Difference between revisions of "2003 AMC 12B Problems/Problem 1"
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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 == |
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Which of the following is the same as | Which of the following is the same as | ||
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</math> | </math> | ||
| − | ==Solution== | + | == Solution 1 == |
| − | + | <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> | |
| − | + | == Solution 2 == | |
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| − | + | 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}} | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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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.
Contents
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/d/d/a/ddaa33db347793c4318c6123185c74195fe22c74.png)
Solution 1
![\[\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}=\frac{2+(-4+6)+(-8+10)+(-12+14)}{3+(-6+9)+(-12+15)+(-18+21)}=\frac{2\cdot4}{3\cdot4}=\text {(C) } \frac{2}{3}\]](http://latex.artofproblemsolving.com/c/0/3/c0305053f7e67176f9a4dfc8d0885136c1f778d8.png)
Solution 2
Each term in the numerator is 
of a corresponding term in the denominator, so this must be the quotient.
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 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
