2002 AMC 10A Problem 16

Five expressions are all equal, so give their common value a name and write everything in terms of it.

1. Name it. Let $x$ be the common value. Then
$$a + 1 = x,\quad b + 2 = x,\quad c + 3 = x,\quad d + 4 = x,$$
so $a = x - 1$, $b = x - 2$, $c = x - 3$, $d = x - 4$.

2. Add them up:
$$a + b + c + d = 4x - (1 + 2 + 3 + 4) = 4x - 10.$$

3. Use the last expression. It says $a + b + c + d + 5 = x$, so
$$a + b + c + d = x - 5.$$

4. Set the two descriptions equal.
$$4x - 10 = x - 5 \quad\Longrightarrow\quad 3x = 5 \quad\Longrightarrow\quad x = \frac{5}{3}.$$

5. Answer the actual question. We want the sum, not $x$:
$$a + b + c + d = x - 5 = \frac{5}{3} - \frac{15}{3} = -\frac{10}{3}.$$

The answer is $\boxed{\textbf{(B)}\ -10/3}$.

A long chain of equal expressions becomes easy once the shared value has a name: each unknown is that value shifted by a number. The last expression, which contains the sum itself, provides the one equation needed. Always finish by answering the question asked; here $x$ itself is a trap choice.

Compare each of the first four expressions with the last one. From $a + 1 = a + b + c + d + 5$, the $a$'s cancel, giving $b + c + d = -4$. In the same way, $a + c + d = -3$, $a + b + d = -2$, and $a + b + c = -1$. Adding these four equations, each letter appears three times:
$$3(a + b + c + d) = -10, \quad\text{so}\quad a + b + c + d = -\frac{10}{3}.$$

Let $x$ be the common value of all five expressions. Then
$$a = x - 1,\quad b = x - 2,\quad c = x - 3,\quad d = x - 4,$$
so $a + b + c + d = 4x - 10$.

The last expression says $a + b + c + d + 5 = x$, i.e. $a + b + c + d = x - 5$. Equating the two descriptions of the sum:
$$4x - 10 = x - 5 \quad\Longrightarrow\quad 3x = 5 \quad\Longrightarrow\quad x = \frac{5}{3}.$$

Therefore $a + b + c + d = x - 5 = \frac{5}{3} - 5 = -\frac{10}{3}$.

The answer is $\boxed{\textbf{(B)}\ -10/3}$.