9. A chemistry student conducted an experiment: from a bottle filled with syrup solution, he poured out one liter of liquid, refilled the bottle with water, then poured out one liter of liquid again and refilled the bottle with water. As a result, the percentage content of syrup decreased from 36 to 1 percent. Determine the volume of the bottle in liters.
# Solution.

1) Let $x$ be the volume of the bottle in liters.
2) Then after pouring out one liter of liquid, there are $(x-1)$ liters of solution left, and in it $(x-1) \cdot \frac{36}{100}$ liters of syrup and $(x-1) \cdot \frac{64}{100}$ liters of water.
3) After adding one liter of water to the bottle, there are $x$ liters of solution, and in it $(x-1) \cdot \frac{36}{100}$ liters of syrup and $(x-1) \cdot \frac{64}{100}+1$ liters of water.
4) At the end of all the pouring, there are $x$ liters of solution in the bottle, and in it $x \cdot \frac{1}{100}$ liters of syrup and $x \cdot \frac{99}{100}$ liters of water.
5) Then before the last addition of one liter of water, there were $(x-1)$ liters of solution in the bottle, and in it $x \cdot \frac{1}{100}$ liters of syrup and $x \cdot \frac{99}{100}-1$ liters of water. This is the same liquid as in point 3) of the solution, so the ratio of syrup to water in it is the same.

We can set up the equation:

$$
\begin{aligned}
& \frac{(x-1) \cdot \frac{36}{100}}{x \cdot \frac{1}{100}}=\frac{(x-1) \cdot \frac{64}{100}+1}{x \cdot \frac{99}{100}-1} \Leftrightarrow \frac{(x-1) \cdot 36}{x \cdot 1}=\frac{(x-1) \cdot 64+100}{x \cdot 99-100} \Leftrightarrow \\
& \frac{36 x-36}{x}=\frac{64 x+36}{99 x-100} \Leftrightarrow(36 x-36)(99 x-100)=x(64 x+36) \Leftrightarrow \\
& 35 x^{2}-72 x+36=0 \Leftrightarrow \quad \Leftrightarrow \quad x \in\left\{1,2 ; \frac{6}{7}\right\} \cdot \frac{6}{7} \quad \text { liters does not satisfy the condition of the problem. }
\end{aligned}
$$

## Answer: 1,2 .