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The ‘moving walkway’ is a 300-foot long walkway consisting of a convey

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Manager
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B
Joined: 07 Jun 2015
Posts: 85
WE: Design (Aerospace and Defense)
Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey [#permalink]

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New post 24 Jan 2016, 12:25
average speed = total distance covered by bill/ total time taken.

total distance covered by bill=300

total distance covered by those people in-front of him=300-120=180

total time taken by bill=total time taken by those people in-front of him=(300-120)/3=60sec ( since they move at const speed)

so avg speed of bill=300/60=5ft/sec
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Joined: 27 Sep 2015
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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey [#permalink]

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New post 02 Oct 2017, 09:26
prome2 wrote:
Here's what I think the answer should be
-------

Time Bill takes to catchup with group = 120/3
= 40 secs
Time Group has moved in 40 secs = 40 x 3
= 120 feet
Remaining Distance on walkway = 300 - 120 -120
= 60
Time taken to cover the last 60 ft = 60/3
= 20sec

Total time Bill spends on walkway = 40 + 20 = 60 secs
Total length of walkway = 300 feet

Bill's Average speed = 300/60 = 5 ft/sec






This problem may also be solved with a shortcut. Consider that Bill’s journey will end when the crowd reaches the end of the walkway (as long as he catches up with the crowd before the walkway ends). When he steps on the walkway, the crowd is 180 feet from the end. The walkway travels this distance in (180 feet)/(3 feet per second) = 60 seconds, and Bill’s average rate of movement is (300 feet)/(60 seconds) = 5 feet per second
Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey   [#permalink] 02 Oct 2017, 09:26

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