AMC 8 · 2013 · #16

Grade 6 rate-rationumber-theory

Archetype Multiplicative Ratio with Integrality Constraints

ratio-proportionlcmmultiples identify-subproblemsratio-proportion ↑ Prerequisites: ratio-proportionlcmmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

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Problem

A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of $8^\text{th}$ -graders to $6^\text{th}$ -graders is $5:3$ , and the the ratio of $8^\text{th}$ -graders to $7^\text{th}$ -graders is $8:5$ . What is the smallest number of students that could be participating in the project?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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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

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

#7 Identify Subproblems 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 $\text{lcm}(5, 8) = 40$.

$$\text{lcm}(5, 8) = 40$$

💡 Finding the least common multiple of two small numbers is a Grade 6 number-system skill.

#7 Identify Subproblems 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 $40$.

$$5 : 3 \;=\; (5 \times 8) : (3 \times 8) \;=\; 40 : 24$$

💡 Multiplying both parts of a ratio by the same number gives an equivalent ratio — the Grade 6 ratio-reasoning move.

#7 Identify Subproblems 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 $40$.

$$8 : 5 \;=\; (8 \times 5) : (5 \times 5) \;=\; 40 : 25$$

💡 Same Grade 6 move on the other ratio so the two ratios now speak the same language about $8^\text{th}$-graders.

#7 Identify Subproblems 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.

$$8^\text{th} : 7^\text{th} : 6^\text{th} \;=\; 40 : 25 : 24$$

💡 Extending a two-term ratio into a three-term ratio by aligning the shared term is the Grade 6 ratio-language skill.

#7 Identify Subproblems 6.RP.A.3 Step 5

The smallest total is the sum of those smallest whole-number counts.

$$40 + 25 + 24 = 89 \;\Rightarrow\; \textbf{(E)}$$

💡 Once the ratio is in lowest whole-number form, adding the parts gives the smallest possible total.

[1] #7 6.NS.B.4 Subproblem 1: make the $8^\text{th}$-grader number the same in both ratios. The

[2] #7 6.RP.A.3 Rescale ratio $1$ ($8^\text{th} : 6^\text{th} = 5 : 3$) by multiplying both side

[3] #7 6.RP.A.3 Rescale ratio $2$ ($8^\text{th} : 7^\text{th} = 8 : 5$) by multiplying both side

[4] #7 6.RP.A.1 Subproblem 2: combine the two rescaled ratios into one three-part ratio. Both no

[5] #7 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.4 Find 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.1 Understand 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.3 Use 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 6)

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