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

2000 AMC 10 Problems/Problem 21

Revision as of 01:11, 5 December 2019 by Anmol04 (talk | contribs) (added Problem)

Problem

If all alligators are ferocious creatures and some creepy crawlers are alligators, which statement(s) must be true?

\[\textrm{I. All alligators are creepy crawlers.}\] \[\textrm{II. Some ferocious creatures are creepy crawlers.}\] \[\textrm{III. Some alligators are not creepy crawlers.}\]

$\mathrm{(A)}\ \text{I only} \qquad\mathrm{(B)}\ \text{II only} \qquad\mathrm{(C)}\ \text{III only} \qquad\mathrm{(D)}\ \text{II and III only} \qquad\mathrm{(E)}\ \text{None must be true}$

Solution

We interpret the problem statement as a query about three abstract concepts denoted as "alligators", "creepy crawlers" and "ferocious creatures". In answering the question, we may NOT refer to reality -- for example to the fact that alligators do exist.

To make more clear that we are not using anything outside the problem statement, let's rename the three concepts as $A$, $C$, and $F$.

We got the following information:

  • If $x$ is an $A$, then $x$ is an $F$.
  • There is some $x$ that is a $C$ and at the same time an $A$.

We CAN NOT conclude that the first statement is true. For example, the situation "Johnny and Freddy are $A$s, but only Johnny is a $C$" meets both conditions, but the first statement is false.

We CAN conclude that the second statement is true. We know that there is some $x$ that is a $C$ and at the same time an $A$. Pick one such $x$ and call it Bobby. Additionally, we know that if $x$ is an $A$, then $x$ is an $F$. Bobby is an $A$, therefore Bobby is an $F$. And this is enough to prove the second statement -- Bobby is an $F$ that is also a $C$.

We CAN NOT conclude that the third statement is true. For example, consider the situation when $A$, $C$ and $F$ are equivalent (represent the same set of objects). In such case both conditions are satisfied, but the third statement is false.

Therefore the answer is $\boxed{\text{(E) \, II only}}$.

See Also

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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=2000_AMC_10_Problems/Problem_21&oldid=112571"