Difference between revisions of "1989 AHSME Problems/Problem 25"
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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 == | ||
| + | {{AHSME box|year=1989|num-b=24|num-a=26}} | ||
| + | |||
| + | [[Category: Intermediate Combinatorics Problems]] | ||
{{MAA Notice}} | {{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 
th position contributes 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 
. So a winning score is anything between 
and 
inclusive. It is easy to check that this range is covered by considering 
, 
and 
, so the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
