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

Question

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

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

pkm9995 · Answer

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

bimalr9 · Answer

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?

rupak007 · Answer

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

keerthinutakki · Answer

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.

hero_with_1000_faces · Answer

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

Nikhil30 · Answer

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.

Fdambro294 · Answer

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

Subhrajyoti · Answer

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 .

Fdambro294 · Answer

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

onlyformohit1122 · Answer

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

Nistha86 · Answer

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

fadrek · Answer

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

nivivacious · Answer

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?

whatsarc · Answer

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.

Veee17 · Answer

I solved this using the following approach. Could a tutor please check whether my solution is correct?

Belt's speed = 3ft/sec
Bill's speed = 6ft/sec

Total speed combined to catch up with the group = 9ft/sec (since Belt's speed will increase Bill's speed to catch up with the group, same case as a moving escalator)
Group's speed = 3ft/sec

Hence, relative speed = (9-3)ft/sec = 6ft/sec

So, the time taken to cover the initial gap of 120ft for Bill will be 20sec.

Hence, he will meet the group after 20 seconds at the 180 ft mark on the belt.

Remaining distance = 120ft, time taken by Bill to cover that will be 120ft/3ft per second = 40sec.

Now, average speed will be total distance/total time = 300ft/(40+20)sec = 300ft/60sec
= 5ft/sec

Edskore · Answer

Your solution is correct, and honestly the way you set it up is exactly the clean way to do it. Answer is E, 5 ft/sec.

The whole problem hinges on realizing Bill's speed relative to the ground is 6+3=9 (he walks AND rides the belt), while the group only gets the belt's 3. So the gap closes at 9-3=6 ft/sec. That's the part people usually trip on, they forget the group is also moving the whole time, not standing still waiting for Bill.

Once you have 120/6=20 seconds to close the gap, the rest is just bookkeeping. Bill's already covered 20x9=180 ft of the 300 when he catches up, then he rides the remaining 120 ft with the group at the belt's 3 ft/sec for 40 more seconds. Total 300 ft over 60 seconds is 5 ft/sec.

Nice clean solve. The trap here isn't really the math, it's forgetting the group keeps moving while Bill closes in.

SVaidyaraman · Answer

it's a two-phase average-rate problem, and the trap is that average rate is total distance ÷ total time, never the average of the two speeds.

Key idea: Split the trip into phases with different speeds, find the time for each, then divide total distance by total time.

Phase 1 — Bill walks to catch the group.

Both Bill and the group are carried by the belt at 3 ft/s, so the belt cancels out for the catching-up — relative to the group, Bill closes at his walking speed of 3 ft/s.

Time to close 120 feet: 120 ÷ 3 = 40 seconds

But his ground speed during this phase is belt + walking = 3 + 3 = 6 ft/s.

Distance covered on the walkway: 6 × 40 = 240 feet

Phase 2 — Bill stands with the group.

Remaining distance: 300 − 240 = 60 feet
Speed = belt only = 3 ft/s
Time: 60 ÷ 3 = 20 seconds

Average rate:

Total distance = 300 feet
Total time = 40 + 20 = 60 seconds

$$	ext{Average} = \frac{300}{60} = 5 	ext{ ft/s}$$

Answer: E. 5 feet per second

The trap: averaging the two speeds — (6 + 3)/2 = 4.5, or picking 4 — is wrong, because he spends *different amounts of time* at each speed. He's at 6 ft/s for 40 seconds but only at 3 ft/s for 20 seconds, so the faster phase dominates. Average rate is always total distance over total time.

The transferable move: when speed changes mid-journey, never average the speeds — accumulate distance and time separately, then divide. Same structure as any weighted-average problem: the weights are the times, not the segments.

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=89290 --- Harsha GMAT-Club-Forum-qxz1vz7b.png

egmat · Answer

Hi Veee17,

Good instinct to reach for relative speed, and you're right that the group drifts along with the belt at  3 ft/sec . But there's one broken step, and the fact that you still landed on  5 ft/sec  is exactly the coincidence you sensed.

The broken step: you added the belt speed twice

Watch how the belt gets counted:

- Bill walks at  3  ft/sec.
- The belt carries him an extra  3  ft/sec.
- So Bill's speed over the ground is  3 + 3 = 6  ft/sec. That's it - the belt is added  once .

You then added the belt a  second  time ( 6 + 3 = 9 ), as if the walkway sped him up twice. It doesn't. His ground speed is  6 , not  9 .

Because of that, your relative speed should be  6 - 3 = 3  ft/sec, not  6 , so the real catch-up time is  120 / 3 = 40 sec , and he meets the group at the  240-ft  mark - not  20 sec  at the  180-ft  mark.

Why the answer came out right anyway

Here's the neat part. The trip ends the instant the  group  reaches the end of the walkway. When Bill steps on, the group is  180 ft  from the end and rolls there at  3 ft/sec , taking  180 / 3 = 60 sec  - a fixed number that doesn't depend on Bill at all.

Once Bill joins them, he finishes with them, so his total time is also  60 sec . That means:

Average = 300 / 60 = 5 ft/sec

no matter what catch-up speed you assume -  6 ,  9 , anything - as long as he catches them before the end. So your final answer was locked in by the group's fixed  60 seconds , not by your catch-up math. The  9 ft/sec  step is genuinely wrong; it just got bailed out.

Stick with  ground speed = 6 , relative speed =  3 , and you'll be safe on any version of this that  doesn't  have the group conveniently reaching the end.

Answer: E

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards