Difference between revisions of "2015 AMC 8 Problems/Problem 13"
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<math>\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}</math> | <math>\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}</math> | ||
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==Solutions== | ==Solutions== | ||
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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 21:40, 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.