Difference between revisions of "2002 AMC 10A Problems/Problem 9"

Revision as of 21:19, 26 December 2008

There are 3 numbers A, B, and C, such that $1001C - 2002A = 4004$ , and $1001B + 3003A = 5005$ . What is the average of A, B, and C?

$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}$ More than 1

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 $1001(A+B+C)=1001(9)$ so $A+B+C=9$ and the average is $\frac{9}{3}=3$ . Our answer is $\boxed{\text{(B)}\ 3}$ .

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

@@ Line 9: / Line 9: @@
 Adding up the equations gives <math>1001(A+B+C)=1001(9)</math> so <math>A+B+C=9</math> and the average is <math>\frac{9}{3}=3</math>. Our answer is <math>\boxed{\text{(B)}\ 3}</math>.
+==See Also==
+{{AMC10 box|year=2002|ab=A|before=Problem 8|num-a=10}}
+[[Category:Introductory Algebra Problems]]