Problem 23 · 1993 AJHSME
Stretch
Logic & Word Problems
orderingconstraints
Five runners, P, Q, R, S, T, have a race. 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?
Show answer
Answer: C — P and S.
Show hints
Hint 1 of 2
Don't try to build one full finishing order — there are many. Instead, link the clues into a chain (who-beats-whom) and ask of each runner: how many people MUST be ahead of them, and how many MUST be behind?
Still stuck? Show hint 2 →
Hint 2 of 2
To be exactly third, a runner needs room: at least 2 people forced ahead AND at least 2 forced behind. Anyone with too many forced ahead (or too many forced behind) is locked out of 3rd.
Show solution
Approach: count forced-ahead and forced-behind for each runner
- Chain the clues. 'T after P and before Q' plus 'Q beats S' gives P < T < Q < S, and separately P < R. Read positions as left-to-right finishing order.
- P sits ahead of T, Q, R, and S — all four others — so P is forced into 1st. With 0 people who can be ahead of it, P can never be 3rd.
- S sits behind Q, which is behind both P and T, so at least P, T, Q finish ahead of S: that's 3 runners, pinning S at 4th or 5th. S can never be 3rd either.
- Everyone else (Q, R, T) has enough wiggle room to land 3rd in some valid order, so the runners who could NOT be third are P and S.
- Why this transfers: for ranking-with-clues, you rarely need the full order. Turn the clues into one chain, then count must-be-ahead / must-be-behind. A position is impossible for anyone whose forced-ahead count is too big (can't move up) or too small (forced even higher).
Mark:
· log in to save