Difference between revisions of "2007 AMC 8 Problems/Problem 13"
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The answer is <math>\boxed{\textbf{(C)}\ 1504}</math> | The answer is <math>\boxed{\textbf{(C)}\ 1504}</math> | ||
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+ | == Solution 2 == | ||
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+ | First find the number of elements in <math>A</math> without including the intersection. There are 2007 elements in total, so there are <math>1006</math> elements in <math>A</math> and <math>B</math> excluding the intersection (<math>2007-1001</math>). There are <math>503</math> elements in set A after dividing <math>1006</math> by <math>2</math>. Add the intersection (<math>1001</math>) to get <math>\boxed{\textbf{(C)}\ 1504}</math> | ||
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+ | - spoamath321 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=12|num-a=14}} | {{AMC8 box|year=2007|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:57, 23 March 2020
Contents
[hide]Problem
Sets and , shown in the Venn diagram, have the same number of elements. Their union has elements and their intersection has elements. Find the number of elements in .
Solution
Let be the number of elements in and .
Since the union is the sum of all elements in and ,
and and have the same number of elements then,
.
The answer is
Solution 2
First find the number of elements in without including the intersection. There are 2007 elements in total, so there are elements in and excluding the intersection (). There are elements in set A after dividing by . Add the intersection () to get
- spoamath321
See Also
2007 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.