13.443 In a flask, there is a salt solution. From the flask, $1 / n$ part of the solution is poured into a test tube, and the solution remaining in the flask is evaporated until the percentage of salt content doubles. After that, the solution from the test tube is poured back into the flask. As a result, the salt content in the solution increased by $p \%$ compared to the initial. Determine the percentage of salt in the initial solution. What part of the initial solution should have been poured out so that, as a result of the described procedure, the percentage of salt content increased by one and a half times?
Solution. Let $x$ be the original percentage of salt, and $y$ be the volume of the solution. After removing $\frac{1}{n}$ of the solution, $\frac{n-1}{n} \cdot y$ of the solution remains in the flask. Since the percentage of salt in the flask increased by two times after the water evaporated, the volume of the solution was halved, i.e., it became $\frac{n-1}{2 n} \cdot y$. After returning the solution from the test tube, the volume became $\left(\frac{n-1}{2 n}+\frac{1}{n}\right) y=\frac{n+1}{2 n} \cdot y$. However, the amount of salt remained the same. Therefore, $y \cdot \frac{x}{100}=\frac{n+1}{2 n} \cdot y \cdot \frac{(x+p)}{100} \Rightarrow x=\frac{n+1}{n-1} \cdot p$.

Similarly, if the percentage of salt in the test tube became $\frac{3}{2} x$ after the same actions, then $x y=\frac{n+1}{2 n} \cdot y \cdot \frac{3}{2} x \Rightarrow n=3$.

Answer: $\frac{n+1}{n-1} \cdot p \% ; \frac{1}{3}$ part.