AMC 8 · 2016 · #12

Grade 6 rate-ratio

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Percent / Fraction Recovery

fraction-arithmeticfraction-multiplicationratio-proportion easier-related-problemidentify-subproblems ↑ Prerequisites: fraction-multiplicationfraction-arithmetic

📏 Medium solution 💡 3 insights

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Problem

Jefferson Middle School has the same number of boys and girls.

\frac{3}{4}

of the girls and

\frac{2}{3}

of the boys went on a field trip. Of the students who went on the trip, what fraction were girls?

Pick an answer.

(A)

$frac{1}{2}$

(B)

$frac{9}{17}$

(C)

$frac{7}{13}$

(D)

$frac{2}{3}$

(E)

$frac{14}{15}$

AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

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 $\frac{3}{4}$ and $\frac{2}{3}$ 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.

1STEP 1

Solve an easier version: use 12 girls and 12 boys, since 12 is the smallest count both 4 and 3 divide evenly.

girls = 12, boys = 12

2STEP 2

Subproblem 1 — girls on the trip: three-quarters of 12 is 9.

\frac{3}{4}

× 12 = 9 girls

3STEP 3

Subproblem 2 — boys on the trip: two-thirds of 12 is 8.

\frac{2}{3}

× 12 = 8 boys

4STEP 4

Subproblem 3 — combine: 9 girls plus 8 boys make 17 on the trip.

9 + 8 = 17 students on the trip

5STEP 5

Form the asked ratio: girls on the trip over everyone on the trip — the result is already in lowest terms.

\frac{\text{girls on trip}}{\text{students on trip}}

\frac{9}{17}

→ (B)

Answer

frac{9}{17}

More than half of the girls (

\frac{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

\frac{1}{2}

. Indeed

\frac{9}{17}

≈ 0.529, just above

\frac{1}{2}

. Choice (A)

\frac{1}{2}

would mean equal numbers, (D)

\frac{2}{3}

would mean twice as many girls as boys on the trip — both are wrong magnitudes. Only (B)

\frac{9}{17}

has the right size.

💡Key takeaway

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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