Competition · AMC preparation · step 4 of 4

AMC 10 · 2021B · #10

Grade 8 geometry-3d

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

volume-conevolume-cylinderformula-substitutionfraction-arithmetic identify-subproblemsdimensional-analysis ↑ Prerequisites: volume-cylindervolume-cone

📏 Short solution 💡 2 insights

Problem

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.

Pick an answer.

(A)

1.5

(B)

(C)

(D)

4.5

(E)

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Subproblems): split into (a) compute the cone's water volume, (b) compute the cylinder's base area, (c) solve volume = base area × height for the height. Tool #8 (Units): cm³ ÷ cm² = cm — confirms the formula gives a length. Tool #9 (Easier Related Problem): the cylinder's radius is exactly 2 × the cone's, so its base area is 4 × the cone's base. A pure scaling shortcut: if the radii were equal the cone-to-cylinder height ratio would be 1/3; doubling the cylinder radius further divides by 4, so the height becomes h_c/(3 · 4) = 18/12 = 1.5. Tool #3 (Eliminate): the heights below h_c / 3 = 6 are the only physically sensible candidates.

1STEP 1

Find the water's volume

Cone water volume V_c = $\frac{1}{3}$ π·12²·18 = 864π cm³.

V_c = 1/3 π · 144 · 18 = π · 144 · 6 = 864 π cm³

2STEP 2

Find the cylinder's base area

Cylinder base area π·24² = 576π cm² — 4 × the cone's base, since doubling the radius quadruples area.

π r_y² = π · 576 = 576 π cm²

3STEP 3

Set the two volumes equal

Water is conserved: 864π = 576π·h_y, so h_y = $\frac{864}{576}$ .

h_y = (864 π)/(576 π) = 864/576

4STEP 4

Simplify the fraction

Simplify $\frac{864}{576}$ by dividing top and bottom by 288 to get h_y = 1.5 cm.

h_y = 864/576 = 3/2 = 1.5 cm

5STEP 5

Match against the choices

1.5 matches answer choice (A).

h_y = 1.5 → (A)

Answer

1.5

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

\frac{1}{3}

· 18 = 6 cm (a cone is exactly

\frac{1}{3}

of a cylinder with same base and height). Now the cylinder radius is doubled, so its base area is 4 × larger and the same water spreads 4 × thinner:

\frac{6}{4}

= 1.5 cm. Matches our answer and confirms (A). Also 1.5 ≪ 18 matches intuition — pouring narrow-cone water into a wider cylinder gives a shallow puddle.

💡Key takeaway

This AMC 10 problem only needs Grade 8 cone-volume formula $\frac{1}{3}$ π r² h you already know — the cone holds 864π cm³ of water, the wider cylinder has base 576π cm², so the water height is $\frac{864}{576}$ = 1.5 cm, choice (A).

Find the water's volume
Find the cylinder's base area
Set the two volumes equal
Simplify the fraction
Match against the choices

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Find the water's volume

Find the cylinder's base area

Set the two volumes equal

Simplify the fraction

Match against the choices

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