Difference between revisions of "2021 AMC 12B Problems/Problem 2"
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| + | {{duplicate|[[2021 AMC 10B Problems#Problem 4|2021 AMC 10B #4]] and [[2021 AMC 12B Problems#Problem 2|2021 AMC 12B #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 <math>132</math> 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> | <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== | ==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 Punxsutawney Phil== | ||
| + | https://youtube.com/watch?v=qpvS2PVkI8A&t=55s | ||
| + | |||
| + | == Video Solution by OmegaLearn (System of Equations) == | ||
| + | https://youtu.be/hyYg62tT0sY | ||
| + | |||
| + | ~ pi_is_3.14 | ||
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| + | ==Video Solution by Hawk Math== | ||
| + | https://www.youtube.com/watch?v=VzwxbsuSQ80 | ||
| + | |||
| + | ==Video Solution by TheBeautyofMath== | ||
| + | https://youtu.be/gLahuINjRzU?t=626 | ||
| + | |||
| + | https://youtu.be/EMzdnr1nZcE?t=154 | ||
| + | |||
| + | ~IceMatrix | ||
| + | |||
| + | ==Video Solution by Interstigation== | ||
| + | https://youtu.be/DvpN56Ob6Zw?t=286 | ||
| + | |||
| + | ~Interstigation | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/JOc-5A9Q9Ic | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==Video Solution (Under 2 min!)== | ||
| + | https://youtu.be/5yCyXUnTu5Y | ||
| + | |||
| + | ~Education, the Study of Everything | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021|ab=B|num-b=1|num-a=3}} | {{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}} | {{MAA Notice}} | ||
Latest revision as of 22:49, 18 July 2023
- The following problem is from both the 2021 AMC 10B #4 and 2021 AMC 12B #2, so both problems redirect to this page.
Contents
Problem
At a math contest, 
students are wearing blue shirts, and another 
students are wearing yellow shirts. The 
students are assigned into 
pairs. In exactly 
of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?
Solution
There are 
students paired with a blue partner. The other 
students wearing blue shirts must each be paired with a partner wearing a shirt of the opposite color. There are 
students remaining. Therefore the requested number of pairs is 
~Punxsutawney Phil
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=qpvS2PVkI8A&t=55s
Video Solution by OmegaLearn (System of Equations)
~ pi_is_3.14
Video Solution by Hawk Math
https://www.youtube.com/watch?v=VzwxbsuSQ80
Video Solution by TheBeautyofMath
https://youtu.be/gLahuINjRzU?t=626
https://youtu.be/EMzdnr1nZcE?t=154
~IceMatrix
Video Solution by Interstigation
https://youtu.be/DvpN56Ob6Zw?t=286
~Interstigation
Video Solution by WhyMath
~savannahsolver
Video Solution (Under 2 min!)
~Education, the Study of Everything
See Also
| 2021 AMC 12B (Problems • Answer Key • Resources) | |
| 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 (Problems • Answer Key • Resources) | ||
| 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. 
