Difference between revisions of "2015 AMC 8 Problems/Problem 13"

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==Problem==
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How many subsets of two elements can be removed from the set <math>\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}</math> so that the mean (average) of the remaining numbers is 6?
 
How many subsets of two elements can be removed from the set <math>\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}</math> so that the mean (average) of the remaining numbers is 6?
  
 
<math>\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}</math>
 
<math>\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}</math>
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==Solutions==
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===Solution 1===
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Since there will be <math>9</math> elements after removal, and their mean is <math>6</math>, we know their sum is <math>54</math>. We also know that the sum of the set pre-removal is <math>66</math>. Thus, the sum of the <math>2</math> elements removed is <math>66-54=12</math>. There are only <math>\boxed{\textbf{(D)}~5}</math> subsets of <math>2</math> elements that sum to <math>12</math>: <math>\{1,11\}, \{2,10\}, \{3, 9\}, \{4, 8\}, \{5, 7\}</math>.
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===Solution 2===
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We can simply remove <math>5</math> subsets of <math>2</math> numbers, while leaving only <math>6</math> behind. The average of this one-number set is still <math>6</math>, so the answer is <math>\boxed{\textbf{(D)}~5}</math>.
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-tryanotherangle
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==See Also==
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{{AMC8 box|year=2015|num-b=12|num-a=14}}
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{{MAA Notice}}

Latest revision as of 16:40, 16 January 2021

Problem

How many subsets of two elements can be removed from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ so that the mean (average) of the remaining numbers is 6?

$\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}$

Solutions

Solution 1

Since there will be $9$ elements after removal, and their mean is $6$, we know their sum is $54$. We also know that the sum of the set pre-removal is $66$. Thus, the sum of the $2$ elements removed is $66-54=12$. There are only $\boxed{\textbf{(D)}~5}$ subsets of $2$ elements that sum to $12$: $\{1,11\}, \{2,10\}, \{3, 9\}, \{4, 8\}, \{5, 7\}$.

Solution 2

We can simply remove $5$ subsets of $2$ numbers, while leaving only $6$ behind. The average of this one-number set is still $6$, so the answer is $\boxed{\textbf{(D)}~5}$.

-tryanotherangle

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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 AJHSME/AMC 8 Problems and Solutions

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