## Solution.

Let $c$ be the proportion of juice in $x$ liters of the mixture.

| first container |  |  |  |  |
| :---: | :---: | :---: | :---: | :---: |
| transfer | proportion of juice | quantity $/ \mathrm{L}$ | proportion of juice | quantity $/ \mathrm{L}$ |
| start | $c_{0}=1$ | $x$ | $c_{0}=1$ | $30-x$ |
| 1. | $c_{1}=\frac{x}{30}$ | 30 | $c_{0}=1$ | $30-x$ |
| 2. | $c_{1}=\frac{x}{30}$ | $30-x$ | $c_{2}=\frac{c_{0} \cdot(30-x)+c_{1} \cdot x}{30}$ | 30 |
| 3. | $c_{3}=\frac{c_{1} \cdot(30-x)+12 c_{2}}{42-x}$ | $42-x$ | $c_{2}=\frac{30-x+c_{1} \cdot x}{30}$ | 18 |

After all the transfers, the amount of juice in the first container is $c_{3} \cdot(42-x)=c_{1} \cdot(30-x)+12 c_{2}$, and in the second container $c_{2} \cdot 18=18 c_{2}$.

According to the conditions of the problem, for the final amount of juice, we have:

$c_{1} \cdot(30-x)+12 c_{2}=2+18 c_{2}$

$c_{1} \cdot(30-x)=2+6 c_{2}$

$\frac{x}{30} \cdot(30-x)=2+6 \cdot \frac{30-x+c_{1} \cdot x}{30}$

$x(30-x)=60+6\left(30-x+\frac{x}{30} \cdot x\right)$

$30 x-x^{2}=60+180-6 x+\frac{x^{2}}{5}$

$x^{2}-30 x+200=0$

$x_{1}=10 \quad x_{2}=20$

There cannot be 10 liters of juice in the first container, as this would result in 32 L of mixture in the first container at the end of the transfer.

Therefore, there were $20 \mathrm{~L}$ in the first container and $10 \mathrm{~L}$ of orange juice in the second container at the start.
## Task B-2.3.

Two equal containers, each of 30 liters, contain a total of 30 liters of freshly squeezed orange juice. In the first container, we add water to the top, and then we fill the second container to the top with the resulting mixture of juice and water. Then, we pour 12 liters of the new mixture of juice and water from the second container into the first container. Now, there are 2 liters less of freshly squeezed juice in the second container than in the first container. How much juice was there in each container at the beginning?