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Difference between revisions of "2015 AMC 8 Problems/Problem 13"
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===Solution 2===
 
===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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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>.
  
 
-tryanotherangle
 
-tryanotherangle
 +
 +
==Video Soluti)==
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https://youtu.be/rojCJ9OSI2o
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~Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==
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~savannahsolver
 
~savannahsolver
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== Video Solution by OmegaLearn ==
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https://youtu.be/51K3uCzntWs?t=68
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 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Latest revision as of 20:40, 4 August 2024

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 Soluti)

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 (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

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