Let $n$ be the smaller integer. The sum of the reciprocals of the consecutive integers can be expressed as $\frac{1}{n} + \frac{1}{n+1} = \frac{11}{30}$. Clearly, a denominator of 30 is convenient, so we multiply both sides of the equation by 30 to get $30 \cdot \left(\frac{1}{n} + \frac{1}{n+1}\right) = 11$ and \[\frac{30}{n} + \frac{30}{n+1} = 11.\]It is easy to see by guess-and-check that $\boxed{\textbf{(A) } 5 \text{ and } 6}$ are the two integers.
