In a certain mega-shopping mall, Carla and Daniel start from the beginning of a 200-meter-long moving walkway and move in the same direction as the walkway, which travels at a constant speed of 0.5 meters per second. Carla starts running at a constant rate of 1 meter per second relative to the walkway, continues running for exactly 40 seconds, and then stops running for x seconds. While not running, Carla stands still and is carried only by the walkway. Afterward, she resumes running at 1.5 meters per second relative to the walkway for exactly 5 seconds, at which point she reaches the end of the walkway. Daniel walks at a constant rate of 0.5 meter per second relative to the walkway, from start to finish without stopping, and reaches the end in y seconds. Select for x and y the value of of x and y, that are consistent with the information provided. Make only two selections, one in each column.

GMAT Club Official Explanation: The walkway moves at a constant speed of 0.5 meters per second. Step 1: Distance Carla runs in each phase – In the first phase, Carla runs at 1 meter per second relative to the walkway. Her total speed = 1 + 0.5 = 1.5 meters per second Time = 40 seconds Distance = 1.5 * 40 = 60 meters – In the last phase, Carla runs at 1.5 meters per second relative to the walkway. Her total speed = 1.5 + 0.5 = 2.0 meters per second Time = 5 seconds Distance = 2.0 * 5 = 10 meters Step 2: Distance and time while Carla is stationary Total distance = 200 meters Distance already covered while running = 60 + 10 = 70 meters Remaining distance = 200 - 70 = 130 meters While standing still, she moves only with the walkway: Speed = 0.5 meters per second Time = 130 / 0.5 = 260 seconds Step 3: Daniel's total time Daniel walks at 0.5 meters per second relative to the walkway. His total speed = 0.5 + 0.5 = 1.0 meter per second Distance = 200 meters Time = 200 / 1.0 = 200 seconds Final Answers: x = 260 seconds y = 200 seconds

Length of walkway = 200 meters Walkway speed = 0.5 m/s (both Carla and Daniel are moving with the walkway) Carla's running speed relative to walkway: First run: 1 m/s (→ absolute speed = 1 + 0.5 = 1.5 m/s) for 40 seconds Stops for x seconds (→ speed = walkway speed = 0.5 m/s) Final run: 1.5 m/s relative to walkway → 1.5 + 0.5 = 2.0 m/s, for 5 seconds Daniel walks at 0.5 m/s relative to walkway → total speed = 0.5 + 0.5 = 1.0 m/s We must find values for: x = time Carla stands still y = total time Daniel takes to finish 🟨 Step 1: Carla’s Movement 🟡 First Phase (Running for 40 sec at 1.5 m/s): Distance = speed × time = 1.5 × 40 = 60 meters 🟡 Second Phase (Standing still for x seconds, moving at 0.5 m/s): Distance = 0.5 × x = 0.5x meters 🟡 Third Phase (Running for 5 seconds at 2.0 m/s): Distance = 2.0 × 5 = 10 meters 🟧 Total distance Carla covers: 60+0.5x+10=200 70+0.5x=200 ⇒0.5x=130 ⇒x=260 🟨 Step 2: Daniel’s Movement Speed = 1.0 m/s (0.5 walking + 0.5 walkway) Total distance = 200 meters Time = distance / speed = 200 / 1 = 200 seconds y=200 ✅ Final Answers: x=260 ; y=200

In a certain mega-shopping mall, Carla and Daniel start from the beginning of a 200-meter-long moving walkway and move in the same direction as the walkway, which travels at a constant speed of 0.5 meters per second. Carla starts running at a constant rate of 1 meter per second relative to the walkway, continues running for exactly 40 seconds, and then stops running for x seconds. While not running, Carla stands still and is carried only by the walkway. Afterward, she resumes running at 1.5 meters per second relative to the walkway for exactly 5 seconds, at which point she reaches the end of the walkway. Daniel walks at a constant rate of 0.5 meter per second relative to the walkway, from start to finish without stopping, and reaches the end in y seconds. 40(1+.5) + .5x + 5(1.5+.5) = 200 60 + .5x + 10 = 200 x = 260 y(.5+.5) = 200 y = 200 Select for x and y the value of of x and y, that are consistent with the information provided. Make only two selections, one in each column. x y 260 200

