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Difference between revisions of "1989 AHSME Problems/Problem 25"

(Created page with "== Problem == In a certain cross country meet between 2 teams of 5 runners each, a runner who finishes in the <math>n</math>th position contributes <math>n</math> to his teams s...")
 
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The scores of all ten runners must sum to <math>55</math>. So a winning score is anything between <math>1+2+3+4+5=15</math> and <math>\lfloor\tfrac{55}{2}\rfloor=27</math> inclusive. It is easy to check that this range is covered by considering <math>1+2+3+4+x</math>, <math>1+2+3+x+10</math> and <math>1+2+x+9+10</math>, so the answer is <math>\boxed{13}</math>.
 
The scores of all ten runners must sum to <math>55</math>. So a winning score is anything between <math>1+2+3+4+5=15</math> and <math>\lfloor\tfrac{55}{2}\rfloor=27</math> inclusive. It is easy to check that this range is covered by considering <math>1+2+3+4+x</math>, <math>1+2+3+x+10</math> and <math>1+2+x+9+10</math>, so the answer is <math>\boxed{13}</math>.
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== See also ==
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{{AHSME box|year=1989|num-b=24|num-a=26}} 
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[[Category: Intermediate Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 07:04, 22 October 2014

Problem

In a certain cross country meet between 2 teams of 5 runners each, a runner who finishes in the $n$th position contributes $n$ to his teams score. The team with the lower score wins. If there are no ties among the runners, how many different winning scores are possible?

(A) 10 (B) 13 (C) 27 (D) 120 (E) 126

Solution

The scores of all ten runners must sum to $55$. So a winning score is anything between $1+2+3+4+5=15$ and $\lfloor\tfrac{55}{2}\rfloor=27$ inclusive. It is easy to check that this range is covered by considering $1+2+3+4+x$, $1+2+3+x+10$ and $1+2+x+9+10$, so the answer is $\boxed{13}$.


See also

1989 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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