A truncated cone-shaped vase, widening upwards, can be filled to the brim with 2.6 liters of water, and to half its height with 0.7 liters of water. How many times larger is the radius of the top circle compared to the radius of the base circle?
Let's write down the volume:

$$
\begin{aligned}
& 2.6=\frac{m \pi\left(R^{2}+R r+r^{2}\right)}{3} \\
& 0.7=\frac{m}{2} \cdot \frac{\pi\left(x^{2}+r x+r^{2}\right)}{3}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_f23ebc606fde2aca1b6bg-1.jpg?height=458&width=570&top_left_y=497&top_left_x=772)

Since we filled the vase with water to half, the radius of the smaller frustum's base is \( x=\frac{R+r}{2} \). Substitute this into the second equality, then take the ratio of the two volumes. After rearranging, we get

$$
\frac{2.6}{0.7}=\frac{8\left(r^{2}+R r+R^{2}\right)}{7 r^{2}+4 R r+R^{2}}
$$

Divide every term in the numerator and the denominator by \( r^{2} \), and introduce the new variable \( \frac{R}{r}=a \); thus, we arrive at the following quadratic equation in \( a \):

$$
5 a^{2}-8 a-21=0
$$

The roots of the equation are: \( a_{1}=3 ; a_{2}=-1.4<0 \) (not a solution).

Our answer: \( R=3 r \).