Difference between revisions of "2003 AMC 12B Problems/Problem 1"
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==Solution== | ==Solution== | ||
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− | + | \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> | |
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==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}} | ||
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+ | [[Category:Introductory Algebra Problems]] |
Revision as of 18:08, 5 February 2008
Problem
Which of the following is the same as
?
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 |