13.093. An alloy of copper and silver contains 1845 g more silver than copper. If a certain amount of pure silver, equal to $1 / 3$ of the mass of pure silver originally contained in the alloy, were added to it, then a new alloy would be obtained, containing $83.5\%$ silver. What is the mass of the alloy and what was the original percentage of silver in it?
## Solution.

Let $x$ g be the mass of silver, $x-1845$ g be the mass of copper. Then $2x - 1845$ g is the mass of the entire alloy. After adding silver, its mass will become $x + \frac{1}{3}x$ g, and the mass of the alloy will be $2x - 1845 + \frac{1}{3}x$ g; $x + \frac{1}{3}x$ g make up $83.5\%$. Therefore, the mass of the alloy is $\left(x + \frac{1}{3}x\right) : 0.835$. Then $\left(x + \frac{1}{3}x\right) : 0.835 = 2x - 1845 + \frac{1}{3}x$, from which $x = 2505$ g is the mass of silver. The mass of the entire alloy is $2 \cdot 2505 - 1845 = 3165$ g; 3165 g make up $100\%$, 2505 g of silver make up $y\%$; $\frac{3165}{2505} = \frac{100}{y}$, hence $y \approx 79.1\%$.

Answer: $3165$ g; $\approx 79.1\%$.