1991 AMC 8 Problem 14

Below is the professionally curated solution for Problem 14 of the 1991 AMC 8, 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.

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Concepts:extremal argument

Difficulty rating: 1090

14.

Several students are competing in a series of three races. A student earns 55 points for winning a race, 33 points for finishing second, and 11 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?

99

1010

1111

1313

1515

Solution:

A total of 1111 (for example 5+5+15+5+1) does not guarantee first place, since another student could also reach 11.11.

But if one student scores 5+5+3=13,5+5+3 = 13, the remaining places give every other student at most 3+3+5=11.3+3+5 = 11. So 1313 points guarantees the lead.

Thus, the correct answer is D .

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