2012 AMC 10A Problem 8

1. Call the three numbers $x$, $y$, $z$, with $x + y = 12$, $x + z = 17$ and $y + z = 19$.
2. Every number shows up in exactly two of the three sums. So adding all three sums counts every number twice:
$$12 + 17 + 19 = 48 = 2(x + y + z).$$
3. Half of $48$ is $24$, so $x + y + z = 24$.
4. Each number is the total minus the pair that leaves it out:
- $z = 24 - (x + y) = 24 - 12 = 12$,
- $y = 24 - (x + z) = 24 - 17 = 7$,
- $x = 24 - (y + z) = 24 - 19 = 5$.
5. The numbers are $5$, $7$ and $12$. Check: $5 + 7 = 12$, $5 + 12 = 17$, $7 + 12 = 19$. The middle number is $7$.

The answer is $\boxed{\textbf{(D)}\ 7}$.

When every unknown appears the same number of times, adding all the equations gives the total in one step. After that, each unknown is the total minus a sum you already know. This is faster than substituting one equation into another, and it leaves less room for mistakes.

Think about which pair leaves out the middle number. The two smaller numbers make the smallest sum, $12$, and the two larger numbers make the largest sum, $19$. The pair that leaves out the middle number is the smallest plus the largest, which gives the sum in between, $17$. So the middle number is $24 - 17 = 7$.

Let the numbers be $x, y, z$ with $x + y = 12$, $x + z = 17$, $y + z = 19$.

Adding the three equations counts every number twice:
$$2(x + y + z) = 12 + 17 + 19 = 48, \quad\text{so}\quad x + y + z = 24.$$

Each number is the total minus the pair sum that excludes it:
$$z = 24 - 12 = 12, \qquad y = 24 - 17 = 7, \qquad x = 24 - 19 = 5.$$

The numbers are $5, 7, 12$, and the middle one is $7$.

The answer is $\boxed{\textbf{(D)}\ 7}$.