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

(Created page with "18. 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...")
 
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<math>(A)\mbox{ }36\mbox{          }(B)\mbox{ }42\mbox{          }(C)\mbox{ }43\mbox{          }(D)\mbox{ }60\mbox{          }(E)\mbox{ }72\mbox{          }</math>
 
<math>(A)\mbox{ }36\mbox{          }(B)\mbox{ }42\mbox{          }(C)\mbox{ }43\mbox{          }(D)\mbox{ }60\mbox{          }(E)\mbox{ }72\mbox{          }</math>
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==Solution==
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From any nth 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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Starting with the first race:
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<math>
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\frac{216}{6}=36 \\
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\frac{36}{6}=6 \\
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\frac{6}{6}=1</math>
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Adding all of the numbers in the second column yields <math>43 \rightarrow \boxed{C}</math>
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{{AMC8 box|year=2016|num-b=17|num-a=19}}
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{{MAA Notice}}

Revision as of 12:21, 23 November 2016

18. 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?


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

Solution

From any nth 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 $43 \rightarrow \boxed{C}$

2016 AMC 8 (ProblemsAnswer KeyResources)
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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