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

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Updated on: 03 Oct 2017, 01:48
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60% (02:35) correct 40% (02:41) wrong based on 750 sessions

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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?

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

Originally posted by Gmatter111 on 02 Dec 2009, 20:34.
Last edited by Bunuel on 03 Oct 2017, 01:48, edited 2 times in total.
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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey  [#permalink]

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28 May 2015, 20:20
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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

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

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30 May 2012, 19:24
16
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This is how I broke it down:

Total Distance = 300 feet
default rate is 3fps (feet per second)
walking rate is 6fps (default + walking)
catch up rate: is walking rate - default rate = 6-3 = 3fps

Part 1: Time for Catch up
As he gets on the walk way he instantly starts "catching up" to the group that is 120 feet ahead of him. How long does this take?

D/R = Time; so 120feet/3fps [ note I'm using the catch up rate here ] = 40 seconds of catch up

After 40 seconds, how much of the 300 feet of the moonwalk has been used? well if he was moving for 40 seconds at 6 feet per second (he was walking) then he covered 240 feet of actual walkway.

300-240 feet = 60 feet.

There are only 60 feet left of walkway.

Part 2: Standing with the crowd
Now bill just idles with the crowd for the last 60 feet. 60ft/3fps [ note this is the default rate ] = 20 seconds.

Part 3: Average rate

6 fps for 40 seconds = 240 feet
3 fps for 20 seconds = 60 feet
X fps for 60 seconds = 300 feet

300/60 = 5fps average rate. Answer is E.
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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey  [#permalink]

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02 Dec 2009, 23:35
1
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?

3.75feet per second?

total distance is 300 feets
120 feet is covered at a speed of (Bills' speed + Belt Speed) = (3+3) = 6feet/sec. Time taken 20sec
remaining 180 feet is covered at a speed of (Bills' speed + Belt Speed) = (0+3) = 3feet/sec. Time taken 60sec

so total time taken to cover 300 foot long walkway is 80 sec. Avg Speed = 300/80 = 3.75 feet/sec
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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey  [#permalink]

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Updated on: 03 Dec 2009, 02:31
14
2
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

Originally posted by prome2 on 03 Dec 2009, 02:11.
Last edited by prome2 on 03 Dec 2009, 02:31, edited 1 time in total.
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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey  [#permalink]

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03 Dec 2009, 02:25
1
kp1811 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?

3.75feet per second?

total distance is 300 feets
120 feet is covered at a speed of (Bills' speed + Belt Speed) = (3+3) = 6feet/sec. Time taken 20sec
remaining 180 feet is covered at a speed of (Bills' speed + Belt Speed) = (0+3) = 3feet/sec. Time taken 60sec

so total time taken to cover 300 foot long walkway is 80 sec. Avg Speed = 300/80 = 3.75 feet/sec

I could be wrong, but the average of 2 average speeds is not the average speed of the entire journey.

Let's take a man cycling 60 miles to work, at 10 miles/hour there and 15 miles/hour back.
The average of the two average speeds will be 12.5 miles/hour

However, the computed average speed is total dist over total time.
If he cycles 10 miles/hour there, he will take 6 hours, and at 15 miles/hour back he will take 4 hours.

120miles/10 hours = 12 miles/hour

----

Although I believe weighted average could be used to compute average speed.
But you would need to know time travelled at each speed.

((AverageSpeed1*Time@Speed1) + (AverageSpeed2*Time@Speed2) / (TotalTime)
=((6*10)+(4*15)) / 10
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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey  [#permalink]

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03 Dec 2009, 04:40
kp1811 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?

3.75feet per second?

total distance is 300 feets
120 feet is covered at a speed of (Bills' speed + Belt Speed) = (3+3) = 6feet/sec. Time taken 20sec
remaining 180 feet is covered at a speed of (Bills' speed + Belt Speed) = (0+3) = 3feet/sec. Time taken 60sec

so total time taken to cover 300 foot long walkway is 80 sec. Avg Speed = 300/80 = 3.75 feet/sec

120 feet is covered at a speed of (Bills' speed + Belt Speed) = (3+3) = 6feet/sec. Time taken 20sec
but in this 20sec the group moves a further of 20*3= 60feet so Bill will take another 60/6 =10sec to catch the group. Now remaining 120 feet will be covered in 120/3 = 40 sec. So total time to cover is 20+10+40 = 70sec
Avg speed = 300/70 = 30/7 feet/sec
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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey  [#permalink]

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03 Dec 2009, 08:14
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 is correct. Average speed is total distance/total time.
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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey  [#permalink]

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07 May 2012, 02:06
kp1811 wrote:
kp1811 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?

