Difference between revisions of "2003 AMC 10B Problems/Problem 1"

Revision as of 15:24, 31 May 2011

Which of the following is the same as $\dfrac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}$ ?

$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ -\frac23 \qquad \textbf{(C)}\ \frac23 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ \frac{14}{3}$

This problem needs a solution. If you have a solution for it, please help us out by adding it. We can rewrite the given fraction as:

$\frac{-3(2)+2(7)}{3(-3)+3(7)}=\frac{2(4)}{3(4)}=\frac{2}{3} \longrightarrow \boxed{\textbf{(C)}}$

2003 AMC 10B (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 10 Problems and Solutions

@@ Line 14: / Line 14: @@
 We can rewrite the given fraction as:
-\[\frac{-3(2)+2(7)}{3(-3)+3(7)}=\frac{2(4)}{3(4)}=\frac{2}{3} \longrightarrow \boxed{\textbf{(C)}}$
+<cmath>\frac{-3(2)+2(7)}{3(-3)+3(7)}=\frac{2(4)}{3(4)}=\frac{2}{3} \longrightarrow \boxed{\textbf{(C)}}</cmath>
 ==See Also==
 {{AMC10 box|year=2003|ab=B|before=First Question|num-a=2}}
 [[Category: Introductory Algebra Problems]]