AMC 10 · 2021 · #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 $\frac{1}{3}$ ; doubling the cylinder radius further divides by 4, so the height becomes $\frac{h_c}{3 · 4}$ = $\frac{18}{12}$ = 1.5. Tool #3 (Eliminate): the heights below $\frac{h_c}{3}$ = 6 are the only physically sensible candidates.

1STEP 1

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

V_c =

\frac{1}{3}

π · 144 · 18 = π · 144 · 6 = 864 π cm³

2STEP 2

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

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

h_y =

\frac{864 π}{576 π}

\frac{864}{576}

4STEP 4

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

h_y =

\frac{864}{576}

\frac{3}{2}

= 1.5 cm

5STEP 5

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).

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