6. A sector is cut out of a paper circle, as shown. The two resulting pieces of paper are "curled" to make two cones. The sum of the circumferences of the bases of these cones equals the circumference of the original paper circle. If the ratio of the lateral surface areas of the cones is $2: 1$, what is the ratio of the volumes of the cones?
(The lateral surface area of a cone is the area of the external surface not including the circular base. A cone with radius $r$, height $h$ and slant height $s$ has lateral surface area equal to $\pi r s$ and volume equal to $\frac{1}{3} \pi r^{2} h$.)
6. Suppose that the original circle has radius $R$.
Thus, the circumference of this circle is $2 \pi R$.
When this circle is cut into two pieces and each piece is curled to make a cone, the ratio of the lateral surface areas of these cones is $2: 1$.
This means that the ratio of the areas of the two pieces into which the circle is cut is $2: 1$ since it is these pieces that become the lateral surfaces of the cones.
In other words, the sector cut out is $\frac{1}{3}$ of the area of the circle, which means that its central angle is $\frac{1}{3}$ of the total angle around the centre, or $\frac{1}{3} \times 360^{\circ}=120^{\circ}$.
Since the central angles of the two pieces are $240^{\circ}$ and $120^{\circ}$, which are in the ratio $2: 1$, then the circumference is split in the ratio $2: 1$ when the circle is cut.
Since the circumference of the original circle is $2 \pi R$, then the lengths of the pieces of the circumference are $\frac{4}{3} \pi R$ and $\frac{2}{3} \pi R$.
These pieces become the circumferences of the circular bases of the two cones.
Since the ratio of the circumference to the radius of a circle is $2 \pi: 1$, then the radii of the bases of the two cones are $\frac{\frac{4}{3} \pi R}{2 \pi}=\frac{2}{3} R$ and $\frac{\frac{2}{3} \pi R}{2 \pi}=\frac{1}{3} R$.
Since the radius of the original circle becomes the slant height of each cone, then the slant height in each cone is $R$.
In a cone, the radius and the height are perpendicular forming a right-angled triangle with the slant height as its hypotenuse.
Therefore, the height of a cone with slant height $R$ and radius $\frac{2}{3} R$ is
$$
\sqrt{R^{2}-\left(\frac{2}{3} R\right)^{2}}=\sqrt{R^{2}-\frac{4}{9} R^{2}}=\sqrt{\frac{5}{9} R^{2}}=\frac{\sqrt{5}}{3} R
$$

Also, the height of a cone with slant height $R$ and radius $\frac{1}{3} R$ is
$$
\sqrt{R^{2}-\left(\frac{1}{3} R\right)^{2}}=\sqrt{R^{2}-\frac{1}{9} R^{2}}=\sqrt{\frac{8}{9} R^{2}}=\frac{\sqrt{8}}{3} R
$$

The volume of a cone with radius $\frac{2}{3} R$ and height $\frac{\sqrt{5}}{3} R$ is $\frac{1}{3} \pi\left(\frac{2}{3} R\right)^{2}\left(\frac{\sqrt{5}}{3} R\right)$ which equals $\frac{4 \sqrt{5}}{81} \pi R^{3}$. The volume of a cone with radius $\frac{1}{3} R$ and height $\frac{\sqrt{8}}{3} R$ is $\frac{1}{3} \pi\left(\frac{1}{3} R\right)^{2}\left(\frac{\sqrt{8}}{3} R\right)$ which equals $\frac{\sqrt{8}}{81} \pi R^{3}$. Dividing the first volume by the second, we obtain
$$
\frac{\frac{4 \sqrt{5}}{81} \pi R^{3}}{\frac{\sqrt{8}}{81} \pi R^{3}}=\frac{4 \sqrt{5}}{\sqrt{8}}=\frac{4 \sqrt{5}}{2 \sqrt{2}}=\frac{4 \sqrt{5} \sqrt{2}}{4}=\sqrt{5} \sqrt{2}=\sqrt{10}
$$

Therefore, the ratio of the larger volume to the smaller volume is $\sqrt{10}: 1$.
ANSWER: $\sqrt{10}: 1$