2002 AMC 10A Problem 9

Below is the professionally curated solution for Problem 9 of the 2002 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10A solutions, or check the answer key.

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Concepts:system of equations mean

Difficulty rating: 1170

Suppose $A,$ $B,$ and $C$ are three numbers for which $1001C-2002A=4004$ and $1001B+3003A=5005.$ The average of the three numbers $A,$ $B,$ and $C$ is

$1$

$3$

$6$

$9$

not uniquely determined

Solution:

Adding the equations, $1001C-2002A+1001B+3003A=1001A+1001B+1001C=9009.$

So $A+B+C=9$ and the average is $\dfrac{9}{3}=3.$

Thus, the correct answer is B.

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