Difference between revisions of "2011 AMC 10A Problems/Problem 6"

Latest revision as of 09:29, 17 October 2025

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)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 13: / Line 13: @@
 ~MrThinker
+==Solution 3 (Same Approach as Solution 1)==
+If <math> A \supset B </math> (<math>\supset</math> in this case is the superset), then <math> A \cup B </math> will contain 15 of the same elements in <math> A </math> and <math> B </math> because <math> B \subset A </math> (<math>\subset</math> in this case is the subset). Therefore our new set must contain 15 elements already variant in both <math> A </math> and <math> B </math> as well as 5 additional elements only the superset <math> A </math> contains, and so our answer is <math> 15+5= \boxed{\textbf{(C) }20}</math>.
+~Pinotation
 == See Also ==
 {{AMC10 box|year=2011|ab=A|num-b=5|num-a=7}}
 {{MAA Notice}}

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