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> | ||
| − | == Solution == | + | == Solution 1 == |
<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>. | <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>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | ~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 == | == See Also == | ||
{{AMC10 box|year=2011|ab=A|num-b=5|num-a=7}} | {{AMC10 box|year=2011|ab=A|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 09:29, 17 October 2025
Contents
Problem 6
Set 
has 
elements, and set 
has 
elements. What is the smallest possible number of elements in 
?
Solution 1
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 
.
Solution 2
Assume WLOG that 
, and 
. Then, all the integers 
through would be redundant in , so 
.
~MrThinker
Solution 3 (Same Approach as Solution 1)
If 
(
in this case is the superset), then will contain 15 of the same elements in and because 
(
in this case is the subset). Therefore our new set must contain 15 elements already variant in both and as well as 5 additional elements only the superset contains, and so our answer is 
.
~Pinotation
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
