AMC 8 · 2015 · #16
Grade 6 rate-ratioProblem
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 31 of all the ninth graders are paired with 52 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
Pick an answer.
AMC 8 2015 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: In a buddy program, $\tfrac{1}{3}$ of the ninth graders are paired one-to-one with $\tfrac{2}{5}$ of the sixth graders. What fraction of all the sixth and ninth graders put together have a buddy?
Givens: Each pair has exactly one ninth grader and one sixth grader (one-to-one pairing); $\tfrac{1}{3}$ of the ninth graders are in a pair; $\tfrac{2}{5}$ of the sixth graders are in a pair; Answer choices: (A) $\tfrac{2}{15}$, (B) $\tfrac{4}{11}$, (C) $\tfrac{11}{30}$, (D) $\tfrac{3}{8}$, (E) $\tfrac{11}{15}$
Unknowns: The fraction $\dfrac{\text{students with a buddy}}{\text{total sixth + ninth graders}}$
Understand
Restated: In a buddy program, $\tfrac{1}{3}$ of the ninth graders are paired one-to-one with $\tfrac{2}{5}$ of the sixth graders. What fraction of all the sixth and ninth graders put together have a buddy?
Givens: Each pair has exactly one ninth grader and one sixth grader (one-to-one pairing); $\tfrac{1}{3}$ of the ninth graders are in a pair; $\tfrac{2}{5}$ of the sixth graders are in a pair; Answer choices: (A) $\tfrac{2}{15}$, (B) $\tfrac{4}{11}$, (C) $\tfrac{11}{30}$, (D) $\tfrac{3}{8}$, (E) $\tfrac{11}{15}$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #3 Eliminate Possibilities
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 $\tfrac{1}{3}$ of the ninth graders equal $\tfrac{2}{5}$ 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.
Execute — Answer: B
5.NF.B.4 Step 1 - Pick the smallest whole numbers that fit the pairing.
- We need $\tfrac{1}{3}$ of the ninth graders to be a whole number, so the number of ninth graders should be a multiple of $3$.
- We need $\tfrac{2}{5}$ of the sixth graders to be a whole number, so the number of sixth graders should be a multiple of $5$.
- Try ninth graders $= 3$ and sixth graders $= 5$ first.
💡 Multiplying a fraction by a whole number is Grade 5 fraction work; we pick numbers small enough to count on our fingers.
5.NF.B.4 Step 2 - Check the one-to-one rule.
- With $3$ ninth graders and $5$ sixth graders we would have $1$ paired ninth grader but $2$ paired sixth graders, which breaks the rule that each pair uses one of each.
- So these sizes do not work — we must scale up until the two paired counts match.
💡 If the picture does not match the story, change the numbers, not the story.
6.NS.B.4 Step 3 - Double the ninth graders so the paired counts agree.
- With $6$ ninth graders, $\tfrac{1}{3}$ of them is $2$.
- We still need $\tfrac{2}{5}$ of the sixth graders to equal $2$, so the sixth graders must be $5$.
- Now the paired counts match: $2$ ninth graders pair with $2$ sixth graders, giving $2$ pairs.
💡 We are quietly using a least common multiple ($\text{lcm}(1,2) = 2$ paired students from each side) — Grade 6 number sense.
4.OA.A.3 Step 4 - Count the students with a buddy and the total students.
- Buddied $= 2$ paired ninth graders $+ \, 2$ paired sixth graders $= 4$ students.
- Total $= 6$ ninth graders $+ \, 5$ sixth graders $= 11$ students.
💡 Just add the two sides — Grade 4 multi-step word-problem arithmetic.
6.RP.A.3 Step 5 Form the requested fraction and match it to a choice (Tool #3, eliminate by matching).
💡 A part-to-whole comparison is a ratio — Grade 6 ratio reasoning.
5.NF.B.4 Pick the smallest whole numbers that fit the pairing. We need $\tfrac{1}{3}$ of 5.NF.B.4 Check the one-to-one rule. With $3$ ninth graders and $5$ sixth graders we would 6.NS.B.4 Double the ninth graders so the paired counts agree. With $6$ ninth graders, $\t 4.OA.A.3 Count the students with a buddy and the total students. Buddied $= 2$ paired nin 6.RP.A.3 Form the requested fraction and match it to a choice (Tool #3, eliminate by matc Review
Reasonableness: Try a bigger size to be sure the answer does not depend on the counts. Take $12$ ninth graders and $10$ sixth graders: $\tfrac{1}{3} \times 12 = 4$ and $\tfrac{2}{5} \times 10 = 4$, so $4$ pairs. Buddied $= 4 + 4 = 8$ out of $12 + 10 = 22$ total, giving $\tfrac{8}{22} = \tfrac{4}{11}$ — the same fraction. The answer is stable, which is exactly what we expected from a problem that never gave us specific counts.
Alternative: Tool #13 (Convert to Algebra) also works: let $N$ ninth graders and $S$ sixth graders satisfy $\tfrac{1}{3}N = \tfrac{2}{5}S$. Calling that common value $P$, we get $N = 3P$ and $S = \tfrac{5}{2}P$, so the fraction is $\dfrac{2P}{3P + \tfrac{5}{2}P} = \dfrac{2P}{\tfrac{11}{2}P} = \dfrac{4}{11}$. Same answer, but it needs fraction-in-a-fraction arithmetic that we avoided by picking small numbers.
CCSS standards used (min grade 6)
4.OA.A.3Solve multistep word problems using the four operations (Adding the paired ninth graders and paired sixth graders ($2 + 2 = 4$), and adding total ninth and sixth graders ($6 + 5 = 11$).)5.NF.B.4Multiply a fraction by a whole number or by another fraction (Computing $\tfrac{1}{3} \times 6 = 2$ and $\tfrac{2}{5} \times 5 = 2$ to find the number of paired students in each grade.)6.NS.B.4Find common multiples and the least common multiple of whole numbers (Choosing the smallest grade sizes ($6$ ninth graders, $5$ sixth graders) that make the two paired counts equal — an lcm of $1$ and $2$ argument.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Expressing the final answer as the part-to-whole ratio $\tfrac{4}{11}$ of students with a buddy out of all students.)
⭐ 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.
⭐ 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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