$ and the numerator is $S$.
The probability equals $\tfrac{2}{5}$.

$$\dfrac{S}{S + W} = \dfrac{2}{5} \;\Rightarrow\; 5S = 2(S+W) \;\Rightarrow\; 3S = 2W$$

💡 Grade 7 probability of a simple event: favorable count over total count, then cross-multiply to clear the fraction.

#4 Introduce a Variable 7.EE.B.4 Step 4

Pick one letter to carry everything.
From step 2, $W = M$, so substitute $M$ for $W$ in $3S = 2W$ to get $3S = 2M$, i.e., $S = \tfrac{2}{3}M$.
Now $S$, $W$, and $M$ are all written in terms of $M$.

$$W = M, \quad S = \dfrac{2}{3}M$$

💡 With two equations linking three unknowns, expressing everything in one variable lets the unknown letter cancel in the final ratio.

#4 Introduce a Variable 7.RP.A.3 Step 5

Compute the fraction of people who are married men.
Substitute $S = \tfrac{2}{3}M$ and $W = M$ into $\tfrac{M}{S + W + M}$, then add the denominator.

$$\dfrac{M}{\tfrac{2}{3}M + M + M} = \dfrac{M}{\tfrac{8}{3}M} = \dfrac{3}{8} \;\Rightarrow\; \textbf{(B)}$$

💡 The $M$ cancels — proof that the answer never depended on the actual head count, only on the ratios.

[1] #4 6.EE.A.2 Name the three groups. Let $S$ = number of single women, $W$ = number of wives (

[2] #4 6.EE.B.6 Use the "married men come with their wives" clue. Every married man at the party

[3] #4 7.SP.C.5 Turn the probability into an equation. "A random woman is single" means we sampl

[4] #4 7.EE.B.4 Pick one letter to carry everything. From step 2, $W = M$, so substitute $M$ for

[5] #4 7.RP.A.3 Compute the fraction of people who are married men. Substitute $S = \tfrac{2}{3}

Review

Reasonableness: 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 $\tfrac{2}{5}$ probability. Total people $= 2 + 3 + 3 = 8$, and the married men are $3$ of them, so the fraction is $\tfrac{3}{8}$. 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 — $\tfrac{3}{8}$ (a bit less than half) fits.

Alternative: Tool #9 (Try a Simpler Case): the $\tfrac{2}{5}$ probability is the cleanest if we just decree $5$ women total. Then $2$ are single and $3$ are wives. Each wife brings a husband, so there are $3$ married men. Total $= 5 + 3 = 8$ people, of whom $3$ are married men: $\tfrac{3}{8}$. Same answer (B), reached without solving any equations.

CCSS standards used (min grade 7)

6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Naming the single women, wives, and married men as $S$, $W$, $M$ so the problem becomes algebra.)
6.EE.B.6 Use variables to represent numbers and write expressions when solving real-world problems (Translating the one-to-one pairing of married men with wives into the equation $M = W$.)
7.SP.C.5 Understand that the probability of a chance event is a number between 0 and 1 expressing the likelihood (Reading $P(\text{single}) = \tfrac{2}{5}$ as $\tfrac{S}{S+W} = \tfrac{2}{5}$ and turning it into the linear equation $3S = 2W$.)
7.EE.B.4 Use variables to represent quantities and construct simple equations to solve problems (Combining $M = W$ with $3S = 2W$ to express $S$ and $W$ in terms of $M$ alone.)
7.RP.A.3 Use proportional relationships to solve multistep ratio and percent problems (Computing the final ratio $\tfrac{M}{S+W+M} = \tfrac{3}{8}$ after substitution and cancellation.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 7)

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