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

Latest revision as of 23:53, 25 July 2024

Problem

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?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E) }\text{Not uniquely determined}$

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 $1001(A+B+C)=9009=1001(9)$ so $A+B+C=9$ and the average is $\frac{9}{3}=3$ . Our answer is $\boxed{\textbf{(B) }3}$ .

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 $B$ and $C$ in both of the equations, in order to represent the variables $C$ and $B$ in terms of A. Ending up with the two equations $C = 2A +4$ and $B = -3A + 5$ , we have to calculate the value of the expression $\frac{A+B+C}{3}$ . Plugging in $2A + 4$ for $C$ and $-3A + 5$ for $B$ , 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 $\boxed{\textbf{(B) }3}$ .

~Darth_Cadet

Video Solution by Daily Dose of Math

https://youtu.be/0k8f5Y5ciSU

~Thesmartgreekmathdude

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.

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