1993 AMC 8 Problem 23

Below is the professionally curated solution for Problem 23 of the 1993 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1993 AMC 8 solutions, or check the answer key.

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Concepts:logical deduction

Difficulty rating: 1070

23.

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

PP and QQ

PP and RR

PP and SS

PP and TT

P,SP, S and TT

Solution:

Since PP beats Q,Q, R,R, and T,T, and no one beats P,P, runner PP finishes first and so cannot be third.

The clues give the chain PP before TT before QQ before S.S. So P,T,P, T, and QQ all finish ahead of S,S, meaning SS is no better than fourth and cannot be third either.

Each of Q,R,TQ, R, T can finish third: for example P,T,Q,R,SP, T, Q, R, S puts QQ third; P,R,T,Q,SP, R, T, Q, S puts TT third; and P,T,R,Q,SP, T, R, Q, S puts RR third. So only PP and SS cannot be third.

Thus, the correct answer is C .

Problem 23 in Other Years

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