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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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== Solution 1 ==
  
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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>.
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==Solution 2==
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Assume WLOG that <math>A={1, 2, 3, 4, \cdots , 20}</math>, and <math>B={6, 7, 8, 9, 10, \cdots , 20}</math>. Then, all the integers <math>6</math> through <math>20</math> would be redundant in <math>A \cup B</math>, so <math>A \cup B = 1, 2, 3, 4, \cdots, 20 \implies \boxed{\textbf{(C) }20}</math>.
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~MrThinker
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== See Also ==
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{{AMC10 box|year=2011|ab=A|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 19:19, 21 August 2023

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 1

$A \cup B$ 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 $\boxed{20 \ \mathbf{(C)}}$.

Solution 2

Assume WLOG that $A={1, 2, 3, 4, \cdots , 20}$, and $B={6, 7, 8, 9, 10, \cdots , 20}$. Then, all the integers $6$ through $20$ would be redundant in $A \cup B$, so $A \cup B = 1, 2, 3, 4, \cdots, 20 \implies \boxed{\textbf{(C) }20}$.

~MrThinker

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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