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

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{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #1]] and [[2003 AMC 10B Problems|2003 AMC 10B #1]]}}
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==Problem==
 
==Problem==
 
Which of the following is the same as
 
Which of the following is the same as
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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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{{AMC10 box|year=2003|ab=B|before=First Question|num-a=2}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:02, 4 January 2014

The following problem is from both the 2003 AMC 12B #1 and 2003 AMC 10B #1, so both problems redirect to this page.

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*}

Alternatively, notice that each term in the numerator is $\frac{2}{3}$ of a term in the denominator, so the quotient has to be $\frac{2}{3}$.

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
2003 AMC 10B (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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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