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 14:40, 4 January 2012
Problem
Which of the following is the same as
![\[\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}\]](http://latex.artofproblemsolving.com/6/d/8/6d8019823e919fab14bd496a543fe92645572845.png)
?
Solution
Alternatively, notice that each term in the numerator is 
of a term in the denominator, so the quotient has to be .
See also
| 2003 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
