AMC 8 · 2019 · #9

Grade 8 geometry-3drate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

volume-cylinderratio-proportionformula-substitution identify-subproblemsformula-substitution ↑ Prerequisites: area-circlesratio-proportion

📏 Short solution 💡 2 insights

Problem

Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are $6$ cm in diameter and $12$ cm high. Felicia buys cat food in cylindrical cans that are $12$ cm in diameter and $6$ cm high. What is the ratio of the volume of one of Alex's cans to the volume of one of Felicia's cans?

Pick an answer.

(A)

1:4

(B)

1:2

(C)

1:1

(D)

2:1

(E)

4:1

AMC 8 2019 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.

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Toolkit + CCSS Solution

Understand

Restated: Alex's cat food can is a cylinder with diameter $6$ cm and height $12$ cm (tall and skinny). Felicia's can is a cylinder with diameter $12$ cm and height $6$ cm (short and wide). Find the ratio of Alex's volume to Felicia's volume and match it to one of the five answer choices.

Givens: Alex's can: diameter $= 6$ cm, height $= 12$ cm $\Rightarrow$ radius $r_A = 3$ cm; Felicia's can: diameter $= 12$ cm, height $= 6$ cm $\Rightarrow$ radius $r_F = 6$ cm; Volume of a cylinder: $V = \pi r^2 h$; Answer choices: (A) $1:4$, (B) $1:2$, (C) $1:1$, (D) $2:1$, (E) $4:1$

Unknowns: The ratio $V_A : V_F$ of Alex's can's volume to Felicia's can's volume

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #8 Analyze the Units, #3 Eliminate Possibilities

The question "ratio of two volumes" naturally splits into three subproblems (Tool #7): (i) compute $V_A$, (ii) compute $V_F$, (iii) form and simplify the ratio. Tool #8 (Analyze the Units) keeps us honest about diameter-vs-radius (the formula wants $r$, not $d$) and reminds us that both volumes are in cm$^3$, so $\pi$ and cm$^3$ cancel in the ratio. Tool #3 (Eliminate Possibilities) is the multiple-choice safety net: once we see Felicia's can is wider where it counts (radius is squared) but only half as tall, we expect $V_F > V_A$, which already eliminates (C), (D), (E).

Execute — Answer: B

#8 Analyze the Units 4.MD.A.1 Step 1

Convert each diameter to a radius.
The volume formula $V = \pi r^2 h$ uses the radius, so halve each diameter before plugging in.

$$r_A = \dfrac{6}{2} = 3 \text{ cm}, \quad r_F = \dfrac{12}{2} = 6 \text{ cm}$$

💡 Knowing that the radius is half the diameter is a Grade 4 measurement fact.

#7 Identify Subproblems 8.G.C.9 Step 2

Compute Alex's volume by plugging $r_A = 3$ and $h_A = 12$ into $V = \pi r^2 h$.
Square the radius first, then multiply by the height.

$$V_A = \pi (3)^2 (12) = \pi \cdot 9 \cdot 12 = 108\pi \text{ cm}^3$$

💡 Applying the cylinder volume formula $V = \pi r^2 h$ is exactly the Grade 8 "volumes of cylinders" standard.

#7 Identify Subproblems 8.G.C.9 Step 3

Compute Felicia's volume the same way with $r_F = 6$ and $h_F = 6$.
Notice the radius is bigger but the height is smaller.

$$V_F = \pi (6)^2 (6) = \pi \cdot 36 \cdot 6 = 216\pi \text{ cm}^3$$

💡 Same cylinder-volume formula, second application — Grade 8 again.

#8 Analyze the Units 6.RP.A.1 Step 4

Form the ratio $V_A : V_F$ as a fraction.
Because both volumes share the factor $\pi$ and the unit cm$^3$, both cancel cleanly, leaving a pure number ratio.

$$\dfrac{V_A}{V_F} = \dfrac{108\pi}{216\pi} = \dfrac{108}{216}$$

💡 Setting up a part-to-part comparison as a fraction is the Grade 6 ratio concept.

#3 Eliminate Possibilities 4.NF.A.1 Step 5

Simplify $\tfrac{108}{216}$ to lowest terms.
Since $216 = 2 \times 108$, the fraction equals $\tfrac{1}{2}$, so the ratio is $1:2$ — choice $\textbf{(B)}$.

$$\dfrac{108}{216} = \dfrac{1}{2} \;\Rightarrow\; V_A : V_F = 1 : 2 \;\Rightarrow\; \textbf{(B)}$$

💡 Recognizing $\tfrac{108}{216}$ as equivalent to $\tfrac{1}{2}$ is Grade 4 equivalent-fractions reasoning, which then pinpoints choice (B).

[1] #8 4.MD.A.1 Convert each diameter to a radius. The volume formula $V = \pi r^2 h$ uses the r

[2] #7 8.G.C.9 Compute Alex's volume by plugging $r_A = 3$ and $h_A = 12$ into $V = \pi r^2 h$.

[3] #7 8.G.C.9 Compute Felicia's volume the same way with $r_F = 6$ and $h_F = 6$. Notice the r

[4] #8 6.RP.A.1 Form the ratio $V_A : V_F$ as a fraction. Because both volumes share the factor

[5] #3 4.NF.A.1 Simplify $\tfrac{108}{216}$ to lowest terms. Since $216 = 2 \times 108$, the fra

Review

Reasonableness: Sanity check by scaling: Felicia's radius is doubled compared to Alex's ($6$ vs $3$), so $r^2$ becomes $4\times$ larger, but her height is halved ($6$ vs $12$), giving an overall factor of $4 \times \tfrac{1}{2} = 2$ in Felicia's favor. So $V_F = 2 V_A$, meaning $V_A : V_F = 1:2$ — exactly choice (B). The squared-radius effect outweighing the halved height matches intuition: short-and-wide cans usually hold more than tall-and-skinny ones with these flipped dimensions.

Alternative: Tool #5 (Look for a Pattern) via dimensional scaling: write the ratio symbolically before plugging numbers, $\tfrac{V_A}{V_F} = \tfrac{\pi r_A^2 h_A}{\pi r_F^2 h_F} = \tfrac{3^2 \cdot 12}{6^2 \cdot 6} = \tfrac{9 \cdot 12}{36 \cdot 6} = \tfrac{108}{216} = \tfrac{1}{2}$. Cancelling $\pi$ up front saves a step and reinforces that the answer can't depend on $\pi$.

CCSS standards used (min grade 8)

4.MD.A.1 Know relative sizes of measurement units and convert larger to smaller units (Converting each diameter to its radius ($r = d/2$) before applying the volume formula.)
4.NF.A.1 Explain why a fraction is equivalent to another fraction (Simplifying $\tfrac{108}{216}$ to $\tfrac{1}{2}$ in lowest terms to match the answer choice $1:2$.)
6.RP.A.1 Understand the concept of a ratio and use ratio language (Forming the part-to-part comparison $V_A : V_F$ as the fraction $\tfrac{V_A}{V_F}$ so that the common factor $\pi$ cancels.)
8.G.C.9 Know the formulas for volumes of cones, cylinders, and spheres (Applying $V = \pi r^2 h$ to compute $V_A = 108\pi$ and $V_F = 216\pi$ cm$^3$ for the two cylindrical cans.)

⭐ This AMC 8 problem really only needs the Grade 8 cylinder volume formula $V = \pi r^2 h$ — once you plug in, $\pi$ cancels and the ratio simplifies in one step!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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