Difference between revisions of "2021 AMC 12B Problems/Problem 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 <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? | 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? |
Revision as of 08:28, 2 March 2021
- 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/EMzdnr1nZcE?t=154
~IceMatrix
Video Solution by Interstigation
https://youtu.be/DvpN56Ob6Zw?t=286
~Interstigation
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.