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

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28 Aug 2010, 22:51

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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?

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

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29 Aug 2010, 08:30

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I would follow the following steps to solve this problem:

1. Determine the amount of time it takes Bill to reach the group based on the difference in speed between Bill and the group (difference in speed is 3ft/s) => 120 ft / (6-3)ft/s = 40s

2. Determine the distance Bill travelled at a speed of 6ft/s and 3ft/s, respectively => 40s * 6ft/s = 240ft => Bill travelled 60ft at the lower speed of 3ft/s, which took him 60ft / 3ft/s = 20s

3. Determine Bill's average speed by dividing total distance by total time for the trip => 300 ft / (40s + 20s) = 5ft/s

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

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18 Sep 2016, 01:44

Ans 5 feet/sec. D=120 feet, moving walkway speed, S1 = 3 feet/sec, Bill's speed, S2 = 3 feet/sec. Total speed of Bill = 3+3 = 6 feet/sec (relative speed) time taken to reach group = 120/3 = 40 sec (Bill's relative speed - speed at which group is moving) So in 40 sec he covers 240 feet to catch up with group. Now 60 feet left which will b covered at 3 feet per sec. Time taken = 60/3 =20 sec Total time = 40+20 = 60 sec Total distance = 300 feet Average Speed = Total distance/total time = 300/60 = 5 feet/sec

gmatclubot

Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey
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18 Sep 2016, 01:44