AMC 10 · 2021 · #10
Grade 8 geometry-3dProblem
An inverted cone with base radius 12cm and height 18cm is full of water. The water is poured into a tall cylinder whose horizontal base has radius of 24cm. What is the height in centimeters of the water in the cylinder?
Pick an answer.
AMC 10 2021 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: An inverted cone (point down) with base radius $12$ cm and height $18$ cm is full of water. All the water is poured into a tall cylinder whose horizontal circular base has radius $24$ cm. Find the height (cm) of water in the cylinder.
Givens: Cone radius $r_c = 12$ cm, cone height $h_c = 18$ cm — cone is full of water; Cylinder radius $r_y = 24$ cm — note this is twice the cone radius; Water is conserved: volume in cone $=$ volume in cylinder; Cone volume formula $V_c = \tfrac{1}{3}\pi r_c^2 h_c$; Cylinder volume formula $V_y = \pi r_y^2 h_y$; Answer choices: (A) $1.5$, (B) $3$, (C) $4$, (D) $4.5$, (E) $6$
Unknowns: The water height $h_y$ in the cylinder (cm)
Understand
Restated: An inverted cone (point down) with base radius $12$ cm and height $18$ cm is full of water. All the water is poured into a tall cylinder whose horizontal circular base has radius $24$ cm. Find the height (cm) of water in the cylinder.
Givens: Cone radius $r_c = 12$ cm, cone height $h_c = 18$ cm — cone is full of water; Cylinder radius $r_y = 24$ cm — note this is twice the cone radius; Water is conserved: volume in cone $=$ volume in cylinder; Cone volume formula $V_c = \tfrac{1}{3}\pi r_c^2 h_c$; Cylinder volume formula $V_y = \pi r_y^2 h_y$; Answer choices: (A) $1.5$, (B) $3$, (C) $4$, (D) $4.5$, (E) $6$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #8 Analyze the Units, #9 Easier Related Problem, #3 Eliminate Possibilities
Tool #7 (Subproblems): split into (a) compute the cone's water volume, (b) compute the cylinder's base area, (c) solve volume $=$ base area $\times$ height for the height. Tool #8 (Units): cm$^3$ $\div$ cm$^2$ $=$ cm — confirms the formula gives a length. Tool #9 (Easier Related Problem): the cylinder's radius is exactly $2 \times$ the cone's, so its base area is $4 \times$ the cone's base. A pure scaling shortcut: if the radii were equal the cone-to-cylinder height ratio would be $\tfrac{1}{3}$; doubling the cylinder radius further divides by $4$, so the height becomes $\tfrac{h_c}{3 \cdot 4} = \tfrac{18}{12} = 1.5$. Tool #3 (Eliminate): the heights below $h_c / 3 = 6$ are the only physically sensible candidates.
Execute — Answer: A
8.G.C.9 Step 1 - Compute the water's volume in the cone using $V_c = \tfrac{1}{3}\pi r_c^2 h_c$.
- Plug in $r_c = 12$, $h_c = 18$: $V_c = \tfrac{1}{3} \pi (12)^2 (18) = \tfrac{1}{3} \pi \cdot 144 \cdot 18$.
💡 Grade 8 cone volume: $\tfrac{1}{3} \pi r^2 h$ — same as a cylinder of the same base and height, divided by $3$.
7.G.B.4 Step 2 - Compute the cylinder's base area: $\pi r_y^2 = \pi (24)^2 = 576 \pi$ cm$^2$.
- Note this is $4 \times$ the cone's base area $\pi \cdot 12^2 = 144\pi$, because radius doubled $\Rightarrow$ area $\times 4$.
💡 Grade 7 circle area: $\pi r^2$, and doubling $r$ multiplies area by $4$.
6.EE.B.7 Step 3 - Conservation of water: $V_c = V_y = \pi r_y^2 \cdot h_y$.
- Substitute: $864 \pi = 576 \pi \cdot h_y$.
