2003 AMC 12B Problems/Problem 1: Difference between revisions
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==Solution== | ==Solution== | ||
< | <cmath> | ||
\begin{align*} | |||
2-4+6-8+10-12+14=-2-2-2+14&=8\\ | |||
3-6+9-12+15-18+21=-3-3-3+21&=12\\ | |||
\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}&=\frac{8}{12}=\frac{2}{3} \Rightarrow \text {(C)} | |||
\end{align*}</cmath> | |||
==See also== | ==See also== | ||
{{AMC12 box|year=2003|ab=B|before=First Question|num-a=2}} | {{AMC12 box|year=2003|ab=B|before=First Question|num-a=2}} | ||
[[Category:Introductory Algebra Problems]] | |||
Revision as of 17:08, 5 February 2008
Problem
Which of the following is the same as
![\[\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}\]](http://latex.artofproblemsolving.com/6/d/8/6d8019823e919fab14bd496a543fe92645572845.png)
?
Solution
See also
| 2003 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by First Question |
Followed by Problem 2 |
| 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 12 Problems and Solutions | |
