AMC 8 · 2011 · #14
Easy mode Grade 6Problem
Colfax Middle School has 270 students. At Colfax, the ratio of boys to girls is 5:4.
Winthrop Middle School has 180 students. At Winthrop, the ratio of boys to girls is 4:5.
The two schools hold one dance together, and every student from both schools shows up.
Out of all the students at the dance, what fraction are girls?
Pick an answer.
AMC 8 2011 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.
Toolkit + CCSS Solution
Understand
Restated: 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?
Givens: Colfax: $270$ students, boys $:$ girls $= 5 : 4$; Winthrop: $180$ students, boys $:$ girls $= 4 : 5$; Every student from both schools attends the dance; Answer choices: (A) $\tfrac{7}{18}$, (B) $\tfrac{7}{15}$, (C) $\tfrac{22}{45}$, (D) $\tfrac{1}{2}$, (E) $\tfrac{23}{45}$
Unknowns: The fraction $\dfrac{\text{total girls at the dance}}{\text{total students at the dance}}$ in lowest terms
Understand
Restated: 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?
Givens: Colfax: $270$ students, boys $:$ girls $= 5 : 4$; Winthrop: $180$ students, boys $:$ girls $= 4 : 5$; Every student from both schools attends the dance; Answer choices: (A) $\tfrac{7}{18}$, (B) $\tfrac{7}{15}$, (C) $\tfrac{22}{45}$, (D) $\tfrac{1}{2}$, (E) $\tfrac{23}{45}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #8 Analyze the Units
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 $\tfrac{4}{9}$ of the school, ratio $4 : 5$ means girls are $\tfrac{5}{9}$ of the school — a Grade 6 "part of a whole" use of ratios.
Execute — Answer: C
6.RP.A.3 Step 1 - Find the number of girls at Colfax.
- The ratio $5 : 4$ splits the school into $5 + 4 = 9$ equal parts, and girls are $4$ of those parts, so girls make up $\tfrac{4}{9}$ of the $270$ students.
💡 Turning a ratio $5 : 4$ into the fraction $\tfrac{4}{9}$ of the whole is the Grade 6 ratio-reasoning move.
6.RP.A.3 Step 2 - Find the number of girls at Winthrop the same way.
- The ratio $4 : 5$ also splits the school into $4 + 5 = 9$ equal parts, but now girls are $5$ of them, so girls make up $\tfrac{5}{9}$ of the $180$ students.
💡 Same Tool #7 subproblem move applied to the second school — the ratio flips, so the girls' share flips from $\tfrac{4}{9}$ to $\tfrac{5}{9}$.
4.NBT.B.4 Step 3 - Add to get the totals at the dance.
- Total students is $270 + 180$, and total girls is $120 + 100$.
💡 Combining the two schools is just Grade 4 multi-digit addition — the units ("students", "girls") line up, so we add directly.
4.NF.A.1 Step 4 - Form the requested fraction $\dfrac{\text{girls}}{\text{students}}$ and reduce to lowest terms.
- Divide top and bottom by $\gcd(220, 450) = 10$.
💡 Dividing numerator and denominator by the same factor leaves the fraction's value unchanged — the Grade 4 equivalent-fractions rule.
6.RP.A.3 Find the number of girls at Colfax. The ratio $5 : 4$ splits the school into $5 6.RP.A.3 Find the number of girls at Winthrop the same way. The ratio $4 : 5$ also splits 4.NBT.B.4 Add to get the totals at the dance. Total students is $270 + 180$, and total gir 4.NF.A.1 Form the requested fraction $\dfrac{\text{girls}}{\text{students}}$ and reduce t Review
Reasonableness: Sanity check the size. Colfax has more boys than girls ($\tfrac{5}{9}$ vs $\tfrac{4}{9}$) and Winthrop has more girls than boys ($\tfrac{5}{9}$ vs $\tfrac{4}{9}$), but Colfax is the bigger school ($270 > 180$), so boys should slightly outnumber girls overall. The girls' fraction $\tfrac{22}{45} \approx 0.489$ is just under $\tfrac{1}{2}$, which matches that intuition perfectly. Choices $\tfrac{7}{18} \approx 0.389$ and $\tfrac{7}{15} \approx 0.467$ are too low, $\tfrac{1}{2}$ would require equal boys and girls, and $\tfrac{23}{45} \approx 0.511$ would flip the imbalance the wrong way.
Alternative: Tool #5 (Find a Pattern / Use Symmetry): both schools split into $9$ equal parts, so count parts directly. Colfax contributes $\tfrac{270}{9} = 30$ students per part $\times 4$ girl-parts $= 120$ girls; Winthrop contributes $\tfrac{180}{9} = 20$ students per part $\times 5$ girl-parts $= 100$ girls. Same totals, same fraction $\tfrac{22}{45}$, but the matching $9$-part structure makes the arithmetic faster.
CCSS standards used (min grade 6)
4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the two schools' totals: $270 + 180 = 450$ students and $120 + 100 = 220$ girls.)4.NF.A.1Explain equivalence of fractions and generate equivalent fractions (Reducing $\tfrac{220}{450}$ to $\tfrac{22}{45}$ by dividing numerator and denominator by $10$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Turning each ratio ($5 : 4$ and $4 : 5$) into a fraction of the school's total and computing the number of girls at Colfax ($\tfrac{4}{9} \times 270 = 120$) and Winthrop ($\tfrac{5}{9} \times 180 = 100$).)
⭐ This AMC 8 problem only needs Grade 6 ratio reasoning — turn each ratio into a fraction of the whole school — that you already know!
⭐ This AMC 8 problem only needs Grade 6 ratio reasoning — turn each ratio into a fraction of the whole school — that you already know!
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