Competition · AMC preparation · step 4 of 4

AMC 8 · 2011 · #14

Grade 6 rate-ratio

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

ratio-proportionfraction-arithmetic ratio-proportionidentify-subproblems ↑ Prerequisites: ratio-proportionfraction-arithmetic

📏 Medium solution 💡 3 insights

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Problem

Colfax Middle School has 270 students with a boy-to-girl ratio of 5 : 4. Winthrop Middle School has 180 students with a boy-to-girl ratio of 4 : 5. All students from both schools attend a joint dance. What fraction of the dancers are girls?

Pick an answer.

(A)

$dfrac7{18}$

(B)

$dfrac7{15}$

(C)

$\dfrac{22}{45}$

(D)

dfrac12

(E)

$\dfrac{23}{45}$

AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The whole question is one fraction — girls over total — but each school contributes to both pieces with its own ratio. Tool #7 (Identify Subproblems) splits the work into two clean subproblems: "how many girls at Colfax?" and "how many girls at Winthrop?" Once those numbers are in hand, the final fraction is just (sum of girls) / (sum of students). Tool #8 (Analyze the Units) is the bookkeeping move that keeps the part-to-whole structure straight: ratio 5 : 4 means girls are 4/9 of the school, ratio 4 : 5 means girls are 5/9 of the school — a Grade 6 "part of a whole" use of ratios.

1STEP 1

Count Colfax girls

Colfax's 5 : 4 splits into 9 equal parts, so girls are $\frac{4}{9}$ of 270 = 120.

Colfax girls = 4/9 × 270 = 4 × 30 = 120

2STEP 2

Count Winthrop girls

Winthrop flips to 4 : 5, so now girls are $\frac{5}{9}$ of 180 = 100.

Winthrop girls = 5/9 × 180 = 5 × 20 = 100

3STEP 3

Add both totals

Add both schools: 450 students total and 220 girls total.

total students = 270 + 180 = 450, total girls = 120 + 100 = 220

4STEP 4

Reduce the fraction

Form girls/students = $\frac{220}{450}$ and reduce by 10 to $\frac{22}{45}$ .

220/450 = (220 ÷ 10)/(450 ÷ 10) = 22/45 → (C)

Answer

22/45

Sanity check the size. Colfax has more boys than girls (

\frac{5}{9}

\frac{4}{9}

) and Winthrop has more girls than boys (

\frac{5}{9}

\frac{4}{9}

), but Colfax is the bigger school (270 ＞ 180), so boys should slightly outnumber girls overall. The girls' fraction

\frac{22}{45}

≈ 0.489 is just under

\frac{1}{2}

, which matches that intuition perfectly. Choices

\frac{7}{18}

≈ 0.389 and

\frac{7}{15}

≈ 0.467 are too low,

\frac{1}{2}

would require equal boys and girls, and

\frac{23}{45}

≈ 0.511 would flip the imbalance the wrong way.

💡Key takeaway

This AMC 8 problem only needs Grade 6 ratio reasoning — turn each ratio into a fraction of the whole school — that you already know!

Count Colfax girls
Count Winthrop girls
Add both totals
Reduce the fraction

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Count Colfax girls

Count Winthrop girls

Add both totals

Reduce the fraction

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