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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>
 
<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==
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===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:
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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>
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Adding all of the numbers in the second column yields <math>43 \rightarrow \boxed{C}</math>
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Adding all of the numbers in the second column yields <math>\boxed{\textbf{(C)}\ 43}</math>
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==Solution 2==
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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>43 \rightarrow \boxed{C}</math>
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===Solution 2===
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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>
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==See Also==
 
{{AMC8 box|year=2016|num-b=17|num-a=19}}
 
{{AMC8 box|year=2016|num-b=17|num-a=19}}
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{{MAA Notice}}\end{align}
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{{MAA Notice}}

Latest revision as of 22:06, 17 May 2024

Problem

In an All-Area track meet, $216$ sprinters enter a $100-$meter dash competition. The track has $6$ lanes, so only $6$ 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?

$\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72$

Solution

Solution 1

From any $n-$th race, only $\frac{1}{6}$ 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: \[\frac{216}{6}=36\] \[\frac{36}{6}=6\] \[\frac{6}{6}=1\] Adding all of the numbers in the second column yields $\boxed{\textbf{(C)}\ 43}$

Solution 2

Every race eliminates $5$ players. The winner is decided when there is only $1$ runner left. You can construct the equation: $216$ - $5x$ = $1$. Thus, $215$ players have to be eliminated. Therefore, we need $\frac{215}{5}$ games to decide the winner, or $\boxed{\textbf{(C)}\ 43}$

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

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