AMC 10 · 2021 · #12
Grade 8 geometry-2dProblem
Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are 3 cm and 6 cm. Into each cone is dropped a spherical marble of radius 1 cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?
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: Two upside-down (apex-down) right cones hold the SAME volume of liquid. The liquid surface in the narrow cone is a circle of radius $3$ cm; in the wide cone it is a circle of radius $6$ cm. Drop one identical marble (radius $1$ cm) into each cone — both fully sink. The liquid level rises by some amount $\Delta_n$ in the narrow cone and $\Delta_w$ in the wide cone. Find the ratio $\Delta_n : \Delta_w$.
Givens: Both cones are right circular, vertex pointing DOWN; Initial liquid volumes are equal in the two cones; Initial top-of-liquid radii: $3$ cm (narrow) and $6$ cm (wide); A radius-$1$ marble of volume $\tfrac{4}{3}\pi$ cm$^3$ drops into each, fully submerged, no spill; Answer choices: $1:1$, $47:43$, $2:1$, $40:13$, $4:1$
Unknowns: Ratio of liquid-level rises $\Delta_n / \Delta_w$
Understand
Restated: Two upside-down (apex-down) right cones hold the SAME volume of liquid. The liquid surface in the narrow cone is a circle of radius $3$ cm; in the wide cone it is a circle of radius $6$ cm. Drop one identical marble (radius $1$ cm) into each cone — both fully sink. The liquid level rises by some amount $\Delta_n$ in the narrow cone and $\Delta_w$ in the wide cone. Find the ratio $\Delta_n : \Delta_w$.
Givens: Both cones are right circular, vertex pointing DOWN; Initial liquid volumes are equal in the two cones; Initial top-of-liquid radii: $3$ cm (narrow) and $6$ cm (wide); A radius-$1$ marble of volume $\tfrac{4}{3}\pi$ cm$^3$ drops into each, fully submerged, no spill; Answer choices: $1:1$, $47:43$, $2:1$, $40:13$, $4:1$
Plan
Primary tool: #9 Easier Related Problem
Secondary: #1 Draw a Diagram, #7 Identify Subproblems, #3 Eliminate Possibilities
Tool #9 (Easier Problem) does the heavy lifting: imagine the cones are SO TALL that near the liquid surface they look like cylinders of radius $3$ and $6$. Then equal added volume $V_m$ raises the narrow cylinder by $V_m / (\pi \cdot 3^2)$ and the wide cylinder by $V_m / (\pi \cdot 6^2)$ — ratio $36/9 = 4$. Tool #1 (Draw) makes the cone-vs-cylinder picture explicit. Tool #7 (Subproblems) cleans up the rigorous version: same initial volume implies the narrow cone's height is $4 \times$ the wide cone's height, and that same factor of $4$ then reappears in the rise ratio. Tool #3 (Eliminate) confirms (E) against the five choices.
Execute — Answer: E
7.G.A.1 Step 1 - Set up the two cones with similar-triangle constants.
- Let the narrow cone's initial liquid height be $h_n$ and the wide cone's be $h_w$.
- Because the cones are similar in their own way, the top-of-liquid radius is proportional to the height: in the narrow cone, $r/h = 3/h_n$ always; in the wide cone, $r/h = 6/h_w$ always.
💡 Grade 7 scale drawings: in a cone, every horizontal slice is a scaled-down copy of the rim.
8.EE.C.7 Step 2 - Equal initial volumes give a clean relation between $h_n$ and $h_w$.
- Volume of a cone is $\tfrac{1}{3}\pi r^2 h$, so $\tfrac{1}{3}\pi (3)^2 h_n = \tfrac{1}{3}\pi (6)^2 h_w$, i.e.
- $9 h_n = 36 h_w$, hence $h_n = 4 h_w$.
- The narrow liquid column is four times as tall as the wide one.
💡 Grade 8 one-variable linear equation: cancel $\tfrac{1}{3}\pi$ on both sides and solve.
7.G.B.6 Step 3 - Try the easier problem first (Tool #9).
- Pretend each cone is locally a CYLINDER of radius $3$ and $6$ near the liquid surface.
- The marble's volume $V_m = \tfrac{4}{3}\pi$ raises the narrow cylinder by $V_m / (\pi \cdot 3^2)$ and the wide cylinder by $V_m / (\pi \cdot 6^2)$.
- The ratio is $(\pi \cdot 36) / (\pi \cdot 9) = 4$.
- This suggests the answer $4 : 1$.
💡 Grade 7 area and volume: a cylinder rise = added volume divided by cross-section area.
8.EE.A.1 Step 4 - Make the cylinder argument exact for cones.
- In the narrow cone, the new height $H_n$ satisfies $\tfrac{1}{3}\pi (r_n^{\text{new}})^2 H_n = \tfrac{1}{3}\pi (3)^2 h_n + V_m$ where $r_n^{\text{new}} = \tfrac{3}{h_n} H_n$.
- Substituting: $\tfrac{1}{3}\pi \tfrac{9}{h_n^2} H_n^3 = 3\pi h_n + V_m$, so $H_n^3 = h_n^3 + \tfrac{V_m \cdot h_n^2}{3\pi}$.
