3.27 An alloy of copper and silver contains 1845 g more silver than copper. If a certain amount of pure silver, equal in mass to $\frac{1}{3}$ of the mass of pure silver originally contained in the alloy, were added to it, 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?

## Group B

## Ratios, Percentages
3.27 Let's construct the following table:

| Alloy | Mass, g |  |
| :--- | :---: | :---: |
|  | Silver | Copper |
| Initial | $x$ | $x-1845$ |
| New | $\frac{4 x}{3}$ | $x-1845$ |

Using the condition, we get the equation

$\frac{4 x}{3}=0.835\left(\frac{4 x}{3}+x-1845\right)$, or $4 x=0.835(7 x-5535)$,

from which

$4 x=5.845 x-0.835 \cdot 5535 ; 1.845 x=0.835 \cdot 5535$.

Thus, $x=\frac{0.835 \cdot 5535}{1.845}=2505(\mathrm{g})$.

Therefore, the mass of the initial alloy is $2505+2505-1845=3165$ (g) and in this alloy, it contains $\frac{2505}{3165} \cdot 100 \% \approx 79 \%$ silver.

Answer: $3165 \mathrm{g} ; \approx 79 \%$.