Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(New page: ==Problem== 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>? <math> \text {(A) } -1 \qquad \text {(B) } -\frac{2}{3} \qquad \text {(C) } \...)
 
(→‎Solution)
Line 10: Line 10:
 
==Solution==
 
==Solution==
 
<math>2-4+6-8+10-12+14=-2-2-2+14=8</math>
 
<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>
 
<math>3-6+9-12+15-18+21=-3-3-3+21=12</math>
 +
 +
  
 
<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>
 
<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>
  
 
==See also==
 
==See also==

Revision as of 09:06, 5 February 2008

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

$2-4+6-8+10-12+14=-2-2-2+14=8$


$3-6+9-12+15-18+21=-3-3-3+21=12$


$\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}=\frac{8}{12}=\frac{2}{3} \Rightarrow \text {(C)}$

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12B_Problems/Problem_1&oldid=22991"