2002 AMC 10A Problem 9

Attempt Problem 9 of the 2002 AMC 10A below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

1

3

6

9

not uniquely determined

Add the two equations together

The sum collapses to a multiple of

A+B+C

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.

Problem 9 in other years