During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

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

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

Revision as of 13:40, 4 January 2012

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
Invalid username
Login to AoPS