2008 AMC 12B Problems/Problem 6: Difference between revisions
New page: ==Problem 6== 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 d... |
No edit summary |
||
| (5 intermediate revisions by 5 users not shown) | |||
| 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? | ||
| Line 5: | Line 6: | ||
==Solution== | ==Solution== | ||
Every time the pedometer flips, Pete has walked <math>100,000</math> | |||
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== | ||
{{AMC12 box|year=2008|ab=B|num-b=5|num-a=7}} | {{AMC12 box|year=2008|ab=B|num-b=5|num-a=7}} | ||
[[Category:Introductory Combinatorics Problems]] | |||
{{MAA Notice}} | |||
Latest revision as of 13:18, 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 | |
These problems are copyrighted © by the Mathematical Association of America.