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== | ||
+ | Let <math>x</math> be the number of elements in <math>A</math> not including the intersection. <math>2007-1001=1006</math> total elements excluding the intersection. Since we know that <math>A=B</math>, we can find that <math>x=\frac{1006}2=503</math>. Now we need to add the intersection. <math>503+1001=\boxed{\textbf{(C)} 1504}</math>. | ||
==Video Solution by WhyMath== | ==Video Solution by WhyMath== | ||
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~savannahsolver | ~savannahsolver | ||
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+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=6F9x1XBOAeo | ||
+ | |||
+ | ==Video Solution by AliceWang== | ||
+ | https://youtu.be/ThBO09fGBgM | ||
==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}} |
Latest revision as of 13:33, 8 July 2024
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 which is equal.
Then we could form equation
.
The answer is
Solution 2
Let be the number of elements in not including the intersection. total elements excluding the intersection. Since we know that , we can find that . Now we need to add the intersection. .
Video Solution by WhyMath
~savannahsolver
Video Solution
https://www.youtube.com/watch?v=6F9x1XBOAeo
Video Solution by AliceWang
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.