2013 AMC 10B Problem 1

The expression looks busy, but each fraction is just one small sum over another. Work those sums out first.

1. The first fraction has $2 + 4 + 6 = 12$ on top and $1 + 3 + 5 = 9$ on the bottom. So it is $\frac{12}{9}$. Dividing top and bottom by $3$ gives $\frac{4}{3}$.
2. The second fraction uses the same two sums, upside down: $\frac{9}{12} = \frac{3}{4}$.
3. Now find $\frac{4}{3} - \frac{3}{4}$. To subtract fractions they need the same bottom number. The smallest number that both $3$ and $4$ divide is $12$:
$$\frac{4}{3} = \frac{16}{12}, \qquad \frac{3}{4} = \frac{9}{12}.$$
4. So the answer is $\frac{16}{12} - \frac{9}{12} = \frac{7}{12}$. Since $7$ and $12$ share no common factor, this does not reduce.

Quick check: the first fraction is more than $1$ and the second is less than $1$, so the answer must be positive. That already rules out (A).

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

Simplifying each piece before combining keeps the numbers small: $\frac{4}{3}$ is much easier to use than $\frac{12}{9}$. A fraction minus its upside-down version has a shortcut worth remembering: for $\frac{4}{3} - \frac{3}{4}$, the top is $4 \times 4 - 3 \times 3 = 7$ and the bottom is $3 \times 4 = 12$.

Evaluate the two sums first: $2 + 4 + 6 = 12$ and $1 + 3 + 5 = 9$. The expression becomes
$$\frac{12}{9} - \frac{9}{12} = \frac{4}{3} - \frac{3}{4}.$$
Over the common denominator $12$:
$$\frac{16}{12} - \frac{9}{12} = \frac{7}{12}.$$

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