13.094. In 500 kg of ore, there is a certain amount of iron. After removing 200 kg of impurities from the ore, which on average contain $12.5\%$ iron, the iron content in the remaining ore increased by $20\%$. How much iron is left in the ore?
Solution.

In 200 kg of impurities, there will be $x=\frac{200 \cdot 12.5}{100}=25$ kg of iron. Initially, $z$ kg of iron constituted $y \%$ of 500 kg of ore, i.e.,
$z=\frac{500 \cdot y}{100}=5 y$ kg. In the remaining ore, $(5 y-25)$ kg of iron constitutes $(y+20) \%$. Thus, $\frac{5 y-25}{300}=\frac{y+20}{100} \Rightarrow 2 y=85, y=\frac{85}{2}$. In the ore, there will be $5 y-25=\frac{425}{2}-25=212.5-25=187.5$ kg of iron left.

Answer: 187.5 kg.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-851.jpg?height=416&width=666&top_left_y=466&top_left_x=121)

Fig. 13.7