GMAT Club Olympics 2025 (Day 1): In a certain mega-shopping mall, Carl : Two-part Analysis (TPA)

Last visit was: 15 May 2026, 16:49

It is currently 15 May 2026, 16:49

DataGuyX

Joined: 23 Apr 2023

Last visit: 02 Apr 2026

Posts: 111

Own Kudos:

[2]

Given Kudos: 161

Location: Brazil

Concentration: Entrepreneurship, Technology

Schools: Booth '26 (A$) Kellogg '26 (A) Sloan '26 (D) Stanford '26 (D) Haas '26 (D) HBS '26 (D)

GMAT Focus 1: 675 Q79 V85 DI86 Verified score

GPA: 7.3

Schools: Booth '26 (A$) Kellogg '26 (A) Sloan '26 (D) Stanford '26 (D) Haas '26 (D) HBS '26 (D)

GMAT Focus 1: 675 Q79 V85 DI86 Verified score

Posts: 111

Kudos: 77

[2]

30 Jun 2025, 08:27

Kudos

Bookmarks

Distance to be traveled: 200 meters
Base velocity (moving walkaway): 0.5 m/s

Daniel: Constantly 0.5m/s relative to the walkway. So, total Daniel's velocity will be 0.5 m/s + 0.5 m/s = 1 m/s (the sum of the moving walkaway + Daniel's relative velocities). Whe can find y by:
(200 meters) / (1 m /s) = 200meters * 1 seconds / 1 meter = 200seconds.

Carla: Here we can make the following approach:
(1) In the beggining, she runs at 1m/s relative to the walkaway per 40 seconds. So, the total distance traveled will be:

( 1m/s + 0.5m/s ) * 40 seconds = 1.5m/s * 40 seconds = 60 meters.

(2) In the final, she runs at 1.5m/s relative to the walkaway per 5 seconds. So, the total distance traveled will be:
( 1.5m/s + 0.5m/s ) * 5 seconds = 2m/s * 5 seconds = 10 meters.

(3) Now, we know that the distance she traveled in the time when she was not running will be 200 - 60 - 10 = 130 meters. Since her speed was only the walkaway speed, whe can find x by:
(130 meters) / (0.5 m /s) = 130 meters * 2 seconds / 1 meter = 260 seconds.

MaxFabianKirchner

Joined: 02 Jun 2025

Last visit: 12 Jul 2025

Posts: 65

Own Kudos:

[1]

Given Kudos: 122

Status:26' Applicant

Location: Denmark

Concentration: Finance, International Business

Schools: ESSEC MiF '26 ESADE HEC Imperial LBS Oxford MFE '26 Sloan (MIT) NYU Stern NUS Bocconi IE'26

GPA: 4.0

WE:Other (Student)

Schools: ESSEC MiF '26 ESADE HEC Imperial LBS Oxford MFE '26 Sloan (MIT) NYU Stern NUS Bocconi IE'26

Posts: 65

Kudos: 63

[1]

30 Jun 2025, 08:27

Kudos

Bookmarks

To find y (Daniel's time):

Daniel's total speed is his walking speed (0.5 m/s) plus the walkway's speed (0.5 m/s), which equals 1.0 m/s.

Time = Distance / Speed

y = 200 meters / 1.0 m/s = 200 seconds.

To find x (Carla's rest time):

We need to sum the distances from her three travel phases to equal 200 meters.

First run: (1.0 m/s + 0.5 m/s) * 40 s = 60 meters.

Resting: (0 m/s + 0.5 m/s) * x s = 0.5x meters.

Second run: (1.5 m/s + 0.5 m/s) * 5 s = 10 meters.

Total Distance Equation:
60 + 0.5x + 10 = 200
70 + 0.5x = 200
0.5x = 130
x = 260 seconds.

