Difference between revisions of "2008 AMC 12B Problems/Problem 6"
| Line 1: | Line 1: | ||
| − | ==Problem | + | ==Problem== |
Postman Pete has a pedometer to count his steps. The pedometer records up to <math>99999</math> steps, then flips over to <math>00000</math> on the next step. Pete plans to determine his mileage for a year. On January <math>1</math> Pete sets the pedometer to <math>00000</math>. During the year, the pedometer flips from <math>99999</math> to <math>00000</math> forty-four times. On December <math>31</math> the pedometer reads <math>50000</math>. Pete takes <math>1800</math> steps per mile. Which of the following is closest to the number of miles Pete walked during the year? | Postman Pete has a pedometer to count his steps. The pedometer records up to <math>99999</math> steps, then flips over to <math>00000</math> on the next step. Pete plans to determine his mileage for a year. On January <math>1</math> Pete sets the pedometer to <math>00000</math>. During the year, the pedometer flips from <math>99999</math> to <math>00000</math> forty-four times. On December <math>31</math> the pedometer reads <math>50000</math>. Pete takes <math>1800</math> steps per mile. Which of the following is closest to the number of miles Pete walked during the year? | ||
Revision as of 12:48, 15 February 2021
Problem
Postman Pete has a pedometer to count his steps. The pedometer records up to 
steps, then flips over to 
on the next step. Pete plans to determine his mileage for a year. On January 
Pete sets the pedometer to . During the year, the pedometer flips from to forty-four times. On December 
the pedometer reads 
. Pete takes 
steps per mile. Which of the following is closest to the number of miles Pete walked during the year?
Solution
Every time the pedometer flips, Pete has walked 
steps. Therefore, Pete has walked a total of 
steps, which is 
miles, which is closest to answer choice 
.
See Also
| 2008 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 5 |
Followed by Problem 7 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
