3.1. [5-6.3 (20 points)] In three flasks, there is concentrated acid: 10 g in the first, 20 g in the second, and 30 g in the third. There is also a fourth flask with water. If some amount of water from the fourth flask is added to the first flask, and the remaining water is poured into the second flask, then in the first flask, the acid will constitute $\frac{1}{20}$ of the mixture, and in the second flask, the fraction of acid will be $\frac{7}{30}$. What fraction of the mixture will the acid constitute in the third flask if all the water from the fourth flask is poured into it?
Answer: $\frac{21}{200}=0.105$.

Solution. If the acid constitutes $\frac{1}{20}$ of the first flask, then it contains $10: \frac{1}{20}=200$ g of the solution, so there are 190 g of water in it. Similarly, for the second flask, we find that the total solution in it is $20: \frac{7}{30}=\frac{600}{7}$ g, and the water is $\frac{600}{7}-20=\frac{460}{7}$ g. Therefore, in the fourth flask, there is $190+\frac{460}{7}=\frac{1790}{7}$ g of water. Then the fraction of acid in the third flask after mixing with this amount of water will be

$$
\frac{30}{30+\frac{1790}{7}}=\frac{210}{2000}=\frac{21}{200}=0.105
$$