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moving walkway ...

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by acecoolan » Sun Nov 09, 2008 8:05 pm
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 combined rate (including both walkway and foot speed) of 6 feet per second relative to the ground. 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?

2 feet per second
2.5 feet per second
3 feet per second
4 feet per second
5 feet per second
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Re: moving walkway ...

by sudhir3127 » Sun Nov 09, 2008 9:02 pm
acecoolan 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 combined rate (including both walkway and foot speed) of 6 feet per second relative to the ground. 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?

2 feet per second
2.5 feet per second
3 feet per second
4 feet per second
5 feet per second
its E 5 seconds

the group has to cover 180 feet at a speed of 3 ft. hence it takes
180/3 = 60 seconds

Bill has cover the whole distance of 300 foot

that means

300 foot/ 60 seconds = 5 ft/seconds
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by stop@800 » Sun Nov 09, 2008 9:31 pm
Good approach Sudhir.

I calculated the point of contact and than remaining distance @ speed of 3.
I agree which is longer :)
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Re: moving walkway ...

by acecoolan » Sun Nov 09, 2008 9:40 pm
sudhir3127 wrote:
acecoolan 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 combined rate (including both walkway and foot speed) of 6 feet per second relative to the ground. 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?

2 feet per second
2.5 feet per second
3 feet per second
4 feet per second
5 feet per second
its E 5 seconds

the group has to cover 180 feet at a speed of 3 ft. hence it takes
180/3 = 60 seconds

Bill has cover the whole distance of 300 foot

that means

300 foot/ 60 seconds = 5 ft/seconds
Thanks Sudhir.

Is there a way this can be calculated using just Bill's and the walkway's speed ?

I had started this problem by calculating Bill's time to cover the first 120 feet (120 / 6 = 20 sec) and then trying to catch up with the group. Well thats pretty much where I got lost.
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by raunekk » Sun Nov 09, 2008 9:51 pm
Bill will walk at a rate = 6-3 =3 feet/ sec

Time taken by bill to travel 120 feet = 120 / 3 = 40 secs.

durin this the group will also travel = 40 * 3 = 120.

Thus Bill will meet the group at 120+120 = 240 feet

the remaining feet will be 60 (300-240)

The time in which Bill will cover it =60/3 = 20 secs

Thus average Rate of Bill will be = 300 / (40+20) = 5 feet/sec

hence E
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by raunekk » Sun Nov 09, 2008 9:59 pm
I have a simililar problem..

If neone wants to try,,,

though i myself couldnt decipher it...

(Its not a GMAT question)


Q)A 100 foot long moving walkway moves at a constant rate of 6 feet per second. Al steps onto the start of the walkway and stands. Bob steps onto the start of the walkway two seconds later and strolls forward along the walkway at a constant rate of 4 feet per second. Two seconds after that, Cy reaches the start of the walkway and walks briskly forward beside the walkway at a constant rate of 8 feet per second. At a certain time, one of these three persons is exactly halfway between the other two. At that time, find the distance in feet between the start of the walkway and the middle person


[spoiler]ans:52feet[/spoiler]
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by sudhir3127 » Mon Nov 10, 2008 6:56 am
raunekk wrote:I have a simililar problem..

If neone wants to try,,,

though i myself couldnt decipher it...

(Its not a GMAT question)


Q)A 100 foot long moving walkway moves at a constant rate of 6 feet per second. Al steps onto the start of the walkway and stands. Bob steps onto the start of the walkway two seconds later and strolls forward along the walkway at a constant rate of 4 feet per second. Two seconds after that, Cy reaches the start of the walkway and walks briskly forward beside the walkway at a constant rate of 8 feet per second. At a certain time, one of these three persons is exactly halfway between the other two. At that time, find the distance in feet between the start of the walkway and the middle person


[spoiler]ans:52feet[/spoiler]
Hi raunekk .. here it goes..

We know that there are 3 people . AI Cy and Bob.
cy starts the last , but highest speed
Bob has the slowest speed among all three
thus we can assume that at some point AI would in between Cy and Bob for sure.

let that time be X

we can form equation using the data we have been given

8(X-4) + 10(X-2)/2 = 6X

solve for X ..
we get X = 26/3
thus
6X will be 26/3*6 = 52

Hope that helps .. do let me know if u have any doubts
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by raunekk » Mon Nov 10, 2008 7:45 am
thanks a lot sudhir...

can u please explain me this equation....

8(X-4) + 10(X-2)/2 = 6X
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by stop@800 » Mon Nov 10, 2008 9:25 am
Speed of A = 6
Speed of B = 6 + 4 = 10
Speed of C = 8


At a certain time, one of these three persons is exactly halfway between the other two
Let that time be X.

Time taken
A: X
B: X-2
C: X-4

A will be between C and B

so
hence distance of A = [dis B + dis C ] / 2

dis A = 6X
dis B = (X-2)10
dis C = (X-4)8

substitute in above equn and solve for X



Sudhir, I have one Qn for you?
I can not understand how you concluded
Thus we can assume that at some point AI would in between Cy and Bob for sure. "

I took two cases and with one I did not get the answer so moved to this one.


You wrote "Bob has the slowest speed among all three"
Which I think is incorrect.

IMO speed of
A < C < B

I could not conclude on meeting point
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