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2002 AMC 10A Problems/Problem 9

Revision as of 21:18, 26 December 2008 by Xpmath (talk | contribs) (New page: == Problem == There are 3 numbers A, B, and C, such that <math>1001C - 2002A = 4004</math>, and <math>1001B + 3003A = 5005</math>. What is the average of A, B, and C? <math>\text{(A)}\ 1...)
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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?

$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}$ More than 1

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 $\boxed{\text{(B)}\ 3}$.

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