2003 AMC 12B Problems/Problem 1

Revision as of 23:09, 13 September 2015 by Ft029 (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
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 numbers in 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS