jmc

The given information can be expressed by writing $x\equiv 9^{-1}(\mathrm{mod}25)$. Thus we wish to compute $11+9^{-1}(\mathrm{mod}25)$.

Modulo $25$, we can write $11$ as $11\cdot (9\cdot 9^{-1})\equiv (11\cdot 9)\cdot 9^{-1}\equiv 99\cdot 9^{-1}$. Thus $$\begin{array}{rl}11+9^{-1}& \equiv 99\cdot 9^{-1}+1\cdot 9^{-1}\\ & \equiv 100\cdot 9^{-1}\\ & \equiv 0\cdot 9^{-1}\\ & \equiv 0(\mathrm{mod}25),\end{array}$$so the remainder when $11+x$ is divided by $25$ is $\boxed{0}$.

Notice that the trick we used here is analogous to using a common denominator to add fractions.