Difference between revisions of "2008 AMC 12B Problems/Problem 6"
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==Solution== | ==Solution== | ||
| − | Every time the pedometer flips, Pete has walked <math>100,000</math> steps. Therefore, Pete has walked a total of <math>100,000 | + | Every time the pedometer flips, Pete has walked <math>100,000</math> steps. Therefore, Pete has walked a total of <math>100,000 \cdot 44 + 50,000 = 4,450,000</math> steps, which is <math>4,450,000/1,800 = 2472.2</math> miles, which is the closest to the answer choice <math>\boxed{A}</math>. |
==See Also== | ==See Also== | ||
Latest revision as of 14:18, 16 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 the closest to the 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. 
