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| − | {{duplicate|[[2021 AMC 10B Problems#Problem 4|2021 AMC 10B #4]] and [[2021 AMC 12B Problems#Problem 2|2021 AMC 12A #2]]}}
| + | #REDIRECT [[2021_AMC_12B_Problems/Problem_2]] |
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| − | ==Problem==
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| − | At a math contest, <math>57</math> students are wearing blue shirts, and another <math>75</math> students are wearing yellow shirts. The 132 students are assigned into <math>66</math> pairs. In exactly <math>23</math> of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?
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| − | <math>\textbf{(A)} ~23 \qquad\textbf{(B)} ~32 \qquad\textbf{(C)} ~37 \qquad\textbf{(D)} ~41 \qquad\textbf{(E)} ~64</math>
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| − | ==Solution==
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| − | 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 OmegaLearn (System of Equations) ==
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| − | https://youtu.be/hyYg62tT0sY
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| − | ~ pi_is_3.14
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| − | ==Video Solution by TheBeautyofMath==
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| − | https://youtu.be/gLahuINjRzU?t=626
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| − | ~IceMatrix
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| − | ==Video Solution by Interstigation==
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| − | https://youtu.be/DvpN56Ob6Zw?t=286
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| − | ~Interstigation
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| − | ==See Also==
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| − | {{AMC12 box|year=2021|ab=B|num-b=1|num-a=3}}
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| − | {{AMC10 box|year=2021|ab=B|num-b=3|num-a=5}}
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| − | {{MAA Notice}}
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