- Solve for $h_y$ — divide both sides by $576 \pi$.
💡 Grade 6 equations: $\pi$ cancels, and units cm$^3$ ÷ cm$^2$ $=$ cm — the answer is a length.
5.NF.B.3 Step 4 - Simplify $\dfrac{864}{576}$.
- Divide top and bottom by $288$: $\dfrac{864}{576} = \dfrac{3}{2} = 1.5$ cm.
💡 Grade 5 fractions: $864 \div 288 = 3$ and $576 \div 288 = 2$, so the quotient simplifies to $\tfrac{3}{2}$.
5.NBT.A.3 Step 5 Match $1.5$ to the answer choices: it is choice (A).
💡 Grade 5 decimals: $1.5$ is exactly choice (A).
8.G.C.9 Compute the water's volume in the cone using $V_c = \tfrac{1}{3}\pi r_c^2 h_c$. 7.G.B.4 Compute the cylinder's base area: $\pi r_y^2 = \pi (24)^2 = 576 \pi$ cm$^2$. Not 6.EE.B.7 Conservation of water: $V_c = V_y = \pi r_y^2 \cdot h_y$. Substitute: $864 \pi = 5.NF.B.3 Simplify $\dfrac{864}{576}$. Divide top and bottom by $288$: $\dfrac{864}{576} = 5.NBT.A.3 Match $1.5$ to the answer choices: it is choice (A). Review
Reasonableness: Sanity by Tool #9 (Easier Related Problem). If the cylinder had the SAME radius $12$ as the cone, the same water would fill a cylinder of height $\tfrac{1}{3} \cdot 18 = 6$ cm (a cone is exactly $\tfrac{1}{3}$ of a cylinder with same base and height). Now the cylinder radius is doubled, so its base area is $4 \times$ larger and the same water spreads $4 \times$ thinner: $\tfrac{6}{4} = 1.5$ cm. Matches our answer and confirms (A). Also $1.5 \ll 18$ matches intuition — pouring narrow-cone water into a wider cylinder gives a shallow puddle.
Alternative: Tool #6 (Guess and Check) on the choices. The cone's water volume is $864\pi$ cm$^3$. For each choice $h_y$, compute the cylinder's water volume $576 \pi \cdot h_y$ and check equality with $864 \pi$: (A) $h_y = 1.5 \Rightarrow 576 \cdot 1.5 = 864$ ✓; (B) $h_y = 3 \Rightarrow 1728 \ne 864$; (C) $h_y = 4 \Rightarrow 2304 \ne 864$; etc. Only (A) works.
CCSS standards used (min grade 8)
8.G.C.9Know the formulas for volumes of cones, cylinders, and spheres (Computing the cone volume $\tfrac{1}{3} \pi r^2 h = 864 \pi$ cm$^3$.)7.G.B.4Know the formulas for area and circumference of a circle (Computing the cylinder's base area $\pi r^2 = 576 \pi$ cm$^2$.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving $576 \pi \cdot h_y = 864 \pi$ for $h_y$.)5.NF.B.3Interpret a fraction as division of the numerator by the denominator (Reducing $\tfrac{864}{576}$ to $\tfrac{3}{2}$.)5.NBT.A.3Read, write, and compare decimals to thousandths (Matching $\tfrac{3}{2} = 1.5$ to choice (A).)
⭐ This AMC 10 problem only needs Grade 8 cone-volume formula $\tfrac{1}{3}\pi r^2 h$ you already know — the cone holds $864\pi$ cm$^3$ of water, the wider cylinder has base $576\pi$ cm$^2$, so the water height is $864 / 576 = 1.5$ cm, choice (A).
⭐ This AMC 10 problem only needs Grade 8 cone-volume formula $\tfrac{1}{3}\pi r^2 h$ you already know — the cone holds $864\pi$ cm$^3$ of water, the wider cylinder has base $576\pi$ cm$^2$, so the water height is $864 / 576 = 1.5$ cm, choice (A).
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