Difference between revisions of "2021 AMC 12A Problems/Problem 4"
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==Video Solution 6== | ==Video Solution 6== |
Revision as of 22:30, 15 February 2021
- The following problem is from both the 2021 AMC 10A #7 and 2021 AMC 12A #4, so both problems redirect to this page.
Contents
Problem
Tom has a collection of snakes,
of which are purple and
of which are happy. He observes that all of his happy snakes can add, none of his purple snakes can subtract, and all of his snakes that can't subtract also can't add. Which of these conclusions can be drawn about Tom's snakes?
Purple snakes can add.
Purple snakes are happy.
Snakes that can add are purple.
Happy snakes are not purple.
Happy snakes can't subtract.
Solution 1
We know that purple snakes cannot subtract, thus they cannot add either. Since happy snakes must be able to add, the purple snakes cannot be happy. Therefore, we know that the happy snakes are not purple and the answer is .
--abhinavg0627
Solution 2 (Explains Solution 1 Using Arrows)
We are given that
Combining and
into
below, we have
Clearly, the answer is
~MRENTHUSIASM
Video Solution by Aaron He (Sets)
https://www.youtube.com/watch?v=xTGDKBthWsw&t=164
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=MUHja8TpKGw&t=259s (Note that there's a slight error in the video I corrected in the description)
Video Solution by Hawk Math
https://www.youtube.com/watch?v=P5al76DxyHY
Video Solution (Using logic to eliminate choices)
~ pi_is_3.14
Video Solution (Simple and Quick)
~ Education the Study of Everything
Video Solution 6
~savannahsolver
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 |
2021 AMC 12A (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.