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

m (Problem)
m (Solution)
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==Solution==
 
==Solution==
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The numerator and denominator can be grouped like this:
 +
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
2-4+6-8+10-12+14=-2-2-2+14&=8\\
+
2+(-4+6)+(-8+10)+(-12+14)&=2*4\\
3-6+9-12+15-18+21=-3-3-3+21&=12\\
+
3+(-6+9)+(-12+15)+(-18+21)&=3*4\\
\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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\frac{2*4}{3*4}&=\frac{2}{3} \Rightarrow \text {(C)}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  

Revision as of 00:08, 14 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 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

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