1993 AMC 8 Problem 23
Attempt Problem 23 of the 1993 AMC 8 below, then check your answer against the professionally curated solution 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.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
23.
Five runners, have a race, and beats beats beats and finishes after and before Who could not have finished third in the race?
and
and
and
and
and
Answer: C
Small hint:
finishes ahead of and while finishes ahead of determine ’s place
Big hint:
If several runners must all finish ahead of a given runner, that runner cannot be as high as third
Solution:
Since beats and finishes ahead of and beats runner finishes first and so cannot be third.
The clues give the chain before before before So and all finish ahead of meaning is no better than fourth and cannot be third either.
Each of can finish third: for example puts third; puts third; and puts third. So only and cannot be third.
Thus, the correct answer is C .
Problem 23 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 AMC 8
- 1990 AMC 8
- 1991 AMC 8
- 1992 AMC 8
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