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The moving walkway is a 300-foot long conveyor belt that

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The moving walkway is a 300-foot long conveyor belt that [#permalink] New post 02 Aug 2007, 16:08
The ‘moving walkway’ is a 300-foot long 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 combined rate (including both walkway and foot speed) of 6 feet per second, reaches the group of people, and then remains stationary 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

I am confused on the rate of Bill. Is it 6 or 3? I assume it is 6 for the first 240 feet and 3 when stationary for the last 60 feet. Please help! Thanks!
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Re: Rate Problem - Moving Walkway [#permalink] New post 02 Aug 2007, 16:21
I get E.

Their rate is 3 ft/sec and his is 6 ft/sec. Their distance is d-120 and his distance is d.

So we get two equations:
3t=d-120
6t=d

Substituting, we get 3t=6t-120, or t=40 which means that d=240 so Bill traveled 240 ft and they traveled 120 ft. (They were at 240 on the conveyor belt when they met.

So, Bill's average rate of movement is:

He traveled 6 ft/sec for 240 ft and 40 seconds. He traveled 3 ft/sec for the remaining 60 ft for 20 seconds.

(240+60)/(20+40)= 5 ft/sec
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 [#permalink] New post 02 Aug 2007, 17:02
Thats correct. The answer is 5 feet per second
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 [#permalink] New post 02 Aug 2007, 20:10
Time taken to catch up with the group = 40 seconds. In 40 seconds, though, the group has already travelled 120 feet from their previous position. So the reamining distance left = 60 feet.

Bill's average speed = 300/60 = 5 ft/sec
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Re: Rate Problem - Moving Walkway [#permalink] New post 02 Aug 2007, 20:46
helpmeongmat wrote:
The ‘moving walkway’ is a 300-foot long 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 combined rate (including both walkway and foot speed) of 6 feet per second, reaches the group of people, and then remains stationary 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

I am confused on the rate of Bill. Is it 6 or 3? I assume it is 6 for the first 240 feet and 3 when stationary for the last 60 feet. Please help! Thanks!



Ok I did this a really strange way and long way, but I think it is correct. Since the walkway is moving the people in front of him will travel x feet. So you cannot say that the people are stationary.

I just made it so Bill is moving at 3ft per sec. and distance is 120ft. the time it takes him is 40seconds. Now since we need average speed. We go back to 6ft per sec. D=RT 6ft/sec*40sec=240feet. This is the distance he had to travel just to catch up to the group of people.

Since its AVERAGE rate. we need Total Distance/Total Time.
Total D=300ft. The first 240feet Bill travels for 40seconds. The last 60feet he travels at 3ft/sec so for the last 60 feet he travels for 20secs. (t=60/3.)

Avg R= 300ft/40sec+20sec --> Avg R= 5ft/sec :-D

Answer E.
Re: Rate Problem - Moving Walkway   [#permalink] 02 Aug 2007, 20:46
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