Difference between revisions of "2016 AMC 8 Problems/Problem 18"
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| + | ==Problem== | ||
In an All-Area track meet, <math>216</math> sprinters enter a <math>100-</math>meter dash competition. The track has <math>6</math> lanes, so only <math>6</math> 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? | In an All-Area track meet, <math>216</math> sprinters enter a <math>100-</math>meter dash competition. The track has <math>6</math> lanes, so only <math>6</math> 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? | ||
| + | <math>\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72</math> | ||
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==Solution== | ==Solution== | ||
| + | ===Solution 1=== | ||
From any <math>n-</math>th race, only <math>\frac{1}{6}</math> will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal. | From any <math>n-</math>th race, only <math>\frac{1}{6}</math> 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: | Starting with the first race: | ||
| Line 9: | Line 11: | ||
<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>43 \ | + | Adding all of the numbers in the second column yields <math>\boxed{\textbf{(C)}\ 43}</math> |
| + | |||
| + | ===Solution 2=== | ||
| + | Every race eliminates <math>5</math> players. The winner is decided when there is only <math>1</math> runner left. You can construct the equation: <math>216</math> - <math>5x</math> = <math>1</math>. 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> | ||
| + | |||
| + | ==See Also== | ||
{{AMC8 box|year=2016|num-b=17|num-a=19}} | {{AMC8 box|year=2016|num-b=17|num-a=19}} | ||
| − | {{MAA Notice} | + | {{MAA Notice}} |
Latest revision as of 22:06, 17 May 2024
Problem
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
Solution 1
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:
![\[\frac{216}{6}=36\]](http://latex.artofproblemsolving.com/7/3/7/7374e5d18f64255ab216a8e82a352a114b66f4fd.png)
![\[\frac{36}{6}=6\]](http://latex.artofproblemsolving.com/a/9/1/a916513079199978782d4485856c6baa38221974.png)
![\[\frac{6}{6}=1\]](http://latex.artofproblemsolving.com/8/8/4/884f0b91255d6b75d42375947ec42a5e307d0bf9.png)
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. You can construct the equation: - 
= . Thus, 
players have to be eliminated. Therefore, we need 
games to decide the winner, or
See Also
| 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. 
