2015 AMC 8 Problems/Problem 13

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

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/rojCJ9OSI2o

~Education, the Study of Everything

Video Solution

https://youtu.be/ZbwdX6sZyQA

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/51K3uCzntWs?t=68

~ pi_is_3.14

See Also

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

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