Difference between revisions of "2008 AMC 10B Problems/Problem 12"
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==Problem== | ==Problem== | ||
− | {{ | + | Postman Pete has a pedometer to count his steps. The pedometer records up to 99999 steps, then flips over to 00000 on the next step. Pete plans to determine his mileage for a year. On January 1 Pete sets the pedometer to 00000. During the year, the pedometer flips from 99999 to 00000 forty-four |
+ | times. On December 31 the pedometer reads 50000. Pete takes 1800 steps per mile. Which of the following is closest to the number of miles Pete walked during the year? | ||
+ | |||
+ | <math>\mathrm{(A)}\ 2500\qquad\mathrm{(B)}\ 3000\qquad\mathrm{(C)}\ 3500\qquad\mathrm{(D)}\ 4000\qquad\mathrm{(E)}\ 4500</math> | ||
==Solution== | ==Solution== | ||
− | {{ | + | Every time the pedometer flips from <math>99999</math> to |
+ | |||
+ | <math>00000</math> Pete has walked <math>100000</math> steps. | ||
+ | |||
+ | So, if the pedometer flipped <math>44</math> times | ||
+ | |||
+ | Pete walked <math>44*100000+50000=4450000</math> steps. | ||
+ | |||
+ | Dividing by <math>1800</math> steps per mile gives <math>2472.\overline{2}</math> | ||
+ | |||
+ | This is closest to answer <math>\boxed{A}</math>. | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|num-b=11|num-a=13}} | {{AMC10 box|year=2008|ab=B|num-b=11|num-a=13}} | ||
+ | {{MAA Notice}} |
Latest revision as of 11:42, 7 June 2021
Problem
Postman Pete has a pedometer to count his steps. The pedometer records up to 99999 steps, then flips over to 00000 on the next step. Pete plans to determine his mileage for a year. On January 1 Pete sets the pedometer to 00000. During the year, the pedometer flips from 99999 to 00000 forty-four times. On December 31 the pedometer reads 50000. Pete takes 1800 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 from to
Pete has walked steps.
So, if the pedometer flipped times
Pete walked steps.
Dividing by steps per mile gives
This is closest to answer .
See also
2008 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.