1991 AMC 8 Problem 14
Attempt Problem 14 of the 1991 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 1991 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
14.
Several students are competing in a series of three races. A student earns points for winning a race, points for finishing second, and point for finishing third. There are no ties. What is the smallest number of points that a student must earn in the three races to be guaranteed of earning more points than any other student?
Answer: D
Small hint:
To be guaranteed the lead, plan for the worst case where another student does as well as possible
Big hint:
Check the attainable three-race totals in increasing order; for each one, see whether the remaining places allow a rival to tie or exceed it
Solution:
A total of (for example ) does not guarantee first place, since another student could also reach
But if one student scores the remaining places give every other student at most So points guarantees the lead.
Thus, the correct answer is D .
Problem 14 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 AMC 8
- 1990 AMC 8
- 1992 AMC 8
- 1993 AMC 8
- 1994 AMC 8
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