AMC 8 · 2015 · #16

Grade 6 rate-ratio

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

ratio-proportionfraction-arithmeticlinear-equations-two-var convert-to-algebraidentify-subproblems ↑ Prerequisites: fraction-arithmeticratio-proportion

📏 Medium solution 💡 3 insights

Problem

In a buddy program,

\frac{1}{3}

of the ninth graders are paired one-to-one with

\frac{2}{5}

of the sixth graders. What fraction of all the sixth and ninth graders put together have a buddy?

Pick an answer.

(A)

$frac{2}{15}$

(B)

$frac{4}{11}$

(C)

$frac{11}{30}$

(D)

$frac{3}{8}$

(E)

$frac{11}{15}$

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

The problem never tells us how many ninth or sixth graders there are, which is a strong hint that the answer does not depend on the actual numbers. Tool #9 (Easier Related Problem) says: pick the smallest whole-number counts that make $\frac{1}{3}$ of the ninth graders equal $\frac{2}{5}$ of the sixth graders, then just count. Once we get a fraction, Tool #3 (Eliminate Possibilities) lets us match it against the five multiple-choice options to confirm. This sidesteps setting up variables and dividing decimals.

1STEP 1

Pick the smallest counts: ninth graders a multiple of 3, sixth graders a multiple of 5, so try 3 ninth and 5 sixth graders.

\frac{1}{3}

× 3 = 1 paired ninth grader,

\frac{2}{5}

× 5 = 2 paired sixth graders

2STEP 2

But 3 ninth graders make only 1 buddy while 5 sixth graders make 2, so the paired counts clash — scale up.

1 ≠ 2, so the pairing fails.

3STEP 3

Double to 6 ninth graders: now $\frac{1}{3}$ ×6 = 2 and $\frac{2}{5}$ ×5 = 2 match, giving 2 pairs from 6 ninth and 5 sixth graders.

\frac{1}{3}

× 6 = 2,

\frac{2}{5}

× 5 = 2

4STEP 4

Buddied students = 2 + 2 = 4; total students = 6 + 5 = 11.

buddied = 2 + 2 = 4, total = 6 + 5 = 11

5STEP 5

Form the ratio buddied/total = $\frac{4}{11}$ and match it to a choice (Tool #3, eliminate by matching).

buddied/total =

\frac{4}{11}

→ (B)

Answer

\frac{4}{11}

Try a bigger size to be sure the answer does not depend on the counts. Take 12 ninth graders and 10 sixth graders:

\frac{1}{3}

× 12 = 4 and

\frac{2}{5}

× 10 = 4, so 4 pairs. Buddied = 4 + 4 = 8 out of 12 + 10 = 22 total, giving

\frac{8}{22}

\frac{4}{11}

— the same fraction. The answer is stable, which is exactly what we expected from a problem that never gave us specific counts.

💡Key takeaway

When a problem hides the totals, plug in the smallest numbers that work — Grade 6 fraction and ratio reasoning is all you need to crack this AMC 8.

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