Last visit was: 19 Nov 2025, 12:45 It is currently 19 Nov 2025, 12:45
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2 
User avatar
pkm9995
User avatar
Joined: 07 Jun 2015
Last visit: 11 Dec 2024
Posts: 294
Own Kudos:
135
 [2]
Given Kudos: 11
WE:Design (Aerospace and Defense)
Posts: 294
Kudos: 135
 [2]
The ‘moving walkway’ is a 300-foot long walkway consisting of a convey 24 Jan 2016, 12:25
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
average speed = total distance covered by bill/ total time taken.

total distance covered by bill=300

total distance covered by those people in-front of him=300-120=180

total time taken by bill=total time taken by those people in-front of him=(300-120)/3=60sec ( since they move at const speed)

so avg speed of bill=300/60=5ft/sec
User avatar
bimalr9
User avatar
Joined: 20 Mar 2015
Last visit: 07 Nov 2017
Posts: 39
Own Kudos:
327
Given Kudos: 9
Posts: 39
Kudos: 327
The ‘moving walkway’ is a 300-foot long walkway consisting of a convey 19 May 2016, 23:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasPrepKarishma
Gmatter111
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

Answer (E)
Show more


Bill and the group start at 180 feet.
Time taken to meet = initial distance/difference of speeds
\(\frac{180}{6-3}\) = 60 sec

What I assumed here is that in this 60 sec Bill will complete the 300 distance.
Thus, I divided the total distance by total time.
=\(\frac{300}{60}\)
=5ft/sec

What, that I am assuming, is wrong? Is the way I got the answer wrong?
avatar
rupak007
avatar
Joined: 27 Sep 2015
Last visit: 04 Oct 2018
Posts: 5
Own Kudos:
3
 [2]
Given Kudos: 299
Products:
My Reviews
Posts: 5
Kudos: 3
 [2]
The ‘moving walkway’ is a 300-foot long walkway consisting of a convey 02 Oct 2017, 09:26
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
prome2
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
Show more





This problem may also be solved with a shortcut. Consider that Bill’s journey will end when the crowd reaches the end of the walkway (as long as he catches up with the crowd before the walkway ends). When he steps on the walkway, the crowd is 180 feet from the end. The walkway travels this distance in (180 feet)/(3 feet per second) = 60 seconds, and Bill’s average rate of movement is (300 feet)/(60 seconds) = 5 feet per second
avatar
keerthinutakki
avatar
Joined: 20 Apr 2014
Last visit: 26 Nov 2018
Posts: 2
Own Kudos:
11
Given Kudos: 33
Posts: 2
Kudos: 11
The ‘moving walkway’ is a 300-foot long walkway consisting of a convey 08 Sep 2018, 05:05
Kudos
Add Kudos
Bookmarks
Bookmark this Post
After going through all explanations given in this thread, I was way too confused why the relative speed of Bill is (6-3) 3 feet/sec. Since both Bill and group of people are moving on 3 feet/sec conveyor Belt, 3 feet/sec gets cancelled when Bill moves to catch up the group of people. Below is my explanation of my understanding.

The rate at which Bill moves is (6+3) 9 feet/sec.

The rate at which the group of people moves is 3 feet/sec.

The distance between Bill and group of people is 120 feet.

Since Bill and group of people are moving in same direction, the relative speed will be (9-3) 6 feet/sec
Total Time taken by Bill to cover 120 feet distance is 120/6 = 20 sec.

When Bill makes his move to catch up the group of people, even the group of people moves at 3 feet/sec. Total distance covered by Bill in 20 sec with his rate of 9 feet/sec is 180 feet, that makes remaining distance = 120 feet.

After Bill catches up with group of people he continues at a rate of 3 feet/sec, and time taken to cover that 120 feet at 3 feet/sec will be 40 sec.

Average rate of Bill = Total distance/Total Time = 300/60 = 5 feet/sec.
User avatar
hero_with_1000_faces
User avatar
Joined: 02 Jan 2016
Last visit: 17 Mar 2025
Posts: 358
Own Kudos:
146
Given Kudos: 314
Status:Studying 4Gmat
Location: India
Concentration: Strategy, Entrepreneurship
Schools: Darden '22 Marshall '22 AGSM '23
GMAT 1: 590 Q37 V33
GPA: 4
WE:Law (Manufacturing)
Products:
My Reviews
Schools: Darden '22 Marshall '22 AGSM '23
GMAT 1: 590 Q37 V33
Posts: 358
Kudos: 146
The ‘moving walkway’ is a 300-foot long walkway consisting of a convey 18 Jan 2020, 21:58
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I did similar to karishma method's

The total distance is 300 foot, it will take 100 seconds if he stands throughout, however he walks for 120 foot.
Hence, he saved time when he walked for 120 foot i.e saved 120/3 = 40 seconds.

