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Conveyor belt and average speed

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Conveyor belt and average speed

by Cybermusings » Tue May 08, 2007 5:16 am
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?
2 feet per second
2.5 feet per second
3 feet per second
4 feet per second
5 feet per second


I got this question right...It's from MGMAT....However, my reasoning was diff from the solution provided...I want to make sure my attacking strategy was fine...So please provide a detailed explanation guys[/b]
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by chitoshia » Tue May 08, 2007 6:06 am
Hi,

My answer doesnt match the options, but i find it logical enough to be correct (hopefully!)

for 120 feet, he moves at 6 f/s and for the remaining distance he moves at 3 f/s.

time in each section, 120/6 = 20 s and 180/3 = 60 s

total distance/total time = 300/80 = 3.75 f/s

please comment. otherwise i will have to review my basics!! :(

regards
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by Cybermusings » Tue May 08, 2007 9:37 am
Nopes correct answer is E

Waiting for more explanation guys!
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by kookoo4tofu » Tue May 08, 2007 12:34 pm
at first i also got 3.75. then i realized that the trick in this question is that the group is still moving as the man approaches them thus, once the man approaches the group he wil no longer be (300-120) 180ft away from the end of the walkway, but in fact closer than 180ft.

so our main equation is R=D/T, where D=300. Once we find the T then we can get our answer R.

first, find the time it takes for the man to reach the group:

t=d/r = 120/6=20sec

next, the distance the group has moved while waiting for the man to reach them:

d=r*t=3*20sec = 60ft

so now we now the group has moved 60 ft closer to the end of the walkway when the man reaches them, so we can do 120ft+60ft=180ft

the amount of walkway left is 300-180 = 120ft

finally, find the time when the man is 120 ft away from the walkway

t=d/r=120/3=40sec

Now we add the 40 sec and the 20 sec from above to get the total time he was on the walkway which is 60sec.

Plug that back into the original equation and we have R=300ft/60s = 5 ft/sec

This is probably the LONGEST way of doing it cuz no way i could have done this under 2 minutes. I would like to see what the solution says
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Re: Conveyor belt and average speed

by jayhawk2001 » Tue May 08, 2007 9:45 pm
Cybermusings 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?
2 feet per second
2.5 feet per second
3 feet per second
4 feet per second
5 feet per second
Relative speed between Bill and the group = 6-3 = 3 feet/sec

120 feet gap can hence be covered by Bill in 40 sec (i.e. 40*3 = 120)

In this 40 sec, Bill covers a total of 40*6 = 240 feet.

The remaining 60 feet, he covers at 3 feet/sec which takes him 20 sec

So, average speed = total distance / total time
= 300 / (40+20) = 300/60 = 5 feet/sec

Hence E
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by Cybermusings » Wed May 09, 2007 3:03 am
I used the same logic as kookoo4tofu

However, the official exp is the same as what Jay stated....Is one approach better than the other?

I hope we didn't get it right by fluke!!!
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