- Similarly, $H_w^3 = h_w^3 + \tfrac{V_m \cdot h_w^2}{12\pi}$.
💡 Grade 8 powers: cone volume in terms of one height variable gives $H^3 - h^3$ relations.
7.RP.A.2 Step 5 - Read off the rise ratio using $h_n = 4 h_w$.
- Both sides of the cubic identities scale the same way: $(H_n / h_n)^3 - 1 = \tfrac{V_m}{3\pi h_n}$, and $(H_w / h_w)^3 - 1 = \tfrac{V_m}{12\pi h_w}$.
- With $h_n = 4 h_w$, both right-hand sides equal $\tfrac{V_m}{12\pi h_w}$, so $H_n / h_n = H_w / h_w$ — call this common ratio $\lambda$.
- Therefore $H_n - h_n = (\lambda - 1) h_n$ and $H_w - h_w = (\lambda - 1) h_w$.
- Divide: $\Delta_n / \Delta_w = h_n / h_w = 4$.
💡 Grade 7 proportional reasoning: both cones scale up by the same factor $\lambda$, so rises are in the ratio of the original heights.
6.EE.B.5 Step 6 - Match $4 : 1$ to the choices.
- It's (E).
- The other ratios ($1:1$, $47:43$, $2:1$, $40:13$) don't survive the equal-volume condition.
💡 Grade 6: compare the computed ratio to the multiple-choice list.
7.G.A.1 Set up the two cones with similar-triangle constants. Let the narrow cone's init 8.EE.C.7 Equal initial volumes give a clean relation between $h_n$ and $h_w$. Volume of a 7.G.B.6 Try the easier problem first (Tool #9). Pretend each cone is locally a CYLINDER 8.EE.A.1 Make the cylinder argument exact for cones. In the narrow cone, the new height $ 7.RP.A.2 Read off the rise ratio using $h_n = 4 h_w$. Both sides of the cubic identities 6.EE.B.5 Match $4 : 1$ to the choices. It's (E). The other ratios ($1:1$, $47:43$, $2:1$, Review
Reasonableness: Two ways to feel that 4:1 is right. (a) The cylinder approximation (Tool #9) already gave exactly $4:1$ — and the marble adds the same volume to each cone, so wider rim = smaller rise. The wide rim is twice the narrow rim, and area scales as radius squared ($6^2 / 3^2 = 4$), so the wide cone needs $4 \times$ less rise. (b) The narrow cone is four times TALLER for the same volume, and the rise scales proportionally with the original height — so 4:1 again. Both arguments converge, and the answer does NOT depend on the marble's actual volume — it could be any small object and the ratio stays $4:1$.
Alternative: Tool #6 (Guess and Check) with concrete numbers: take $h_w = 1$ cm, so $h_n = 4$ cm. The marble volume is $V_m = \tfrac{4}{3}\pi$. Then $H_n^3 = 4^3 + \tfrac{V_m \cdot 16}{3\pi} = 64 + \tfrac{64}{9}$ and $H_w^3 = 1 + \tfrac{V_m}{12\pi} = 1 + \tfrac{1}{9}$. Numerically $H_n \approx 4.073$ and $H_w \approx 1.018$, giving $\Delta_n \approx 0.073$ and $\Delta_w \approx 0.018$. Ratio $\approx 4.06$, very close to $4$ — the small discrepancy is the non-cylinder correction, and exactly equal to $4$ when computed symbolically. Confirms (E).
CCSS standards used (min grade 8)
6.EE.B.5Understand solving an equation or inequality as a process of finding values (Selecting the matching answer choice $4:1$ from the five options.)7.RP.A.2Recognize and represent proportional relationships between quantities (Reading $\Delta_n / \Delta_w = h_n / h_w$ once both cones scale by the same factor $\lambda$.)7.G.A.1Solve problems involving scale drawings of geometric figures (Using similar triangles inside each cone to say $r$ is proportional to $h$.)7.G.B.6Solve real-world and mathematical problems involving area, volume and surface area (The cylinder approximation: rise $= $ volume / cross-section area, giving the $36/9 = 4$ ratio.)8.EE.A.1Know and apply the properties of integer exponents (Expressing each cone's new volume as $\tfrac{1}{3}\pi (r/h)^2 H^3$ to get a $H^3 - h^3$ relation.)8.EE.C.7Solve linear equations in one variable (Cancelling $\tfrac{1}{3}\pi$ in the equal-volume equation $9 h_n = 36 h_w$ to get $h_n = 4 h_w$.)
⭐ This AMC 10 problem only needs Grade 8 volume formulas and proportional reasoning you already know! Same liquid volume + a wider top means the narrow cone is $4 \times$ taller. Drop the SAME marble into both — the wider rim spreads the rise out over $4 \times$ the cross-section area, so the narrow cone's level jumps up $4 \times$ as much. Ratio $4:1$, answer (E).
⭐ This AMC 10 problem only needs Grade 8 volume formulas and proportional reasoning you already know! Same liquid volume + a wider top means the narrow cone is $4 \times$ taller. Drop the SAME marble into both — the wider rim spreads the rise out over $4 \times$ the cross-section area, so the narrow cone's level jumps up $4 \times$ as much. Ratio $4:1$, answer (E).
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