Difference between revisions of "2021 AMC 10B Problems/Problem 4"

(Redirected to the corresponding AMC 12B problem.)
(Tag: New redirect)
(Removed all unnecessary contents. I merged all contents on the corresponding AMC 12 page.)
(Tag: Replaced)
Line 1: Line 1:
#REDIRECT [[2021_AMC_12B_Problems/Problem_2]]
#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>
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) ==
~ pi_is_3.14
==Video Solution by TheBeautyofMath==
==Video Solution by Interstigation==
==See Also==
{{AMC12 box|year=2021|ab=B|num-b=1|num-a=3}}
{{AMC10 box|year=2021|ab=B|num-b=3|num-a=5}}
{{MAA Notice}}

Latest revision as of 05:18, 4 March 2021

Invalid username
Login to AoPS