Art of Problem Solving

2003 AMC 12B Problems/Problem 1: Difference between revisions

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{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #1]] and [[2003 AMC 10B Problems|2003 AMC 10B #1]]}}
{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #1]] and [[2003 AMC 10B Problems|2003 AMC 10B #1]]}}


==Problem==
== Problem ==
 
Which of the following is the same as
Which of the following is the same as


Line 10: Line 11:
</math>
</math>


==Solution==
== Solution 1 ==
<cmath>
 
\begin{align*}
<cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}=\frac{2+(-4+6)+(-8+10)+(-12+14)}{3+(-6+9)+(-12+15)+(-18+21)}=\frac{2\cdot4}{3\cdot4}=\text {(C) } \frac{2}{3}</cmath>
2-4+6-8+10-12+14=-2-2-2+14&=8\\
 
3-6+9-12+15-18+21=-3-3-3+21&=12\\
== Solution 2 ==
\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>


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>.
Each term in the numerator is <math>\text {(C) } \frac{2}{3}</math> of a corresponding term in the denominator, so this must be the quotient.


==See also==
== See also ==
{{AMC12 box|year=2003|ab=B|before=First Question|num-a=2}}
{{AMC12 box|year=2003|ab=B|before=First Question|num-a=2}}
{{AMC10 box|year=2003|ab=B|before=First Question|num-a=2}}
{{AMC10 box|year=2003|ab=B|before=First Question|num-a=2}}
{{MAA Notice}}
[[Category:Introductory Algebra Problems]]
[[Category:Introductory Algebra Problems]]
{{MAA Notice}}

Latest revision as of 21:43, 16 January 2025

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 1

\[\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}=\frac{2+(-4+6)+(-8+10)+(-12+14)}{3+(-6+9)+(-12+15)+(-18+21)}=\frac{2\cdot4}{3\cdot4}=\text {(C) } \frac{2}{3}\]

Solution 2

Each term in the numerator is $\text {(C) } \frac{2}{3}$ of a corresponding term in the denominator, so this must be the quotient.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing