2002 AMC 10A Problem 9

Problem 9 of 25EasierAlgebra

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

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Solution

Adding the equations,

\begin{gathered} 1001C-2002A+1001B \\ {}+3003A \\ = 1001A+1001B+1001C \\ = 9009. \end{gathered}

A+B+C=9

and the average is

\dfrac{9}{3}=3.

Thus, the correct answer is B.

Concepts: system of equations · mean

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.