Difference between revisions of "1962 AHSME Problems/Problem 30"

 
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==Solution==
 
==Solution==
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As long as the statement isn't p and q are both true, it satisfies the negative, so 2, 3, and 4 work, yielding an answer of 3, or \textbf{(D)}.
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By De Morgan's Law, the negation of <math>p</math> and <math>q</math> are both true is that at least one of them is false, with the exception of the statement 1, i.e.;
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<math> \text{p and q are both true}.</math>
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The other three statements state that at least one statement is false. So, 2, 3, and 4 work, yielding an answer of 3, or <math>\textbf{(D)}.</math>

Latest revision as of 01:55, 2 May 2020

Problem

Consider the statements:

$\textbf{(1)}\ \text{p and q are both true}\qquad\textbf{(2)}\ \text{p is true and q is false}\qquad\textbf{(3)}\ \text{p is false and q is true}\qquad\textbf{(4)}\ \text{p is false and q is false.}$

How many of these imply the negative of the statement "p and q are both true?"

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution

By De Morgan's Law, the negation of $p$ and $q$ are both true is that at least one of them is false, with the exception of the statement 1, i.e.;

$\text{p and q are both true}.$

The other three statements state that at least one statement is false. So, 2, 3, and 4 work, yielding an answer of 3, or $\textbf{(D)}.$