Difference between revisions of "2003 AMC 12B Problems/Problem 1"

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==Solution==
 
==Solution==
<math>2-4+6-8+10-12+14=-2-2-2+14=8</math>
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<cmath>
 
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\begin{align*}
 
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2-4+6-8+10-12+14=-2-2-2+14&=8\\
 
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3-6+9-12+15-18+21=-3-3-3+21&=12\\
<math>3-6+9-12+15-18+21=-3-3-3+21=12</math>
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\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}&=\frac{8}{12}=\frac{2}{3} \Rightarrow \text {(C)}
 
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\end{align*}</cmath>
 
 
 
 
<math>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}=\frac{8}{12}=\frac{2}{3} \Rightarrow \text {(C)}</math>
 
  
 
==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

\[\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}\]?

$\text {(A) } -1 \qquad \text {(B) } -\frac{2}{3} \qquad \text {(C) } \frac{2}{3} \qquad \text {(D) } 1 \qquad \text {(E) } \frac{14}{3}$

Solution

\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*}

See also

2003 AMC 12B (ProblemsAnswer KeyResources)
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
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