Difference between revisions of "2017 AMC 10A Problems/Problem 6"
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− | == Solution 2 (Elimination | + | == Solution 2 (Elimination)== |
− | <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 | + | Note: An A is usually 90%-100% of the questions correct. |
+ | |||
+ | <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{(B)}</math> True. If Lewis | + | <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. | + | <math>\textbf{(C)}</math> False. Again, Lewis can get 19/20 or 18/20, which is still an A. |
− | <math>\textbf{(D)}</math> False. | + | <math>\textbf{(D)}</math> False. The above situation can happen. |
− | <math>\textbf{(E)}</math> False. Lewis | + | <math>\textbf{(E)}</math> False. Lewis can get 17/20 or less but it is not an A. |
Therefore, our answer is <math>\boxed{\textbf{(B)}}.</math> | Therefore, our answer is <math>\boxed{\textbf{(B)}}.</math> | ||
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~edits by BakedPotato66 | ~edits by BakedPotato66 | ||
+ | |||
+ | <!-- edits by MrThinker--> | ||
==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 | ||
+ | |||
+ | ~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
Contents
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?
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 .
- 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.
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.
True. If Lewis did not receive an A, then he must have got at least one wrong. Otherwise, Lewis would have gotten an A.
False. Again, Lewis can get 19/20 or 18/20, which is still an A.
False. The above situation can happen.
False. Lewis can get 17/20 or less but it is not an A.
Therefore, our answer is
~minor edits by Terribleteeth
~edits by BakedPotato66
Video Solution
https://youtu.be/pxg7CroAt20 (TheBeautyofMath)
Video Solution 2
~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.