10. (10 points) There is a cup filled with a 15% saltwater solution. There are large, medium, and small iron balls, with volume ratios of 10: 5: 3. First, the small ball is submerged in the saltwater cup, causing 10% of the saltwater to overflow. The small ball is then removed. Next, the medium ball is submerged in the saltwater cup and then removed. Finally, the large ball is submerged in the saltwater cup and then removed. After that, pure water is added to the cup until it is full. What is the concentration of the saltwater in the cup at this point?
【Analysis】The overflow water volume is actually the volume of the large ball, which is $10 \% \times \frac{10}{3}=\frac{1}{3}$ of the entire cup of brine. Therefore, after filling with water, the concentration becomes $\frac{15 \times\left(1-\frac{1}{3}\right)}{100}$. Solve accordingly.
【Solution】Solution: $10 \% \times \frac{10}{3}=\frac{1}{3}$,
$$
\frac{15 \times\left(1-\frac{1}{3}\right)}{100}=\frac{10}{100}=10 \% \text {, }
$$

Answer: The concentration of the brine in the cup at this time is $10 \%$.