The surface area ratio of a certain right circular cone to a sphere is 3:1. How many times larger is the volume of the right circular cone compared to the volume of the sphere?
Let the radius of the base of the right circular cone be $R$, and the radius of the sphere be $r$. The surface area of the cone is $3 R^{2} \pi$; the surface area of the sphere is $4 r^{2} \pi$; according to the problem, therefore,

$$
3 R^{2} \pi: 4 r^{2} \pi=3: 1
$$

from which

$$
R=2 r
$$

The volume of the right circular cone is

$$
K=\frac{\pi R^{3} \sqrt{3}}{3}=\frac{8 \pi r^{3} \sqrt{3}}{3}
$$

The volume of the sphere is

$$
k=\frac{4 r^{3} \pi}{3}
$$

and thus

$$
K: k=2 \sqrt{3}: 1
$$

so

$$
K=2 \sqrt{3} k
$$

(Lajos Kertész.)

The problem was also solved by: A. Erdős, E. Freibauer, B. Friedmann, A. Hrivzák, Gy. Krátky, E. Manheim, E. Orlowszky, G. Petrogalli, Gy. Porde, H. Schiffer, Ö. Spitzer, K. Szabó, J. Weisz.