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

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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]]}}
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==Problem==
 
==Problem==
 
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?
 
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?

Revision as of 05:44, 2 March 2021

The following problem is from both the 2021 AMC 10B #4 and 2021 AMC 12A #2, so both problems redirect to this page.

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

Video Solution by TheBeautyofMath

https://youtu.be/gLahuINjRzU?t=626

~IceMatrix

Video Solution by Interstigation

https://youtu.be/DvpN56Ob6Zw?t=286

~Interstigation

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
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
2021 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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