AMC 8 · 2016 · #12
Grade 6 rate-ratioProblem
Jefferson Middle School has the same number of boys and girls. 43 of the girls and 32
of the boys went on a field trip. What fraction of the students on the field trip were girls?
Pick an answer.
AMC 8 2016 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: Jefferson Middle School has the same number of boys and girls. $\tfrac{3}{4}$ of the girls and $\tfrac{2}{3}$ of the boys went on a field trip. Of the students who went on the trip, what fraction were girls?
Givens: Number of boys = number of girls; $\tfrac{3}{4}$ of the girls went on the trip; $\tfrac{2}{3}$ of the boys went on the trip; Answer choices: (A) $\tfrac{1}{2}$, (B) $\tfrac{9}{17}$, (C) $\tfrac{7}{13}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{14}{15}$
Unknowns: The fraction of the trip-going students who are girls
Understand
Restated: Jefferson Middle School has the same number of boys and girls. $\tfrac{3}{4}$ of the girls and $\tfrac{2}{3}$ of the boys went on a field trip. Of the students who went on the trip, what fraction were girls?
Givens: Number of boys = number of girls; $\tfrac{3}{4}$ of the girls went on the trip; $\tfrac{2}{3}$ of the boys went on the trip; Answer choices: (A) $\tfrac{1}{2}$, (B) $\tfrac{9}{17}$, (C) $\tfrac{7}{13}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{14}{15}$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #7 Identify Subproblems
The school size is never given, which usually means it does not matter — a perfect cue for Tool #9 (Easier Related Problem). Pick a convenient number of girls (and boys) so that both $\tfrac{3}{4}$ and $\tfrac{2}{3}$ produce whole students with no leftover fractions. The least common denominator of $4$ and $3$ is $12$, so choose $12$ girls and $12$ boys. Then Tool #7 (Identify Subproblems) splits the count into three clean pieces: (a) girls on the trip, (b) boys on the trip, (c) combine and form the ratio. This sidesteps Tool #13 (Algebra) entirely.
Execute — Answer: B
6.RP.A.3 Step 1 - Pick an easier related problem.
- Let there be $12$ girls and $12$ boys (so boys $=$ girls, matching the original).
- The number $12$ is chosen because it is the smallest number that both $4$ and $3$ divide evenly, so $\tfrac{3}{4}$ and $\tfrac{2}{3}$ of it will both be whole numbers.
💡 Replacing the unknown school size with a clean number is exactly the Tool #9 move. Because the answer is a fraction of trip-goers, the school size cancels — any equal count works, but $12$ avoids partial students.
4.NF.B.4 Step 2 - Subproblem 1: count the girls on the trip.
- Three-quarters of $12$ girls go on the trip.
💡 Taking a fraction of a whole number — Grade 4 fraction-times-whole skill.
4.NF.B.4 Step 3 - Subproblem 2: count the boys on the trip.
- Two-thirds of $12$ boys go on the trip.
💡 Same fraction-of-a-whole move as the previous step, just with a different fraction.
4.NF.B.4 Step 4 - Subproblem 3: combine.
- The total number of students on the trip is the girls plus the boys.
💡 Adding the two trip subgroups is the final piece of the subproblem split.
6.RP.A.1 Step 5 Form the requested ratio: girls on the trip out of all students on the trip.
💡 Writing a part-to-whole ratio is the heart of Grade 6 ratio reasoning. Note that $9$ and $17$ share no common factor, so $\tfrac{9}{17}$ is already simplified.
6.RP.A.3 Pick an easier related problem. Let there be $12$ girls and $12$ boys (so boys $ 4.NF.B.4 Subproblem 1: count the girls on the trip. Three-quarters of $12$ girls go on th 4.NF.B.4 Subproblem 2: count the boys on the trip. Two-thirds of $12$ boys go on the trip 4.NF.B.4 Subproblem 3: combine. The total number of students on the trip is the girls plu 6.RP.A.1 Form the requested ratio: girls on the trip out of all students on the trip. Review
Reasonableness: More than half of the girls ($\tfrac{3}{4}$) but only two-thirds of the boys went, so the trip should have slightly more girls than boys — the girl-fraction should be a little above $\tfrac{1}{2}$. Indeed $\tfrac{9}{17} \approx 0.529$, just above $\tfrac{1}{2}$. Choice (A) $\tfrac{1}{2}$ would mean equal numbers, (D) $\tfrac{2}{3}$ would mean twice as many girls as boys on the trip — both are wrong magnitudes. Only (B) $\tfrac{9}{17}$ has the right size.
Alternative: Tool #9 again with a different size confirms scale-independence: try $24$ girls and $24$ boys. Then girls on trip $= \tfrac{3}{4} \times 24 = 18$, boys on trip $= \tfrac{2}{3} \times 24 = 16$, total $= 34$, fraction $= \tfrac{18}{34} = \tfrac{9}{17}$. Same answer — confirming the result does not depend on the chosen school size.
CCSS standards used (min grade 6)
4.NF.B.4Apply and extend understandings of multiplication to multiply a fraction by a whole number (Computing $\tfrac{3}{4} \times 12 = 9$ girls on the trip and $\tfrac{2}{3} \times 12 = 8$ boys on the trip — both fraction-of-a-whole calculations.)6.RP.A.1Understand the concept of a ratio and use ratio language to describe a relationship (Writing the answer as the part-to-whole ratio $\tfrac{9}{17}$ (girls on trip out of all students on trip).)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Recognizing that the answer (a ratio of trip-goers) does not depend on the school size, which justifies picking $12$ as a convenient stand-in for the unknown count.)
⭐ If the actual number is never given, pick a friendly one — then the AMC 8 problem becomes simple fraction-of-a-group arithmetic.
⭐ If the actual number is never given, pick a friendly one — then the AMC 8 problem becomes simple fraction-of-a-group arithmetic.
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