|
|
| (4 intermediate revisions by the same user not shown) |
| Line 1: |
Line 1: |
| − | ==Problem==
| + | #REDIRECT [[2021_AMC_12B_Problems/Problem_2]] |
| − | 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?
| |
| − | | |
| − | <math>\textbf{(A)} ~23 \qquad\textbf{(B)} ~32 \qquad\textbf{(C)} ~37 \qquad\textbf{(D)} ~41 \qquad\textbf{(E)} ~64</math>
| |
| − | | |
| − | ==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
| |
| − | | |
| − | == Video Solution by OmegaLearn (System of Equations) ==
| |
| − | https://youtu.be/hyYg62tT0sY
| |
| − | | |
| − | ~ pi_is_3.14
| |
| − | | |
| − | ==Video Solution by TheBeautyofMath==
| |
| − | https://youtu.be/gLahuINjRzU?t=626
| |
| − | | |
| − | ~IceMatrix
| |
| − | | |
| − | ==Video Solution by Interstigation==
| |
| − | https://youtu.be/DvpN56Ob6Zw?t=286
| |
| − | | |
| − | ~Interstigation
| |
| − | {{AMC10 box|year=2021|ab=B|num-b=3|num-a=5}}
| |
