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Difference between revisions of "2016 AMC 8 Problems/Problem 18"

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<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>
 
==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.
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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:
 
<cmath>\frac{216}{6}=36</cmath>
 
<cmath>\frac{216}{6}=36</cmath>
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<cmath>\frac{6}{6}=1</cmath>
 
<cmath>\frac{6}{6}=1</cmath>
 
Adding all of the numbers in the second column yields <math>43 \rightarrow \boxed{C}</math>
 
Adding all of the numbers in the second column yields <math>43 \rightarrow \boxed{C}</math>
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==Solution 2==
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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>43 \rightarrow \boxed{C}</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>43 \rightarrow \boxed{C}</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 17:29, 23 November 2016

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

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 $43 \rightarrow \boxed{C}$

Solution 2

Every race eliminates $5$ players. The winner is decided when there is only $1$ runner left. Thus, $215$ players have to be eliminated. Therefore, we need $\frac{215}{5}$ games to decide the winner, or $43 \rightarrow \boxed{C}$

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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