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Difference between revisions of "2008 AMC 10B Problems/Problem 11"

(New page: ==Problem== {{problem}} ==Solution== {{solution}} ==See also== {{AMC10 box|year=2008|ab=B|num-b=10|num-a=12}})
 
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==Problem==
 
==Problem==
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{{problem}}
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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 mileae 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?
  
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(A) 2500 (B) 3000 (C) 3500 (D) 4000 (E) 4500
 
==Solution==
 
==Solution==
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{{solution}}
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Every time the pedometer flips from <math>99999</math> to
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<math>00000</math> Pete has walked <math>100000</math> steps.
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So, if the pedometer flipped <math>44</math> times
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Pete walked <math>44*100000+50000=4450000</math> steps.
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Dividing by <math>1800</math> gives <math>2472.\overline{2}</math>
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This is closest to answer <math>\boxed{A}</math>.
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=10|num-a=12}}
 
{{AMC10 box|year=2008|ab=B|num-b=10|num-a=12}}

Revision as of 16:33, 10 August 2008

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 mileae 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?

(A) 2500 (B) 3000 (C) 3500 (D) 4000 (E) 4500

Solution

Every time the pedometer flips from $99999$ to

$00000$ Pete has walked $100000$ steps.

So, if the pedometer flipped $44$ times

Pete walked $44*100000+50000=4450000$ steps.

Dividing by $1800$ gives $2472.\overline{2}$

This is closest to answer $\boxed{A}$.

See also

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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
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