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

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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 29 May 2015, 14:31
It took me 2.14 min. May be someone will find my way of solving this task useful.

Av Speed is TotalDistance / TotalTime.
TotalDistance = 300 ft
TotalTime = .... here's the trick. The Bill couldn't make it faster then group of people, because he joined them, so his time is just equal to the time of the group of peopel making the last 180 ft, which is 180/3 = 60 sec.

Av Speed = 300 / 60 = 5 ft/sec.

E.

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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 13 Jan 2016, 10:08
Bunuel wrote:
prasannajeet wrote:
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......Please make me understand whether bill there in this group to cover 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


I was little confused please som1 help me out



No, he is not. Bill takes 40 seconds to catch up the group but in that time the walkway continues to move at 3 feet per second, thus the group (as well as Bill) moves 3*40=120 feet towards the end.

Hope it's clear.


If the group moves 120 feet in 40 sec, doesn't we should consider the time taken by the Bill to catch up to that 120 feet also. Can you please explain why we are not considering time taken for that?

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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 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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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey [#permalink]

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New post 19 May 2016, 23:21
VeritasPrepKarishma wrote:
Gmatter111 wrote:
The ‘moving walkway’ is a 300-foot long walkway consisting of a conveyor belt that moves continuously at 3 feet per second. When Bill steps on the walkway, a group of people that are also on the walkway stands 120 feet in front of him. He walks toward the group at a rate of 3 feet per second. Once Bill reaches the group of people, he stops walking and stands with them until the walkway ends. What is Bill’s average rate of movement for his trip along the moving walkway?

A. 2 feet per second
B. 2.5 feet per second
C. 3 feet per second
D. 4 feet per second
E. 5 feet per second


Using relative speed concepts:

Bill covers 120 feet at a relative speed of 6 - 3 = 3 feet/sec. Time taken to cover 120 feet = 120/3 = 40 sec
Distance covered in 40 secs = 6*40 = 240 feet
So now he just has 300 - 240 = 60 feet leftover to cover at 3 feet/sec
Time taken to cover 60 feet = 60/3 = 20 sec

Average Speed = Total Distance/Total Time = 300/(40 + 20) = 5 feet/sec

Answer (E)



Bill and the group start at 180 feet.
Time taken to meet = initial distance/difference of speeds
\(\frac{180}{6-3}\) = 60 sec

What I assumed here is that in this 60 sec Bill will complete the 300 distance.
Thus, I divided the total distance by total time.
=\(\frac{300}{60}\)
=5ft/sec

What, that I am assuming, is wrong? Is the way I got the answer wrong?

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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey [#permalink]

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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

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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey   [#permalink] 02 Oct 2017, 09:26

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