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Difference between revisions of "2015 AMC 8 Problems/Problem 13"
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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>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>. Therefore, our answer is <math>\textbf{(D)}\text{ 5}</math>.
 
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>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>. Therefore, our answer is <math>\textbf{(D)}\text{ 5}</math>.
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==See Also==
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{{AMC8 box|year=2015|before=12|num-a=13}}
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{{MAA Notice}}

Revision as of 15:13, 25 November 2015

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?

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

Solution

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 $5$ subsets of $2$ elements that sum to $12$: $\{1,11\}, \{2,10\}, \{3, 9\}, \{4, 8\}, \{5, 7\}$. Therefore, our answer is $\textbf{(D)}\text{ 5}$.

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
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12 		Followed by
Problem 13
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All AJHSME/AMC 8 Problems and Solutions

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