1993 AJHSME Problems/Problem 23


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?

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


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 (ProblemsAnswer KeyResources)
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