Difference between revisions of "2011 AMC 10A Problems/Problem 6"
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<math> \textbf{(A)}5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300 </math> | <math> \textbf{(A)}5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300 </math> | ||
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| + | == Solution == | ||
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| + | <math>A \cup B</math> will be smallest if <math>B</math> is completely contained in <math>A</math>, in which case all the elements in <math>B</math> would be counted for in <math>A</math>. So the total would be the number of elements in <math>A</math>, which is <math>\boxed{20 \ \mathbf{(C)}}</math>. | ||
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| + | == See Also == | ||
| + | {{AMC10 box|year=2011|ab=A|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Revision as of 12:00, 4 July 2013
Problem 6
Set 
has 
elements, and set 
has 
elements. What is the smallest possible number of elements in 
?
Solution
will be smallest if is completely contained in , in which case all the elements in would be counted for in . So the total would be the number of elements in , which is 
.
See Also
| 2011 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
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