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?

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

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 $1001(A+B+C)=9009=1001(9)$ so $A+B+C=9$ and the average is $\frac{9}{3}=3$ . Our answer is $\boxed{\textbf{(B) }3}$ .

Solution 2

As there are only 2 equations and 3 variables, just set one of the variables to something convenient, like 0. Setting C as 0, we get A is -2, and B is 11 by substitution. By basic arithmetic the average is 3=>B

-dragoon

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AMC 10 Problems and Solutions

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2002 AMC 10A Problems/Problem 9

Contents

Problem

Solution

Solution 2

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