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Difference between revisions of "2011 AMC 10A Problems/Problem 6"
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== Solution ==
 
== 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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<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>.

Revision as of 14:02, 14 February 2011

Problem 6

Set $A$ has $20$ elements, and set $B$ has $15$ elements. What is the smallest possible number of elements in $A \cup B$?

$\textbf{(A)}5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300$

Solution

$A \cup B$ will be smallest if $B$ is completely contained in $A$, in which case all the elements in $B$ would be counted for in $A$. So the total would be the number of elements in $A$, which is $\boxed{20 \ \mathbf{(C)}}$.

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