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2002 AMC 10A Problem 9

Problem 9 of 25EasierAlgebraProbability & Statistics

Suppose A,A, B,B, and CC are three numbers for which 1001C−2002A=40041001C-2002A=4004 and 1001B+3003A=5005.1001B+3003A=5005. The average of the three numbers A,A, B,B, and CC is

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Solution

Adding the equations, 1001C−2002A+1001B+3003A=1001A+1001B+1001C=9009. \begin{gathered} 1001C-2002A+1001B \\ {}+3003A \\ = 1001A+1001B+1001C \\ = 9009. \end{gathered} So A+B+C=9A+B+C=9 and the average is 93=3.\dfrac{9}{3}=3. Thus, the correct answer is B.
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