2014 AMC 10A Problem 1

The power $-1$ sits on the whole bracket, so the bracket has to become one number before anything is flipped.

1. Rewrite each fraction with denominator $10$, because $10$ is a multiple of $2$ and $5$:
$$\frac{1}{2} = \frac{5}{10}, \qquad \frac{1}{5} = \frac{2}{10}, \qquad \frac{1}{10} = \frac{1}{10}.$$
2. Add them: $\frac{5}{10} + \frac{2}{10} + \frac{1}{10} = \frac{8}{10}$, which simplifies to $\frac{4}{5}$.
3. A power of $-1$ means "take the reciprocal", that is, flip the fraction upside down. So $\left(\frac{4}{5}\right)^{-1} = \frac{5}{4}$.
4. Multiply by $10$: $10 \cdot \frac{5}{4} = \frac{50}{4} = \frac{25}{2}$.

The answer is $\boxed{\textbf{(C)}\ \frac{25}{2}}$.

Flipping a sum is not the same as adding the flipped pieces. For example, $\left(\frac12 + \frac12\right)^{-1} = 1^{-1} = 1$, but $2 + 2 = 4$. So whenever a power of $-1$ sits on a bracket with a sum inside, finish the sum first, then flip once.