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 "1993 AJHSME Problems/Problem 23"
 
(One intermediate revision by one other user not shown)
Line 9: Line 9:
 
Similarly, <math>S</math> is beaten by <math>P</math>, <math>Q</math>, and by transitivity, <math>T</math>, so <math>S</math> is in fourth or fifth, and not in third.
 
Similarly, <math>S</math> is beaten by <math>P</math>, <math>Q</math>, and by transitivity, <math>T</math>, so <math>S</math> is in fourth or fifth, and not in third.
 
All of the others can be in third, as all of the following sequences show. Each follows all of the assumptions of the problem, and they are in order from first to last:
 
All of the others can be in third, as all of the following sequences show. Each follows all of the assumptions of the problem, and they are in order from first to last:
−
PTQRS
+
PTQRS, PTRQS, PRTQS. Thus the answer is <math>\boxed{\text{(C)}\ P\  \text{and}\ S}</math>.
−
PTRQS
+
 
−
PRTQS
+
==See Also==
−
Thus the answer is C.
+
{{AJHSME box|year=1993|num-b=22|num-a=24}}
 +
{{MAA Notice}}

Latest revision as of 23:12, 4 July 2013

Problem

Five runners, $P$, $Q$, $R$, $S$, $T$, have a race, and $P$ beats $Q$, $P$ beats $R$, $Q$ beats $S$, and $T$ finishes after $P$ and before $Q$. Who could NOT have finished third in the race?

$\text{(A)}\ P\text{ and }Q \qquad \text{(B)}\ P\text{ and }R \qquad \text{(C)}\ P\text{ and }S \qquad \text{(D)}\ P\text{ and }T \qquad \text{(E)}\ P,S\text{ and }T$

Solution

First, note that $P$ must beat $Q$, $R$, $T$, and by transitivity, $S$. Thus $P$ is in first, and not in 3rd. Similarly, $S$ is beaten by $P$, $Q$, and by transitivity, $T$, so $S$ is in fourth or fifth, and not in third. All of the others can be in third, as all of the following sequences show. Each follows all of the assumptions of the problem, and they are in order from first to last: PTQRS, PTRQS, PRTQS. Thus the answer is $\boxed{\text{(C)}\ P\ \text{and}\ S}$.

See Also

1993 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 AJHSME/AMC 8 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=1993_AJHSME_Problems/Problem_23&oldid=55668"