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 $\frac{8}{3}$ 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 original solution be $p \%$ and let $\mathcal{X}$ liters of the solution be poured out the first time.
2) Then after pouring out the liquid, there are $(1000-x)$ liters of solution left, 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 are 1000 liters of solution in the tank, 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 are 1000 liters of solution in the tank with a syrup content of $\frac{3 p}{8} \%$, which is $1000 \cdot \frac{\frac{3 p}{8}}{100}=\frac{30 p}{8}$ liters of syrup and $1000-\frac{30 p}{8}$ liters of water.
5) Then before the last addition of $2 x$ liters of water, there were $(1000-2 x)$ liters of solution in the tank, and in it $\frac{30 p}{8}$ liters of syrup and $1000-\frac{30 p}{8}-2 x$ 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 can set up the equation:

$$
\frac{(1000-x) \cdot \frac{p}{100}}{1000}=\frac{\frac{30 p}{8}}{1000-2 x} \Leftrightarrow 2 x^{2}-3000 x+\frac{5}{8} \cdot 1000^{2}=0 \Leftrightarrow
$$

$\Leftrightarrow x \in\{1250 ; 250\}, 1250$ liters does not satisfy the condition of the problem.

Answer: 250.

## Solution variant № 2

1. If a two-digit natural number is decreased by 27, the result is a two-digit number with the same digits but in reverse order. In the answer, indicate the median of the numerical sequence formed from all such numbers.

## Solution.

Let $\overline{x y}=10 x+y$ be the original two-digit number, and $\overline{y x}=10 y+x$ be the number with the digits in reverse order. We get the equation $10 x+y=10 y+x+27$. From the equation, it is clear that the two-digit number is greater than 27. Let's start the investigation with the tens digit equal to 3.

| $\boldsymbol{x}$ | equation | $\boldsymbol{y}$ | number |
| :--- | :--- | :---: | :--- |
| 3 | $30+y=10 y+3+27$ | $y=0$ | 30 does not fit the condition |
| 4 | $40+y=10 y+4+27$ | $y=1$ | 41 |
| 5 | $50+y=10 y+5+27$ | $y=2$ | 52 |
| 6 | $60+y=10 y+6+27$ | $y=3$ | 63 |
| 7 | $70+y=10 y+7+27$ | $y=4$ | 74 |
| 8 | $80+y=10 y+8+27$ | $y=5$ | 85 |
| 9 | $90+y=10 y+9+27$ | $y=6$ | 96 |

These can be the numbers $41,52,63,74,85,96$. The median of the sequence is 68.5.

Answer: 68.5.