Difference between revisions of "2016 AMC 8 Problems/Problem 18"
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<cmath>\frac{36}{6}=6</cmath> | <cmath>\frac{36}{6}=6</cmath> | ||
<cmath>\frac{6}{6}=1</cmath> | <cmath>\frac{6}{6}=1</cmath> | ||
− | Adding all of the numbers in the second column yields <math> | + | Adding all of the numbers in the second column yields <math>\boxed{\textbf{(C)}\ 43}</math> |
==Solution 2== | ==Solution 2== | ||
− | Every race eliminates <math>5</math> players. The winner is decided when there is only <math>1</math> runner left. Thus, <math>215</math> players have to be eliminated. Therefore, we need <math>\frac{215}{5}</math> games to decide the winner, or <math> | + | Every race eliminates <math>5</math> players. The winner is decided when there is only <math>1</math> runner left. Thus, <math>215</math> players have to be eliminated. Therefore, we need <math>\frac{215}{5}</math> games to decide the winner, or <math>\boxed{\textbf{(C)}\ 43}</math> |
{{AMC8 box|year=2016|num-b=17|num-a=19}} | {{AMC8 box|year=2016|num-b=17|num-a=19}} | ||
{{MAA Notice}}\end{align} | {{MAA Notice}}\end{align} |
Revision as of 00:44, 28 November 2016
In an All-Area track meet, sprinters enter a meter dash competition. The track has lanes, so only sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?
Solution
From any th race, only will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal. Starting with the first race: Adding all of the numbers in the second column yields
Solution 2
Every race eliminates players. The winner is decided when there is only runner left. Thus, players have to be eliminated. Therefore, we need games to decide the winner, or
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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