Difference between revisions of "2002 AMC 10A Problems/Problem 9"
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Adding up the equations gives <math>1001(A+B+C)=9009=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{\textbf{(B) }3}</math>. | Adding up the equations gives <math>1001(A+B+C)=9009=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{\textbf{(B) }3}</math>. | ||
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| + | ==Solution 2== | ||
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| + | As there are only 2 equations and 3 variables, just set one of the variables to something convenient, like 0. Setting C as 0, we get A is -2, and B is 11 by substitution. By basic arithmetic the average is 3=>B | ||
| + | |||
| + | -dragoon | ||
==See Also== | ==See Also== | ||
Revision as of 15:10, 18 September 2022
Contents
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 
and 
actually are, just their average. In other words, if we can find 
, we will be done.
Adding up the equations gives 
so 
and the average is 
. Our answer is 
.
Solution 2
As there are only 2 equations and 3 variables, just set one of the variables to something convenient, like 0. Setting C as 0, we get A is -2, and B is 11 by substitution. By basic arithmetic the average is 3=>B
-dragoon
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
