2002 AMC 10A Problems/Problem 9

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

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
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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