Difference between revisions of "2008 AMC 12B Problems/Problem 6"

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==Problem 6==
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==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?
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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 * 44 + 50,000 = 4,450,000</math> steps, which is <math>4,450,000/1,800 = 2472.2</math> miles, which is closest to answer choice <math>\boxed{A}</math>.
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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==
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{{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]]
 
[[Category:Introductory Combinatorics Problems]]
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{{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 $99999$ 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?

$\textbf{(A)}\ 2500 \qquad \textbf{(B)}\ 3000 \qquad \textbf{(C)}\ 3500 \qquad \textbf{(D)}\ 4000 \qquad \textbf{(E)}\ 4500$

Solution

Every time the pedometer flips, Pete has walked $100,000$ steps. Therefore, Pete has walked a total of $100,000 \cdot 44 + 50,000 = 4,450,000$ steps, which is $4,450,000/1,800 = 2472.2$ miles, which is the closest to the answer choice $\boxed{A}$.

See Also

2008 AMC 12B (ProblemsAnswer KeyResources)
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

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