Thus, 100 secs - 40 secs = 60 seconds total time taken.

Therefore 300/60 = 5 fps
User avatar
Nikhil30
User avatar
Current Student
Joined: 03 Jan 2019
Last visit: 29 Sep 2021
Posts: 28
Own Kudos:
42
Given Kudos: 13
Schools: Ross '23 (A)
Schools: Ross '23 (A)
Posts: 28
Kudos: 42
The ‘moving walkway’ is a 300-foot long walkway consisting of a convey 28 Sep 2020, 19:30
Kudos
Add Kudos
Bookmarks
Bookmark this Post
The concept here tested is "relative speed " and "Average speed"
relative speed of objest and person moving in same direction Sa-Sb
and avg speed= total distance / total time.
1. RS= 6-3=3feet/sec
time take to cover 120 feet to touch crowd= t=d/s
120/3=40 second
but in that 40 seconds crowd would have moved = 40*3 d=s*t= 120feet
so , ben gona catch them at 240 feet coz ben D=speed * time = 40*3=120 feet
now remaing 60 feet he gona travel with crowd @ 3feet/sec. i.e time=d/s
60/3=20 seconds
now , total time for ben = 40+20=60 seconds
and total distance =300 feet avg speed=300/60=5 feet/sec.
hope its clear. :)
User avatar
Fdambro294
User avatar
Joined: 10 Jul 2019
Last visit: 20 Aug 2025
Posts: 1,350
Own Kudos:
742
 [1]
Given Kudos: 1,656
Posts: 1,350
Kudos: 742
 [1]
The ‘moving walkway’ is a 300-foot long walkway consisting of a convey 06 Oct 2020, 10:40
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Average Rate of Movement seems like a different way to try to say “average Speed”


Avg Speed = (Total Distance) / (Total Time)


1st) Bill walks with the elevator until he Catches up with the People


Bill’s Effective Speed = (3 + 3) = 6 ft/sec ———as the moving walkways helps push him along and extra 3 ft/sec faster

The Gap Distance that Bill must Catch Up and Close = 120 ft

The group of people are moving with the walkway at 3 ft/sec in the Same Direction

Time = Gap Distance / Relative Speed = 120 / (6 - 3) = 40 seconds

In these 40 seconds, Bill travels: 6 * 40 = 240 feet



The 2nd Part is where Bill moves with the moving walkway and stands still.

60 feet remains

Bill’s Speed at this point = Walkway Speed of 3 ft/sec

Time = 60 / 3 = 20 seconds




Average Speed = (Total Distance of 300 ft) / (Total Time of 40 seconds + 20 seconds)

Avg Speed = 300 / 60 =


5 ft/sec


-E- is the Correct Answer

Posted from my mobile device
User avatar
Subhrajyoti
User avatar
Joined: 03 Jun 2013
Last visit: 11 Jan 2022
Posts: 70
Own Kudos:
15
Given Kudos: 1
Posts: 70
Kudos: 15
The ‘moving walkway’ is a 300-foot long walkway consisting of a convey 08 Oct 2020, 08:52
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I am not sure that i understood the explainations provided. Below is my question:
Bill speed wil be (3ft/sec (walking speed)+ 3ft/sec (conveyor belt speed))= 6ft/sec, now at this rate he will cover 120 ft @ 20sec. Now in 20 sec the group would have traveled additional 60ft i.e 3ft/sec conveyor belt speed* 20sec , so Bill will take 10 more sec, that makes 30sec for Bill to catch with group now the rest 120 (180-120) is what is left will be covered @ 40 sec so the total 300 ft will be covered in 70 sec. I guess my doubt is why is there additional feet which is getting added. Can somebody please explain .
User avatar
Fdambro294
User avatar
Joined: 10 Jul 2019
Last visit: 20 Aug 2025
Posts: 1,350
Own Kudos:
742
Given Kudos: 1,656
Posts: 1,350
Kudos: 742
The ‘moving walkway’ is a 300-foot long walkway consisting of a convey 08 Oct 2020, 09:31
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Subhrajyoti
I am not sure that i understood the explainations provided. Below is my question:
Bill speed wil be (3ft/sec (walking speed)+ 3ft/sec (conveyor belt speed))= 6ft/sec, now at this rate he will cover 120 ft @ 20sec. Now in 20 sec the group would have traveled additional 60ft i.e 3ft/sec conveyor belt speed* 20sec , so Bill will take 10 more sec, that makes 30sec for Bill to catch with group now the rest 120 (180-120) is what is left will be covered @ 40 sec so the total 300 ft will be covered in 70 sec. I guess my doubt is why is there additional feet which is getting added. Can somebody please explain .
Show more


It will take Bill 40 seconds to catch up with the group.

