Difference between revisions of "2002 AMC 10A Problems/Problem 9"
(New page: == Problem == 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>\text{(A)}\ 1...) |
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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== | ||
| − | 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. | + | Notice that we don't need to find what <math>A, B,</math> and <math>C</math> actually are, just their average. In other words, if we can find <math>A+B+C</math>, we will be done. |
| − | 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{\ | + | 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>. |
| + | |||
| + | ==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 | ||
| + | |||
| + | ==Solution 3== | ||
| + | |||
| + | Start by isolating <math> B </math> and <math> C </math> in both of the equations, in order to represent the variables <math> C </math> and <math> B </math> in terms of A. Ending up with the two equations <math> C = 2A +4 </math> and <math> B = -3A + 5 </math>, we have to calculate the value of the expression <math>\frac{A+B+C}{3} </math>. Plugging in <math> 2A + 4 </math> for <math> C </math> and <math> -3A + 5 </math> for <math> B </math>, we add them up and end up with a value of 9. Dividing 9 by 3 to compute the average, we get our answer of <math>\boxed{\textbf{(B) }3}</math>. | ||
| + | |||
| + | ~Darth_Cadet | ||
| + | |||
| + | ==Video Solution by Daily Dose of Math== | ||
| + | |||
| + | https://youtu.be/0k8f5Y5ciSU | ||
| + | |||
| + | ~Thesmartgreekmathdude | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2002|ab=A|num-b=8|num-a=10}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:53, 25 July 2024
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
Solution 3
Start by isolating 
and in both of the equations, in order to represent the variables and in terms of A. Ending up with the two equations 
and 
, we have to calculate the value of the expression 
. Plugging in 
for and 
for , we add them up and end up with a value of 9. Dividing 9 by 3 to compute the average, we get our answer of .
~Darth_Cadet
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
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. 
