AMC 8 · 2016 · #12

Easy mode Grade 6

Problem

Jefferson Middle School has the same number of boys as girls. Picture the two groups as equal in size.

On field-trip day, $\frac{3}{4}$ of the girls go on the trip. Also, $\frac{2}{3}$ of the boys go on the trip.

Now look at just the kids who actually went. Out of those kids, what fraction are girls?

$\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{9}{17}\qquad\textbf{(C) }\frac{7}{13}\qquad\textbf{(D) }\frac{2}{3}\qquad \textbf{(E) }\frac{14}{15}$

Pick an answer.

(A)

$frac{1}{2}$

(B)

$frac{9}{17}$

(C)

$frac{7}{13}$

(D)

$frac{2}{3}$

(E)

$frac{14}{15}$

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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

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

#9 Solve an Easier Related Problem 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.

$$\text{girls} = 12, \quad \text{boys} = 12$$

💡 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.

#7 Identify Subproblems 4.NF.B.4 Step 2

Subproblem 1: count the girls on the trip.
Three-quarters of $12$ girls go on the trip.

$$\tfrac{3}{4} \times 12 = 9 \text{ girls}$$

💡 Taking a fraction of a whole number — Grade 4 fraction-times-whole skill.

#7 Identify Subproblems 4.NF.B.4 Step 3

Subproblem 2: count the boys on the trip.
Two-thirds of $12$ boys go on the trip.

$$\tfrac{2}{3} \times 12 = 8 \text{ boys}$$

💡 Same fraction-of-a-whole move as the previous step, just with a different fraction.

#7 Identify Subproblems 4.NF.B.4 Step 4

Subproblem 3: combine.
The total number of students on the trip is the girls plus the boys.

$$9 + 8 = 17 \text{ students on the trip}$$

💡 Adding the two trip subgroups is the final piece of the subproblem split.

#9 Solve an Easier Related Problem 6.RP.A.1 Step 5

Form the requested ratio: girls on the trip out of all students on the trip.

$$\dfrac{\text{girls on trip}}{\text{students on trip}} = \dfrac{9}{17} \;\Rightarrow\; \textbf{(B)}$$

💡 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.

[1] #9 6.RP.A.3 Pick an easier related problem. Let there be $12$ girls and $12$ boys (so boys $

[2] #7 4.NF.B.4 Subproblem 1: count the girls on the trip. Three-quarters of $12$ girls go on th

[3] #7 4.NF.B.4 Subproblem 2: count the boys on the trip. Two-thirds of $12$ boys go on the trip

[4] #7 4.NF.B.4 Subproblem 3: combine. The total number of students on the trip is the girls plu

[5] #9 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.4 Apply 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.1 Understand 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.3 Use 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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