9. A chemistry student conducted an experiment: from a tank filled with syrup solution, he poured out several liters of liquid, refilled the tank with water, then poured out twice as much liquid and refilled the tank with water again. As a result, the amount of syrup in the tank decreased by $\frac{25}{3}$ times. Determine how many liters of liquid the student poured out the first time, if the tank's volume is 1000 liters.

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# Solution.

1) Let the syrup content in the initial solution be $p \%$ and let $x$ liters of the solution were poured out the first time.
2) Then after pouring out the liquid, there remained $(1000-x)$ liters of the solution, and in it $(1000-x) \cdot \frac{p}{100}$ liters of syrup and $(1000-x) \cdot \frac{100-p}{100}$ water.
3) After adding $x$ liters of water, there became: 1000 liters of the solution, and in it $(1000-x) \cdot \frac{p}{100}$ liters of syrup and $(1000-x) \cdot \frac{100-p}{100}+x$ liters of water.
4) At the end of all the pouring, there became: 1000 liters of the solution with a syrup content of $\frac{3 p}{25} \%$,

$$
3 p
$$

that is, $1000 \cdot \frac{\overline{25}}{100}=\frac{30 p}{25}$ liters of syrup and $1000-\frac{30 p}{25}$ liters of water.

5) Then before the last addition of $2 x$ liters of water, there was $(1000-2 x)$ liters of the solution in the tank, and in it $\frac{30 p}{25}$ liters of syrup and $1000-\frac{30 p}{25}-2 x$ water. This is the same liquid as in point 3) of the solution, so the ratio of syrup to liquid in it is the same. We set up the equation:

$$
\frac{(1000-x) \cdot \frac{p}{100}}{1000}=\frac{\frac{30 p}{25}}{1000-2 x} \Leftrightarrow 2 x^{2}-3000 x+\frac{22}{25} \cdot 1000^{2}=0 \Leftrightarrow
$$

$\Leftrightarrow x \in\{1100 ; 400\}, 1100$ liters does not satisfy the condition of the problem.

Answer: 400.

## Solution variant № 2