Difference between revisions of "2002 AMC 10A Problems/Problem 9"
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There are 3 numbers A, B, and C, such that <math>1001C - 2002A = 4004</math>, and <math>1001B + 3003A = 5005</math>. What is the average of A, B, and C? | There are 3 numbers A, B, and C, such that <math>1001C - 2002A = 4004</math>, and <math>1001B + 3003A = 5005</math>. What is the average of A, B, and C? | ||
− | <math>\ | + | <math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E) }\text{Not uniquely determined}</math> |
==Solution== | ==Solution== |
Revision as of 11:51, 8 November 2021
Problem
There are 3 numbers A, B, and C, such that , and . What is the average of A, B, and C?
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 so and the average is . Our answer is .
See Also
2002 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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