smc

The magnitude of the loss after the first sale is $C-100$, which equals $x$% of the selling price, $\text{textdollar}100$. Thus, $C-100=100\ast \frac{x}{100}$, and so $C=x+100$. The profit made after the second sale, $S^{\prime }-100$, is $x$% of the new selling price, and this quantity is represented by $S^{\prime }\ast \frac{x}{100}$. Equating these two expressions, we see that $S^{\prime }-100=S^{\prime }\ast \frac{x}{100}$, and so $S^{\prime }=\frac{10,000}{100-x}$. Because we know that the difference between $S^{\prime }$ and $C$ is $\frac{10}{9}$, we can solve the following equation: $$\begin{array}{rl}& \\ \frac{10,000}{100-x}-(x+100)& =\frac{10}{9}\\ 10,000-(100+x)(100-x)& =\frac{10}{9}(100-x)\\ 10,000-(10,000-x^2)& =\frac{1000}{9}-\frac{10}{9}x\\ x^2+\frac{10}{9}x-\frac{1000}{9}& =0\\ 9x^2+10x-1000& =0\\ (9x+100)(x-10)& =0\end{array}$$ Because $x\ge 0$, $x=10$, and so we choose answer $\boxed{10}$.