Thus, x = 260 and y = 200

sanyuktabhattad

Joined: 11 Aug 2024

Last visit: 20 Feb 2026

Posts: 17

Own Kudos:

[2]

Given Kudos: 30

Location: India

Concentration: Finance, Strategy

Posts: 17

Kudos: 6

[2]

30 Jun 2025, 08:27

Kudos

Bookmarks

40*1.5 + 0.5*+ 2*5 = 200
which gives x =260 seconds

(0.5+0.5)*y = 200
y=200 seconds

Confsdconsultant

Joined: 09 Apr 2021

Last visit: 16 Jul 2025

Posts: 6

Own Kudos:

[1]

Given Kudos: 7

Posts: 6

Kudos: 4

[1]

30 Jun 2025, 08:29

Kudos

Bookmarks

Step 1: Carla's Motion
First Phase (Running for 40s):
Relative to walkway: 1 m/s.
Actual speed: Walkway (0.5 m/s) + Carla (1 m/s) = 1.5 m/s
Time: 40 s
Distance covered =1.5×40=60 m

Second Phase (Standing Still for x seconds):
Only moves with walkway: 0.5 m/s.
Time: x seconds
Distance covered=0.5×x meters

Third Phase (Runs at 1.5 m/s for 5s):
Relative to walkway: 1.5 m/s.
Actual speed: Walkway (0.5 m/s) + Carla (1.5 m/s) = 2.0 m/s
Time: 5 s
Distance covered=2.0×5=10 meters

Total Distance Covered by Carla:60+0.5x+10=70+0.5x
This must equal 200 meters (length of walkway):

70+0.5x=200
x=260 seconds

Step 2: Daniel's Motion
Speed relative to walkway: 0.5 m/s
Actual speed: 0.5 + 0.5 = 1.0 m/s
Distance: 200 m.
Time taken(y)= 200/1=200 seconds

x=260, y=200

anuraagnandi9

Joined: 25 Mar 2025

Last visit: 21 Jan 2026

Posts: 11

Own Kudos:

[1]

Given Kudos: 3

Location: India

GMAT Focus 1: 705 Q87 V84 DI84 Verified score

GPA: 3.34

GMAT Focus 1: 705 Q87 V84 DI84 Verified score

Posts: 11

Kudos: 12

[1]

30 Jun 2025, 08:37

Kudos

Bookmarks

Total Distance to be covered by both = 200m

Starting with Daniel's calculation as its easier,
y = 200 / (0.5+0.5) = 200 seconds

Carla's -
1) travelled - 40* (1+0.5) = 60m
2) stood still, but travelled - x * 0.5 m
3) travelled - 5 * (1.5+0.5) = 10

sum of the above must be 200,
60+ x/2 + 10 = 200
=> x = 260

Harika2024

Joined: 27 Jul 2024

Last visit: 13 May 2026

Posts: 99

Own Kudos:

[1]

Given Kudos: 31

Location: India

Posts: 99

Kudos: 84

[1]

30 Jun 2025, 08:38

Kudos

Bookmarks

length =200 meters
speed=0.5 m/s

Carla's Journey:
Carla's - Running :
Carla's speed relative to walkway=1 m/s
Carla's speed relative to ground=1+0.5=1.5 m/s
t1=40 sec
Distance covered, D1=1.5*40=60 meters

Carla's - Standing Still :
Carla's speed relative to walkway=0 m/s (standing still)
Carla's speed relative to ground=0+0.5=0.5 m/s
t2=x sec
Distance covered, D2=0.5*x=0.5x meters

Carla's - Running (second part)
Carla's speed relative to walkway=1.5 m/s
Carla's speed relative to ground=1.5+0.5=2.0 m/s
t3=5 sec
Distance covered, D3=2.0*5=10 meters

Total Distance for Carla:
The sum of the distances covered in total equal the total length of the walkway, D1 +D2 +D3 = L
60+0.5x+10=200
70+0.5x=200
0.5x=200−70
0.5x=130
x=260 seconds

Daniel's Journey
Daniel's speed relative to walkway=0.5 m/s
Daniel's speed relative to ground=0.5+0.5=1.0 m/s
Distance covered, L=200 meters
Time taken = y seconds

Using the formula, Distance = Speed * Time:
200=1.0*y
y=200 sec

Therefore, Value of x=260 sec and y=200 sec

Duongduong273

Joined: 08 Apr 2023

Last visit: 03 Feb 2026

Posts: 32

Own Kudos:

[2]

Given Kudos: 4

Posts: 32

Kudos: 32

[2]

