Difference between revisions of "1993 AJHSME Problems/Problem 23"
Mrdavid445 (talk | contribs) (Created page with "==Problem== Five runners, <math>P</math>, <math>Q</math>, <math>R</math>, <math>S</math>, <math>T</math>, have a race, and <math>P</math> beats <math>Q</math>, <math>P</math> be...") |
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<math>\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</math> | <math>\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</math> | ||
| + | |||
| + | ==Solution== | ||
| + | First, note that <math>P</math> must beat <math>Q</math>, <math>R</math>, <math>T</math>, and by transitivity, <math>S</math>. Thus <math>P</math> is in first, and not in 3rd. | ||
| + | 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: | ||
| + | PTQRS | ||
| + | PTRQS | ||
| + | PRTQS | ||
| + | Thus the answer is C. | ||
Revision as of 15:09, 22 November 2012
Problem
Five runners, 
, 
, 
, 
, 
, have a race, and beats , beats , beats , and finishes after and before . Who could NOT have finished third in the race?
Solution
First, note that must beat , , , and by transitivity, . Thus is in first, and not in 3rd.
Similarly, is beaten by , , and by transitivity, , so 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 C.
