2002 AMC 10A Problems/Problem 9

Problem

There are 3 numbers A, B, and C, such that $1001C - 2002A = 4004$, and $1001B + 3003A = 5005$. What is the average of A, B, and C?

$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}$ Not uniquely determined

Solution

Notice that we don't need to find what A, B, and C actually are, just their average. In other words, if we can find A+B+C, we will be done.

Adding up the equations gives $1001(A+B+C)=1001(9)$ so $A+B+C=9$ and the average is $\frac{9}{3}=3$. Our answer is $\boxed{\text{(B)}\ 3}$.

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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