3.75feet per second?

total distance is 300 feets
120 feet is covered at a speed of (Bills' speed + Belt Speed) = (3+3) = 6feet/sec. Time taken 20sec
remaining 180 feet is covered at a speed of (Bills' speed + Belt Speed) = (0+3) = 3feet/sec. Time taken 60sec

so total time taken to cover 300 foot long walkway is 80 sec. Avg Speed = 300/80 = 3.75 feet/sec

120 feet is covered at a speed of (Bills' speed + Belt Speed) = (3+3) = 6feet/sec. Time taken 20sec
but in this 20sec the group moves a further of 20*3= 60feet so Bill will take another 60/6 =10sec to catch the group. Now remaining 120 feet will be covered in 120/3 = 40 sec. So total time to cover is 20+10+40 = 70sec
Avg speed = 300/70 = 30/7 feet/sec

is this correct? does the group move further, even so is it required to be calculated for bill?
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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey  [#permalink]

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09 May 2012, 11:36
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

Hi Bunuel, could u please explain how to solve this question...
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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey  [#permalink]

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20 Sep 2013, 11:29
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

Similar question to practice: barry-walks-from-one-end-to-the-other-of-a-30-meter-long-141013.html
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26 Sep 2013, 10:29
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I agree with the above posters about the way to handle the equations (Catona & Prome2) but to me there is an option that hasn't been discussed that just POPS out at me. I don't know if it does for anyone else...

If you look at the options for the choices:

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

A, B, and C don't make ANY sense! we know for some time...(a majority of the 300ft) Bill will walk at a pace of 6ft per second...and then at some later time he will stop and move at 3 ft per second... The average MUST be between 3-6 ft per second or the Earth is FLAT!!!.

So we are down to D & E, D doesn't make sense on a weighted averages level because we know when Bill gets on the walkway the patrons are already 40% of the way down the 300ft walkway AND they are still moving at half the rate of Bill!! So for Bill to average 4 ft per second he would have had to spend a majority of the 300 ft at 3ft per second because 4 is much closer (when you factor in the size of the values we're dealing with 2 is double 1) to 3 than to 6. We know from the information that isn't possible. Bill must have spent the majority of his time at 6 ft per second before he stopped walking. That leaves only answer E as plausible.
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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey  [#permalink]

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22 Feb 2014, 02:20
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
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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey  [#permalink]

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22 Feb 2014, 03:04
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.
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27 Feb 2014, 14:00
Hello,
I just bumped into this question, I solved it this way :
At time t, Bill meets the crowd at the the distance x, the crowd has covered 120+X :
=> X/6=(X+120)/3
=> X=720/3=240 feet (I neglected the sign '-', x being a distance, so an absolute value)
So, Bill spends 240/6 =40 sec to cover the 240 feet

Remain 60 feet that bills covers in : 60/3 = 20 s

Hence total time : 40+20=60 sec.t

Bill s rate is : 300/60 = 5 feet

I think there s sthg wrong with my reasoning but I cant figure it out
Thank you for helping
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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey  [#permalink]

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28 Feb 2014, 02:52
First 240 Feet are covered @ 6 f/sec...... so time taken = 40sec
Last 60 Feet are covered @ 3 f/sec...... so time taken = 20sec

Total time = 60seconds, total distance = 300 Feets

Avg speed = 300/60 = 5 = Answer = E
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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey  [#permalink]

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28 Feb 2014, 03:44
breakfree1 wrote:
Hello,
I just bumped into this question, I solved it this way :
At time t, Bill meets the crowd at the the distance x, the crowd has covered 120+X :
=> X/6=(X+120)/3
=> X=720/3=240 feet (I neglected the sign '-', x being a distance, so an absolute value)
So, Bill spends 240/6 =40 sec to cover the 240 feet

Remain 60 feet that bills covers in : 60/3 = 20 s

Hence total time : 40+20=60 sec.t

Bill s rate is : 300/60 = 5 feet

I think there s sthg wrong with my reasoning but I cant figure it out
Thank you for helping

The equation should be $$\frac{x}{6}=\frac{(x-120)}{3}$$ --> $$x=240$$.

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

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16 Apr 2014, 08:49
I think all info should be one place ; so here goes

let x be total distance covered by Bill till he catches group of people

as we know speed=distance/time; as time will be equal when it will catch

x/6(relative speed bill +group)=total -120/3 (*obviously it will be added only then bill will be able to catch; total x but 120 is already covered)

x/6=x-120/3

x=240 covered by speed of 6 , so time took 240/6=40, rest 60 will be covered by 3 (*speed of wheel)=60/3=20

Avg. speed of bill= 300/(40+20)=5
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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey  [#permalink]

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29 May 2015, 14:31
1
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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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?
Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a convey &nbs [#permalink] 13 Jan 2016, 10:08

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