13.182. Water is poured into two vessels of the same mass, with the mass of vessel $A$ with water being $4 / 5$ of the mass of vessel $B$ with water. If the water from vessel $B$ is poured into vessel $A$, the mass of vessel $A$ with water will become 8 times the mass of vessel $B$. Find the mass of the vessels and the amount of water in them, knowing that vessel $B$ contains 50 g more water than vessel $A$.
Solution.

Let $x$ g be the mass of the vessels. The mass of vessel $B$ with water is $y$ g, and the mass of vessel $A$ with water is $\frac{4}{5} y$ g. The mass of water in vessel $A$ is $\left(\frac{4}{5} y - x\right)$ g, and in $B$ is $(y - x)$ g. According to the problem, we have $\left\{\begin{array}{l}\frac{4}{5} y + y - x = 8 x, \\ y - x - \left(\frac{4}{5} y - x\right) = 50,\end{array}\right.$ from which $x = 50$ g, $y = 250$ g. In vessel $A$ there is 150 g of water, and in $B$ there is 200 g.

Answer: 50, 150, and 200 g.