Difference between revisions of "1959 AHSME Problems/Problem 47"

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Again, concerning the freshmen, all we know is that they are human. Thus, we cannot prove assertion (3).  
 
Again, concerning the freshmen, all we know is that they are human. Thus, we cannot prove assertion (3).  
  
Concerning those who are not students, all that we can say is that some freshmen or humans may not be students. This information is not sufficient to prove assertion (4).
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Concerning those who are not students, all that we can say is that some freshmen or humans might not be students. This information is not sufficient to prove assertion (4).
  
 
Thus, because we can only prove assertion (2), our answer is <math>\fbox{\textbf{(A)}}</math>.
 
Thus, because we can only prove assertion (2), our answer is <math>\fbox{\textbf{(A)}}</math>.

Latest revision as of 13:05, 22 July 2024

Problem

Assume that the following three statements are true: (I). All freshmen are human. (II). All students are human. (III). Some students think.

Given the following four statements:

$\textbf{(1)}\ \text{All freshmen are students.}\qquad \\ \textbf{(2)}\ \text{Some humans think.}\qquad \\ \textbf{(3)}\ \text{No freshmen think.}\qquad \\ \textbf{(4)}\ \text{Some humans who think are not students.}$

Those which are logical consequences of I, II, and III are:

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

Solution

All that we know about the freshmen is that they are all human (by statement I), so we can not prove assertion (1).

Because all students are human (by statement II) and some students think (by statement III), some humans think, which proves assertion (2).

Again, concerning the freshmen, all we know is that they are human. Thus, we cannot prove assertion (3).

Concerning those who are not students, all that we can say is that some freshmen or humans might not be students. This information is not sufficient to prove assertion (4).

Thus, because we can only prove assertion (2), our answer is $\fbox{\textbf{(A)}}$.

See also

1959 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 46
Followed by
Problem 48
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All AHSME Problems and Solutions

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