2021 AMC 10A Problem 12

Let the initial liquid heights in the narrow and wide cones be $h_1$ and $h_2.$ Since the liquid volumes are equal, $$\frac13\pi(3)^2h_1=\frac13\pi(6)^2h_2,$$ so $h_1=4h_2.$ After the identical marbles are dropped in, each cone must contain the same final volume below the liquid surface: the original liquid volume plus the volume of one marble. If the new liquid-surface radii are $3x$ and $6y,$ similarity gives new heights $h_1x$ and $h_2y.$ Thus $$\frac13\pi(3x)^2h_1x=\frac13\pi(6y)^2h_2y.$$ Using $h_1=4h_2,$ this simplifies to $x^3=y^3,$ so $x=y.$ The rise ratio is therefore $$\begin{aligned} h_1(x-1):h_2(y-1) &= h_1:h_2 \\ &= 4:1. \end{aligned}$$ Thus, E is the correct answer.