2002 AMC 10A Problem 9

We are asked for the average, which only needs the sum $A + B + C$. So instead of finding each number, look for a way to get the sum directly.

1. Make the numbers smaller. Every number in the equations is a multiple of $1001$ ($2002 = 2 \times 1001$, $3003 = 3 \times 1001$, and so on). Divide each equation by $1001$:
$$C - 2A = 4, \qquad B + 3A = 5.$$

2. Add the two equations. On the left, the $A$ parts combine: $-2A + 3A = A$. So
$$A + B + C = 4 + 5 = 9.$$
That is exactly the sum we wanted.

1. Average: $\dfrac{9}{3} = 3$.

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

Two equations cannot pin down three separate numbers, but they can still pin down one combination of them, such as their sum. When a question asks for a sum or an average, try adding or subtracting the given equations before solving for anything. Here the problem was built so that simply adding them works.

Since the sum is all that matters, pick one number freely and see what happens. Let $C = 0$. Then $-2A = 4$ gives $A = -2$, and $B + 3(-2) = 5$ gives $B = 11$. The sum is $-2 + 11 + 0 = 9$ and the average is $3$. Any other choice of $C$ gives different numbers but the same sum.

We need only $A + B + C$, not the individual values. Every coefficient is a multiple of $1001$, so divide each equation by $1001$:
$$C - 2A = 4, \qquad B + 3A = 5.$$

Adding these two equations makes the $A$ coefficients combine to $-2 + 3 = 1$:
$$A + B + C = 9.$$

The average is $\frac{9}{3} = 3$.

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