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]]}} | ||
+ | |||
==Problem== | ==Problem== | ||
Which of the following is the same as | 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> | + | <cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?</cmath> |
<math> | <math> | ||
Line 9: | Line 11: | ||
==Solution== | ==Solution== | ||
− | |||
− | |||
− | |||
− | |||
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+ | The numbers in the numerator and denominator can be grouped like this: | ||
+ | <cmath> | ||
+ | \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*}</cmath> | ||
− | <math>\frac{2 | + | 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>. |
==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}} | ||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 00:09, 14 September 2015
- 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
Solution
The numbers in the numerator and denominator can be grouped like this:
Alternatively, notice that each term in the numerator is of a term in the denominator, so the quotient has to be .
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.