Difference between revisions of "2015 AMC 8 Problems/Problem 13"
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===Solution 2=== | ===Solution 2=== | ||
| − | We can simply remove <math>5</math> subsets of <math>2</math> numbers | + | 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)== | ||
| + | https://youtu.be/rojCJ9OSI2o | ||
| + | |||
| + | ~Education, the Study of Everything | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 20:40, 4 August 2024
Contents
Problem
How many subsets of two elements can be removed from the set 
so that the mean (average) of the remaining numbers is 6?
Solutions
Solution 1
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 
: 
.
Solution 2
We can simply remove 
subsets of numbers while leaving only behind. The average of this one-number set is still , so the answer is .
-tryanotherangle
Video Soluti)
~Education, the Study of Everything
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
