2011 AMC 10B Problem 1

Each fraction is built from small sums, so work those out before anything else.

1. The even numbers: $2 + 4 + 6 = 12$.
2. The odd numbers: $1 + 3 + 5 = 9$.
3. Now the problem reads
$$\frac{12}{9} - \frac{9}{12}.$$
4. Simplify each fraction first to keep the numbers small: $\frac{12}{9} = \frac{4}{3}$ and $\frac{9}{12} = \frac{3}{4}$.
5. To subtract, put both over $12$: $\frac{4}{3} = \frac{16}{12}$ and $\frac{3}{4} = \frac{9}{12}$.
6. Then
$$\frac{16}{12} - \frac{9}{12} = \frac{7}{12}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{7}{12}}$.

The second fraction is just the first one flipped over. A number minus its flip, like $\frac{4}{3} - \frac{3}{4}$, is found by writing both over the same bottom number. Doing the small sums first and simplifying right away keeps every step easy to check.

Add the numerators and denominators first: $2+4+6 = 12$ and $1+3+5 = 9$. The expression becomes
$$\frac{12}{9} - \frac{9}{12} = \frac{4}{3} - \frac{3}{4}.$$
With common denominator $12$ this is
$$\frac{16}{12} - \frac{9}{12} = \frac{7}{12}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{7}{12}}$.