Difference between revisions of "2021 AMC 12A Problems/Problem 4"
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MRENTHUSIASM (talk | contribs) (→Solution 3 (Process of Elimination): Changed Sol 2 to Sol 2.1 to be more specific.) |
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==Solution 3 (Process of Elimination)== | ==Solution 3 (Process of Elimination)== | ||
− | From Solution 2, we can also see this through the process of elimination. | + | From Solution 2.1, we can also see this through the process of elimination. |
Statement <math>A</math> is false because purple snakes cannot add. <math>B</math> is false as well because since happy snakes can add and purple snakes can not add, purple snakes are not happy snakes. <math>E</math> is false using the same reasoning, purple snakes are not happy snakes so happy snakes can subtract since purple snakes cannot subtract. <math>C</math> is false since snakes that can add are happy, not purple. That leaves statement D. <math>\boxed{\textbf{(D)}}</math> is the only correct statement. | Statement <math>A</math> is false because purple snakes cannot add. <math>B</math> is false as well because since happy snakes can add and purple snakes can not add, purple snakes are not happy snakes. <math>E</math> is false using the same reasoning, purple snakes are not happy snakes so happy snakes can subtract since purple snakes cannot subtract. <math>C</math> is false since snakes that can add are happy, not purple. That leaves statement D. <math>\boxed{\textbf{(D)}}</math> is the only correct statement. | ||
Revision as of 16:54, 14 May 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
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Comprehensive Explanation Using Arrows)
- 4 Solution 3 (Process of Elimination)
- 5 Solution 4 (Rigorous)
- 6 Video Solution (Simple & Quick)
- 7 Video Solution by Aaron He (Sets)
- 8 Video Solution by Punxsutawney Phil
- 9 Video Solution by Hawk Math
- 10 Video Solution (Using logic to eliminate choices)
- 11 Video Solution 6
- 12 Video Solution by TheBeautyofMath
- 13 See also
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 (Comprehensive Explanation Using Arrows)
We are given that Two solutions follow from here:
Solution 2.1 (Intuitive)
Combining and gives Clearly, the answer is
~MRENTHUSIASM
Solution 2.2 (Rigorous)
Recall that every conditional statement is always logically equivalent to its contrapositive
Combining and gives Applying the hypothetical syllogism to we conclude that whose contrapositive is Therefore, the answer is
Remark
The conclusions in the other choices do not follow from
~MRENTHUSIASM
Solution 3 (Process of Elimination)
From Solution 2.1, we can also see this through the process of elimination. Statement is false because purple snakes cannot add. is false as well because since happy snakes can add and purple snakes can not add, purple snakes are not happy snakes. is false using the same reasoning, purple snakes are not happy snakes so happy snakes can subtract since purple snakes cannot subtract. is false since snakes that can add are happy, not purple. That leaves statement D. is the only correct statement.
~Bakedpotato66
Solution 4 (Rigorous)
We first convert each statement to "If X, then Y" form:
- If a snake is happy, then it can add.
- If a snake is purple, then it can't subtract.
- If a snake can't subtract, then it can't add.
Now, we simply check the truth value for each statement:
- Combining the last two propositions, we have
- If a snake is purple, then it can't add.
- From the last part, we found that
- If a snake is purple, then it can't add.
- If a snake can't add, then it isn't happy.
- If a snake is purple, then it isn't happy. Purple snakes are not happy.
- From part we found that "If a snake is purple, then it can't add." This implies its contrapositive, "If a snake can add, then it is not purple." is true, meaning is NEVER true. [Thanks again to MRENTHUSIASM for pointing this out!]
- From the first statement, we have
- If a snake is happy, then it can add.
- If a snake can add, then it can subtract.
- If a snake can subtract, then it is not purple.
- If a snake is happy, then it is not purple.
- From the first proposition, we have
- If a snake is happy, then it can add.
- If a snake can add, then it can subtract.
- If a snake is happy, then it can subtract.
Therefore, is our answer.
~ Peace09 (My First Wiki Solution!)
~ MRENTHUSIASM (Revision Suggestions and Code Adjustments)
Video Solution (Simple & Quick)
~ Education the Study of Everything
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 6
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/s6E4E06XhPU?t=202 (AMC10A)
https://youtu.be/rEWS75W0Q54?t=353 (AMC12A)
~IceMatrix
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.