Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2 (Elimination of Answer Choices))
(→‎Solution 2 (Elimination of Answer Choices))
Line 21: Line 21:
 
== Solution 2 (Elimination of Answer Choices)==
 
== Solution 2 (Elimination of Answer Choices)==
  
−
<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 have only gotten <math>1</math> multiple-choice question wrong.
+
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 got all the multiple-choice questions right, he would've gotten an A. If he didn't get an A, then he didn't get all of them right, since Lewis would have gotten an A if he got all the multiple-choice section right.
+
<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.
  
−
<math>\textbf{(C)}</math> False. There might be other sections as well. For example, his teacher can dictate that any student who answers all the Writing problems incorrectly will get an A. Or perhaps, a student who answers at least 5/8 of all questions correctly will receive an A. Or perhaps a student who draws a smiley face will receive an A. There may be many other ways to receive an A.
+
<math>\textbf{(C)}</math> False. Again, Lewis can get 19/20 or 18/20, which is still an A.  
  
−
<math>\textbf{(D)}</math> False. He might've missed some number of multiple-choice section but still received an A using the example methods above or any other possible methods.
+
<math>\textbf{(D)}</math> False. The above situation can happen.
 
   
 
   
−
<math>\textbf{(E)}</math> False. Lewis could have gotten 0 multiple-choice questions correct but still received an A using the example methods above or any other possible methods.
+
<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>
Line 36: Line 38:
  
 
~edits by BakedPotato66
 
~edits by BakedPotato66
 +
 +
<!-- edits by MrThinker-->
  
 
==Video Solution==
 
==Video Solution==

Revision as of 09:36, 27 July 2022

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 of Answer Choices)

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 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.

$\textbf{(C)}$ 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

https://youtu.be/7j5JigR0pbs

~savannahsolver

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_6&oldid=176300"