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

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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 OmegaLearn (System of Equations) ==
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https://youtu.be/hyYg62tT0sY
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~ pi_is_3.14

Revision as of 20:53, 11 February 2021

Problem

At a math contest, $57$ students are wearing blue shirts, and another $75$ students are wearing yellow shirts. The 132 students are assigned into $66$ pairs. In exactly $23$ of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?

$\textbf{(A)} ~23 \qquad\textbf{(B)} ~32 \qquad\textbf{(C)} ~37 \qquad\textbf{(D)} ~41 \qquad\textbf{(E)} ~64$

Solution

There are $46$ students paired with a blue partner. The other $11$ students wearing blue shirts must each be paired with a partner wearing a shirt of the opposite color. There are $64$ students remaining. Therefore the requested number of pairs is $\tfrac{64}{2}=\boxed{\textbf{(B)} ~32}$ ~Punxsutawney Phil

Video Solution by OmegaLearn (System of Equations)

https://youtu.be/hyYg62tT0sY

~ pi_is_3.14