10.1. Salt was added to the water twice, with the percentage increase in the mass of the solution being the same each time. What was the initial mass of the water and what was the percentage increase in the mass of the solution each time, if the final mass of the solution was 850 g and the final concentration was $36 \%$?
Answer. The initial mass of water is 544 g, the mass increase each time is $25 \%$.

Solution. Let the initial mass of water be $x$ g, and the mass increase of the solution each time be $p \%$. After the second time, the mass will be $\left(1+\frac{p}{100}\right)^{2} x$, and the mass of salt in it $\left(\left(1+\frac{p}{100}\right)^{2}-1\right) x$. Then $\left(\left(1+\frac{p}{100}\right)^{2}-1\right) x=\frac{36}{100}\left(1+\frac{p}{100}\right)^{2} x$, from which $p=25$. Therefore, $\frac{25}{16} x=850, x=544$.