Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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...")
 
Line 3: Line 3:
  
 
<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>
 +
==Solution==
 +
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.
 +
Starting with the first race:
 +
<math>
 +
\frac{216}{6}=36 \\
 +
\frac{36}{6}=6 \\
 +
\frac{6}{6}=1</math>
 +
Adding all of the numbers in the second column yields <math>43 \rightarrow \boxed{C}</math>
 +
{{AMC8 box|year=2016|num-b=17|num-a=19}}
 +
{{MAA Notice}}

Revision as of 11:21, 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 (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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_18&oldid=81439"