2011 AMC 10A Problems/Problem 6

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

Solution 3 (Same Approach as Solution 1)

If $A \supset B$ ( $\supset$ in this case is the superset), then $A \cup B$ will contain 15 of the same elements in $A$ and $B$ because $B \subset A$ ( $\subset$ in this case is the subset). Therefore our new set must contain 15 elements already variant in both $A$ and $B$ as well as 5 additional elements only the superset $A$ contains, and so our answer is $15+5= \boxed{\textbf{(C) }20}$ .

~Pinotation

2011 AMC 10A (Problems • Answer Key • Resources)
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2011 AMC 10A Problems/Problem 6

Contents

Problem 6

Solution 1

Solution 2

Solution 3 (Same Approach as Solution 1)

See Also