Difference between revisions of "2015 AMC 8 Problems/Problem 13"
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==Solution== | ==Solution== | ||
| − | 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>\boxed{\textbf{(D)}~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>. | + | 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>\boxed{\textbf{(D)}~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>. <math>Amber is AWESOME!!!</math> |
==See Also== | ==See Also== | ||
Revision as of 20:17, 6 November 2019
How many subsets of two elements can be removed from the set 
so that the mean (average) of the remaining numbers is 6?
Solution
Since there will be 
elements after removal, and their mean is 
, we know their sum is 
. We also know that the sum of the set pre-removal is 
. Thus, the sum of the 
elements removed is 
. There are only 
subsets of elements that sum to 
: 
. 
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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