2021 AMC 10A Problem 12

Let the liquid heights be $h_1$ (narrow cone, radius $3$) and $h_2$ (wide cone, radius $6$). Equal volumes give
$$\tfrac13\pi \cdot 9 h_1 = \tfrac13 \pi \cdot 36 h_2 \quad\Longrightarrow\quad h_1 = 4h_2.$$

Each body of liquid is itself a cone similar to its container, so its volume is proportional to the cube of its height. After the marble is added, both liquids have the same new volume $V + \frac{4}{3}\pi$, so both volumes grow by the same factor $t = \frac{V + 4\pi/3}{V}$, and both heights grow by the same factor $k = t^{1/3}$.

The rises are $kh_1 - h_1 = (k-1)h_1$ and $(k-1)h_2$, whose ratio is
$$\frac{h_1}{h_2} = 4.$$

The answer is $\boxed{\textbf{(E)}\ 4:1}$.

Liquid in an inverted cone always forms a cone similar to the container, so height is a function of volume alone, scaling as the cube root. Two similar situations with the same volume before and the same volume after keep the same height ratio throughout, so the rises inherit the original height ratio. The marble's exact volume never matters.

Assign numbers: let the narrow cone hold liquid of height $4$ and the wide cone height $1$ (radii $3$ and $6$ give volumes $12\pi$ and $12\pi$). Adding volume $\frac{4}{3}\pi$ multiplies each volume by $\frac{10}{9}$, so each height becomes $\sqrt[3]{10/9}$ times larger; the rises are $4(\sqrt[3]{10/9} - 1)$ and $\sqrt[3]{10/9} - 1$, ratio $4:1$.

• Trying to compute the rise for each cone separately with the marble's volume, or assuming the rises are inversely proportional to the surface areas (4:1 by luck, but the reasoning fails in general).
• Treating the cones as cylinders; that also happens to give $4:1$ here but for the wrong reason and would fail on a variant with unequal starting volumes.