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Difference between revisions of "2017 AMC 10A Problems/Problem 6"

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~minor edits by BakedPotato66
 
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== Solution 2 (Elimination of Answer Choices)==
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== Solution 2 (Elimination)==
  
 
Note: An A is usually 90%-100% of the questions correct.
 
Note: An A is usually 90%-100% of the questions correct.
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<math>\textbf{(A)}</math> False. If Lewis did not receive an A, that does not mean he got ''all'' the multiple-choice questions wrong. For example, he might get 19/20 or 18/20, which still accounts for an A.
 
<math>\textbf{(A)}</math> False. If Lewis did not receive an A, that does not mean he got ''all'' the multiple-choice questions wrong. For example, he might get 19/20 or 18/20, which still accounts for an A.
  
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<math>\textbf{(B)}</math> True. If Lewis got all the multiple-choice questions right, he would've gotten an A. If he didn't get an A, then he must've got at least one wrong since Lewis would have gotten an A if he got all the multiple-choice sections right.
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<math>\textbf{(B)}</math> True. If Lewis did not receive an A, then he must have got at least one wrong. Otherwise, Lewis would have gotten an A.
  
 
<math>\textbf{(C)}</math> False. Again, Lewis can get 19/20 or 18/20, which is still an A.  
 
<math>\textbf{(C)}</math> False. Again, Lewis can get 19/20 or 18/20, which is still an A.  
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==Video Solution==
 
==Video Solution==
 
https://youtu.be/pxg7CroAt20
 
https://youtu.be/pxg7CroAt20
 +
(TheBeautyofMath)
  
 +
==Video Solution 2==
 
https://youtu.be/7j5JigR0pbs
 
https://youtu.be/7j5JigR0pbs
  
 
~savannahsolver
 
~savannahsolver
 +
 +
==Video Solution 3==
 +
https://www.youtube.com/watch?v=iHlOCfubaq8
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 +
~Math4All999
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=A|num-b=5|num-a=7}}
 
{{AMC10 box|year=2017|ab=A|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:46, 14 September 2024

Problem

Ms. Carroll promised that anyone who got all the multiple choice questions right on the upcoming exam would receive an A on the exam. Which one of these statements necessarily follows logically?

$\textbf{(A)}\ \text{If Lewis did not receive an A, then he got all of the multiple choice questions wrong.}\\\textbf{(B)}\ \text{If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong.}\\\textbf{(C)}\ \text{If Lewis got at least one of the multiple choice questions wrong, then he did not receive an A. }\\\textbf{(D)}\ \text{If Lewis received an A, then he got all of the multiple choice questions right.}\\\textbf{(E)}\ \text{If Lewis received an A, then he got at least one of the multiple choice questions right.}$

Solution 1 (Using the Contrapositive)

Rewriting the given statement: "if someone got all the multiple choice questions right on the upcoming exam then he or she would receive an A on the exam." If that someone is Lewis the statement becomes: "if Lewis got all the multiple choice questions right, then he would receive an A on the exam."

The contrapositive: "If Lewis did not receive an A, then he did not get all of them right"

The contrapositive (in other words): "If Lewis did not get an A, then he got at least one of the multiple choice questions wrong".

B is also equivalent to the contrapositive of the original statement, which implies that it must be true, so the answer is $\boxed{\textbf{(B)}\text{ If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong.}}$.

  • Note that answer choice (B) is the contrapositive of the given statement. (That is, it has been negated as well as reversed.) We know that the contrapositive is always true if the given statement is true.

~minor edits by BakedPotato66

Solution 2 (Elimination)

Note: An A is usually 90%-100% of the questions correct.

$\textbf{(A)}$ False. If Lewis did not receive an A, that does not mean he got all the multiple-choice questions wrong. For example, he might get 19/20 or 18/20, which still accounts for an A.

$\textbf{(B)}$ True. If Lewis did not receive an A, then he must have got at least one wrong. Otherwise, Lewis would have gotten an A.

$\textbf{(C)}$ False. Again, Lewis can get 19/20 or 18/20, which is still an A.

$\textbf{(D)}$ False. The above situation can happen.

$\textbf{(E)}$ False. Lewis can get 17/20 or less but it is not an A.

Therefore, our answer is $\boxed{\textbf{(B)}}.$

~minor edits by Terribleteeth

~edits by BakedPotato66


Video Solution

https://youtu.be/pxg7CroAt20 (TheBeautyofMath)

Video Solution 2

https://youtu.be/7j5JigR0pbs

~savannahsolver

Video Solution 3

https://www.youtube.com/watch?v=iHlOCfubaq8

~Math4All999

See Also

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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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