2015 AMC 8 Problems/Problem 13

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 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

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2015 AMC 8 Problems/Problem 13

Solution 1

Solution 2

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