jmc

We begin by realizing that $0.\overline{1}=0.\overline{11}$, so $0.\overline{1}+0.\overline{01}=0.\overline{11}+0.\overline{01}=0.\overline{12}$. (Note that this can be done because there is no carrying involved.)

To express the number $0.\overline{12}$ as a fraction, we call it $x$ and subtract it from $100x$: \begin{array}{r r c r@{}l} &100x &=& 12&.121212\ldots \\ - &x &=& 0&.121212\ldots \\ \hline &99x &=& 12 & \end{array} This shows that $0.\overline{12}=\frac{12}{99}$.

But that isn't in lowest terms, since $12$ and $99$ share a common factor of $3$. We can reduce $\frac{12}{99}$ to $\boxed{\frac{4}{33}}$, which is in lowest terms.