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 18:32, 12 February 2021
Contents
[hide]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.