Śárka poured juice into a glass and a mug and filled both containers with water. The mug had twice the volume of the glass. The ratio of juice to water in the glass was $2: 1$ and in the mug $4: 1$. Then she poured the contents of the glass and the mug into a jug. What was the ratio of juice to water in the jug?

(L. Hozová)
If we denote the volume of the glass as $V$, then according to the problem, there was $\frac{2}{3} V$ of juice and $\frac{1}{3} V$ of water in the glass. The volume of the mug was twice the volume of the glass, i.e., $2 V$. The volume of juice in it was $\frac{4}{5} \cdot 2 V = \frac{8}{5} V$ and the volume of water in it was $\frac{1}{5} \cdot 2 V = \frac{2}{5} V$. The volume of juice in the jug was then

$\frac{2}{3} V + \frac{8}{5} V = \frac{10 + 24}{15} V = \frac{34}{15} V$. The volume of water in the jug was $\frac{1}{3} V + \frac{2}{5} V = \frac{5 + 6}{15} V = \frac{11}{15} V$. The desired ratio of juice to water in the jug was $\frac{34}{15} V : \frac{11}{15} V$, i.e., after simplification, $34: 11$.

Evaluation. 1 point for expressing the volumes in one container; 2 points for expressing the volumes in the second container using the same unknown as for the first container; 2 points for expressing the volumes in the jug; 1 point for the final ratio.

Note. In the case of the glass and the mug, we can express only the volume of one component. Only in the jug do we need to know the volume of both components. Then we can determine the volume of the second component by subtracting the volume of the first component from the total volume of the mixture, i.e., $\frac{34}{15} \mathrm{~V} = 3 \mathrm{~V} - \frac{11}{15} \mathrm{~V}$.