1. For chemical experiments, two identical test tubes were taken, each containing 200 ml of a liquid substance. From the first test tube, $1 / 4$ of the content was poured out and the same amount of water was added, then this procedure was repeated 3 more times, each time pouring out a quarter of the content of the test tube and adding the same amount of water. A similar procedure was performed twice for the second test tube, each time pouring out a certain amount of the content of the test tube and adding the same amount of water. In the end, the concentration of the resulting mixtures in the first and second test tubes became related to each other as 9/16. Determine how much of the mixture was poured out each time from the second test tube.

(12 points)
Solution. The initial amount of the substance is $-V$. After pouring out $a$ part, the concentration of the substance in the test tube becomes $\frac{V-a}{V}$. After the second time, the concentration is $\left(\frac{V-a}{V}\right)^{2}$, and after the fourth time, it is $\left(\frac{V-a}{V}\right)^{4}$. Substituting the data from the problem, we get the ratio $\left(\frac{200-50}{200}\right)^{4}:\left(\frac{200-x}{200}\right)^{2}=\frac{9}{16}$, or $\left(\frac{3}{4}\right)^{4}:\left(\frac{200-x}{200}\right)^{2}=\frac{9}{16} \Rightarrow\left(\frac{3}{4}\right)^{2}:\left(\frac{200-x}{200}\right)^{2}=1 \Rightarrow \frac{200-x}{200}=\frac{3}{4} \Rightarrow x=50$

Answer: 50 ml.