As he’s moving to catch up at 6 ft/sec, the other people are moving also at a rate of 3ft/sec. The entire time that Bill is walking the people are moving with the escalator.

After 40 seconds, eventually Bill’s faster speed will overcome the Distance between him and the group, as well as the fact that the group is constantly moving away from him.


After 40 seconds of walking, Bill meets up and catches the group, he will have traveled 240 feet.

60 feet remains. He “floats” with the escalator for this last 60 feet.


I hope that helps somewhat and didn’t make it anymore confusing?

Posted from my mobile device
avatar
onlyformohit1122
avatar
Joined: 26 May 2017
Last visit: 26 Oct 2020
Posts: 3
Own Kudos:
1
 [1]
Given Kudos: 13
Posts: 3
Kudos: 1
 [1]
The ‘moving walkway’ is a 300-foot long walkway consisting of a convey 22 Oct 2020, 02:01
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Short-cut:

This problem may also be solved with a shortcut. Consider that Bill’s journey will end when the crowd reaches the end of the walkway (as long as he catches up with the crowd before the walkway ends). When he steps on the walkway, the crowd is 180 feet from the end. The walkway travels this distance in (180 feet)/(3 feet per second) = 60 seconds, and Bill’s average rate of movement is (300 feet)/(60 seconds) = 5 feet per
avatar
Nistha86
avatar
Joined: 14 Mar 2020
Last visit: 08 Feb 2022
Posts: 6
Own Kudos:
4
Given Kudos: 304
Posts: 6
Kudos: 4
The ‘moving walkway’ is a 300-foot long walkway consisting of a convey 21 Nov 2020, 21:49
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Banuel

For objects on movewalk aways we take relative speed as + for same direction and - for opp?or do we calculate the relative speed from ground(stationary)?

here the relative speed till B reaches group of people can it be determined as b+w(w-walkaway speed)?

Thanks
Nistha
User avatar
fadrek
User avatar
Joined: 26 Mar 2022
Last visit: 27 May 2023
Posts: 1
Given Kudos: 9
Posts: 1
Kudos: 0
The ‘moving walkway’ is a 300-foot long walkway consisting of a convey 03 Oct 2022, 21:51
Kudos
Add Kudos
Bookmarks
Bookmark this Post
There is a logical way to think about this question :-)

Savg=Dtotal/Ttotal

Since he is going to meet the pple at a certain point, this means that they will reach to the end at the same time! Hence, we can use the time of pple to reach the end of the trip which is t=180f/ 3f/sec = 60sec. Now S=300/60= 5 sec.

Answer is E
User avatar
nivivacious
User avatar
Joined: 10 Mar 2015
Last visit: 18 Aug 2024
Posts: 240
Own Kudos:
302
 [1]
Given Kudos: 175
Location: India
Concentration: Strategy, Marketing
GPA: 3.5
WE:Advertising (Advertising and PR)
Posts: 240
Kudos: 302
 [1]
The ‘moving walkway’ is a 300-foot long walkway consisting of a convey 04 Jun 2023, 01:19
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Cantona
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.
Show more


I didn't understand why 120/ (6-3)
So this is how I worked towards it:

Imagine a point B
Start of the walkway ------ Point where group is ------ Point B ----- End of the walkway
Distance from where group is to Point B to be X
Therefore: Man's time = group's time
(120+x)/6 = x/3
Which tells us x = 120
Replacing it in the equation
120+120/6 =240/6 =40s

Now we have to look into remaining: 300-240 = 60m
60m/3m/s = 20s

Total : 40+20 = 60s
Therefore 300/60 = 5m/s

brunel: Does this method make sense?
User avatar
whatsarc
User avatar
Joined: 26 Aug 2023
Last visit: 19 Nov 2025
Posts: 109
Own Kudos:
102
 [1]
Given Kudos: 78
Products:
My Reviews
Posts: 109
Kudos: 102
 [1]
The ‘moving walkway’ is a 300-foot long walkway consisting of a convey 14 Jul 2024, 07:17
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
The group covers 180 ft in 60 seconds (3ft/sec)
And since the Bill catches up the group somewhere and finishes the walkway together,
We can assume that it takes bill to cover the distance in 60 seconds as well.

So bill covers 300 ft in 60 seconds and that is 5ft/second.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,588
Own Kudos:
1,079
Posts: 38,588
Kudos: 1,079
The ‘moving walkway’ is a 300-foot long walkway consisting of a convey 31 Jul 2025, 11:48
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
   1   2 
Moderators:
Bunuel
Math Expert
105390 posts
Krunaal
Tuck School Moderator
805 posts