Difference between revisions of "2015 AMC 8 Problems/Problem 13"
m (→Problem) |
m (→Solution 2) |
||
Line 13: | Line 13: | ||
===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 |
Revision as of 08:14, 26 March 2023
Contents
[hide]Problem
How many subsets of two elements can be removed from the set so that the mean (average) of the remaining numbers is 6?
h
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 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.