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

Revision as of 15:46, 14 May 2023

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?

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

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

2015 AMC 8 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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