AMC 8 · 2013 · #16
Grade 6 rate-rationumber-theoryProblem
A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of 8th-graders to 6th-graders is 5:3, and the the ratio of 8th-graders to 7th-graders is 8:5. What is the smallest number of students that could be participating in the project?
Pick an answer.
AMC 8 2013 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: Fibonacci Middle School has $8^\text{th}$-graders, $7^\text{th}$-graders, and $6^\text{th}$-graders in a community-service project. The $8^\text{th}$-to-$6^\text{th}$ ratio is $5:3$, and the $8^\text{th}$-to-$7^\text{th}$ ratio is $8:5$. What is the smallest total number of students that fits both ratios at the same time?
Givens: Ratio $8^\text{th} : 6^\text{th} = 5 : 3$; Ratio $8^\text{th} : 7^\text{th} = 8 : 5$; All three counts must be whole numbers; Answer choices: (A) $16$, (B) $40$, (C) $55$, (D) $79$, (E) $89$
Unknowns: The smallest possible total number of students $= 8^\text{th} + 7^\text{th} + 6^\text{th}$
Understand
Restated: Fibonacci Middle School has $8^\text{th}$-graders, $7^\text{th}$-graders, and $6^\text{th}$-graders in a community-service project. The $8^\text{th}$-to-$6^\text{th}$ ratio is $5:3$, and the $8^\text{th}$-to-$7^\text{th}$ ratio is $8:5$. What is the smallest total number of students that fits both ratios at the same time?
Givens: Ratio $8^\text{th} : 6^\text{th} = 5 : 3$; Ratio $8^\text{th} : 7^\text{th} = 8 : 5$; All three counts must be whole numbers; Answer choices: (A) $16$, (B) $40$, (C) $55$, (D) $79$, (E) $89$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #6 Guess and Check
We are given two ratios that share one term ($8^\text{th}$-graders) but use different numbers for it ($5$ vs. $8$). Tool #7 (Identify Subproblems) splits the work into two clean pieces: (1) rescale each ratio so the $8^\text{th}$-grader count agrees, then (2) merge into a single three-part ratio and add. The bridge between the two subproblems is finding a common value for $8^\text{th}$-graders — namely $\text{lcm}(5, 8)$. Tool #6 (Guess and Check) is held in reserve for the review phase, since the smaller answer choices ($16, 40, 55, 79$) can be ruled out by quick divisibility checks.
Execute — Answer: E
6.NS.B.4 Step 1 - Subproblem 1: make the $8^\text{th}$-grader number the same in both ratios.
- The current $8^\text{th}$-grader values are $5$ and $8$.
- The smallest number that both divide is $\text{lcm}(5, 8) = 40$.
💡 Finding the least common multiple of two small numbers is a Grade 6 number-system skill.
6.RP.A.3 Step 2 Rescale ratio $1$ ($8^\text{th} : 6^\text{th} = 5 : 3$) by multiplying both sides by $8$ so the $8^\text{th}$-grader entry becomes $40$.
💡 Multiplying both parts of a ratio by the same number gives an equivalent ratio — the Grade 6 ratio-reasoning move.
6.RP.A.3 Step 3 Rescale ratio $2$ ($8^\text{th} : 7^\text{th} = 8 : 5$) by multiplying both sides by $5$ so the $8^\text{th}$-grader entry also becomes $40$.
💡 Same Grade 6 move on the other ratio so the two ratios now speak the same language about $8^\text{th}$-graders.
6.RP.A.1 Step 4 - Subproblem 2: combine the two rescaled ratios into one three-part ratio.
- Both now use $40$ for $8^\text{th}$-graders, so they snap together.
- Check that the combined ratio cannot be reduced — $\gcd(40, 25, 24) = 1$, so $40 : 25 : 24$ is already the smallest set of whole numbers.
💡 Extending a two-term ratio into a three-term ratio by aligning the shared term is the Grade 6 ratio-language skill.
6.RP.A.3 Step 5 The smallest total is the sum of those smallest whole-number counts.
💡 Once the ratio is in lowest whole-number form, adding the parts gives the smallest possible total.
6.NS.B.4 Subproblem 1: make the $8^\text{th}$-grader number the same in both ratios. The 6.RP.A.3 Rescale ratio $1$ ($8^\text{th} : 6^\text{th} = 5 : 3$) by multiplying both side 6.RP.A.3 Rescale ratio $2$ ($8^\text{th} : 7^\text{th} = 8 : 5$) by multiplying both side 6.RP.A.1 Subproblem 2: combine the two rescaled ratios into one three-part ratio. Both no 6.RP.A.3 The smallest total is the sum of those smallest whole-number counts. Review
Reasonableness: Check that $40 : 25 : 24$ actually satisfies both original ratios. $40 : 24 = 5 : 3$ (divide by $8$) ✓ and $40 : 25 = 8 : 5$ (divide by $5$) ✓. The total $89$ is the smallest because $\gcd(40, 25, 24) = 1$, so we cannot shrink the ratio further without breaking whole-number counts.
Alternative: Tool #6 (Guess and Check) on the choices: the total must be divisible into parts in the ratio $40 : 25 : 24$, so the total must be a multiple of $89$. Test each choice — (A) $16$, (B) $40$, (C) $55$, (D) $79$ are all less than $89$ and therefore impossible. Only (E) $89$ works. This rules out every wrong choice in one pass.
CCSS standards used (min grade 6)
6.NS.B.4Find the greatest common factor and the least common multiple of two whole numbers (Computing $\text{lcm}(5, 8) = 40$ so that both ratios can be rewritten with the same $8^\text{th}$-grader value.)6.RP.A.1Understand the concept of a ratio and use ratio language (Combining the two rescaled ratios into a single three-part ratio $8^\text{th} : 7^\text{th} : 6^\text{th} = 40 : 25 : 24$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Scaling each ratio by multiplying both parts by the same factor to produce equivalent ratios, then adding the parts to get the smallest possible total.)
⭐ This AMC 8 problem only needs Grade 6 ratio reasoning — match the shared term using $\text{lcm}$, then add the parts!
⭐ This AMC 8 problem only needs Grade 6 ratio reasoning — match the shared term using $\text{lcm}$, then add the parts!
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