AMC 8 · 2016 · #12
Easy mode Grade 6Problem
Jefferson Middle School has the same number of boys as girls. Picture the two groups as equal in size.
On field-trip day, 43 of the girls go on the trip. Also, 32 of the boys go on the trip.
Now look at just the kids who actually went. Out of those kids, what fraction are 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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