AMC 8 · 2011 · #14
Grade 6 rate-ratioProblem
There are 270 students at Colfax Middle School, where the ratio of boys to girls is 5 :4. There are 180 students at Winthrop Middle School, where the ratio of boys to girls is 4 :5. The two schools hold a dance and all students from both schools attend. What fraction of the students at the dance 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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