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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
  
<cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}</cmath>?
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<cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?</cmath>
  
 
<math>
 
<math>
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==Solution==
 
==Solution==
<math>2-4+6-8+10-12+14=-2-2-2+14=8</math>
 
 
 
 
<math>3-6+9-12+15-18+21=-3-3-3+21=12</math>
 
  
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The numbers in the numerator and denominator can be grouped like this:
  
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<cmath>
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\begin{align*}
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2+(-4+6)+(-8+10)+(-12+14)&=2*4\\
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3+(-6+9)+(-12+15)+(-18+21)&=3*4\\
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\frac{2*4}{3*4}&=\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>
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Alternatively, notice that each term in the numerator is <math>\frac{2}{3}</math> of a term in the denominator, so the quotient has to be <math>\frac{2}{3}</math>.
  
 
==See also==
 
==See also==
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{{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}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 23:09, 13 September 2015

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

The numbers in the numerator and denominator can be grouped like this:

\begin{align*} 2+(-4+6)+(-8+10)+(-12+14)&=2*4\\ 3+(-6+9)+(-12+15)+(-18+21)&=3*4\\ \frac{2*4}{3*4}&=\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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