Difference between revisions of "2021 AMC 12B Problems/Problem 2"
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==Solution== | ==Solution== | ||
There are <math>46</math> students paired with a blue partner. The other <math>11</math> students wearing blue shirts must each be paired with a partner wearing a shirt of the opposite color. There are <math>64</math> students remaining. Therefore the requested number of pairs is <math>\tfrac{64}{2}=\boxed{\textbf{(B)} ~32}</math> ~Punxsutawney Phil | There are <math>46</math> students paired with a blue partner. The other <math>11</math> students wearing blue shirts must each be paired with a partner wearing a shirt of the opposite color. There are <math>64</math> students remaining. Therefore the requested number of pairs is <math>\tfrac{64}{2}=\boxed{\textbf{(B)} ~32}</math> ~Punxsutawney Phil | ||
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| + | ==Video Solution by Punxsutawney Phil== | ||
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| + | http://youtube.com/watch?v=qpvS2PVkI8A&t=55s | ||
== Video Solution by OmegaLearn (System of Equations) == | == Video Solution by OmegaLearn (System of Equations) == | ||
Revision as of 19:32, 12 February 2021
Contents
Problem
At a math contest, 
students are wearing blue shirts, and another 
students are wearing yellow shirts. The 132 students are assigned into 
pairs. In exactly 
of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?
Solution
There are 
students paired with a blue partner. The other 
students wearing blue shirts must each be paired with a partner wearing a shirt of the opposite color. There are 
students remaining. Therefore the requested number of pairs is 
~Punxsutawney Phil
Video Solution by Punxsutawney Phil
http://youtube.com/watch?v=qpvS2PVkI8A&t=55s
Video Solution by OmegaLearn (System of Equations)
~ pi_is_3.14
Video Solution by Hawk Math
https://www.youtube.com/watch?v=VzwxbsuSQ80
See Also
| 2021 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
