During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

2017 AMC 10A Problems/Problem 6

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?

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

Solution

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 got an A on the exam." The contrapositive: "If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong (did not get all of them right)" must also be true leaving B as the correct answer. B is also equivalent to the contrapositive of the original statement, which implies that it must be true, so the answer is $\boxed{\textbf{(B)}\text{ If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong.}}$.

See Also

 2017 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 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
 2017 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 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.

Invalid username
Login to AoPS