30 Jun 2025, 08:40

Kudos

Bookmarks

Carla: Runs for 40 s at 1 m/s relative to walkway -> Her total speed = = 0.5 + 1 = 1.5m/s, distance covered: 40×1.5=60m. Stops running for x seconds -> Distance = 0.5x. Runs 5 s at 1.5 m/s relative to walkway -> Her total speed = 1.5 + 0.5 = 2m/s, distance: 5 x 2 = 10m. Total distance: s=200= 60 + 0.5x + 10 => x=260. Daniel: Walks at a constant rate of 0.5 meter per second relative to the walkway -> His speed: v=0.5+0.5 =1 m/s => Time: y=s/v=200/1=200s

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

In a certain mega-shopping mall, Carla and Daniel start from the beginning of a 200-meter-long moving walkway and move in the same direction as the walkway, which travels at a constant speed of 0.5 meters per second.

Carla starts running at a constant rate of 1 meter per second relative to the walkway, continues running for exactly 40 seconds, and then stops running for x seconds. While not running, Carla stands still and is carried only by the walkway.
Afterward, she resumes running at 1.5 meters per second relative to the walkway for exactly 5 seconds, at which point she reaches the end of the walkway.
Daniel walks at a constant rate of 0.5 meter per second relative to the walkway, from start to finish without stopping, and reaches the end in y seconds.

Select for x and y the value of of x and y, that are consistent with the information provided. Make only two selections, one in each column.

iamchinu97

Joined: 14 Dec 2020

Last visit: 09 May 2026

Posts: 136

Own Kudos:

146

[1]

Given Kudos: 35

Products:

Posts: 136

Kudos: 146

[1]

30 Jun 2025, 08:45

Kudos

Bookmarks

Daniel = (0.5 + 0.5)y = 200 so y => 200
Carla =(1+0.5) * 40 + 0.5*x + 2*5 = 200
60+ 0.5x+10 = 200
0.5x=130
x=> 260

Kinshook

Major Poster

Joined: 03 Jun 2019

Last visit: 15 May 2026

Posts: 6,003

Own Kudos:

5,877

[1]

Given Kudos: 163

Location: India

Schools: HBS '26 CBS '26 Wharton '26

GMAT 1: 690 Q50 V34 Verified score

WE:Engineering (Transportation)

Products:

Schools: HBS '26 CBS '26 Wharton '26

GMAT 1: 690 Q50 V34 Verified score

Posts: 6,003

Kudos: 5,877

[1]

30 Jun 2025, 08:45

Kudos

Bookmarks

Carla starts running at a constant rate of 1 meter per second relative to the walkway, continues running for exactly 40 seconds, and then stops running for x seconds. While not running, Carla stands still and is carried only by the walkway.
Afterward, she resumes running at 1.5 meters per second relative to the walkway for exactly 5 seconds, at which point she reaches the end of the walkway.
Daniel walks at a constant rate of 0.5 meter per second relative to the walkway, from start to finish without stopping, and reaches the end in y seconds.

40(1+.5) + .5x + 5(1.5+.5) = 200
60 + .5x + 10 = 200
x = 260

y(.5+.5) = 200
y = 200

Select for x and y the value of of x and y, that are consistent with the information provided. Make only two selections, one in each column.

x	y
260	200

Sh3333

Joined: 27 Jun 2025

Last visit: 14 May 2026

Posts: 2

Given Kudos: 1

Posts: 2

Kudos: 0

30 Jun 2025, 08:46

Kudos

Bookmarks

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

Carla starts running at a constant rate of 1 meter per second relative to the walkway, continues running for exactly 40 seconds, and then stops running for x seconds. While not running, Carla stands still and is carried only by the walkway.
Afterward, she resumes running at 1.5 meters per second relative to the walkway for exactly 5 seconds, at which point she reaches the end of the walkway.
Daniel walks at a constant rate of 0.5 meter per second relative to the walkway, from start to finish without stopping, and reaches the end in y seconds.

Select for x and y the value of of x and y, that are consistent with the information provided. Make only two selections, one in each column.

romanmoonshine

Joined: 26 Apr 2025

Last visit: 04 Sep 2025

Posts: 12

Own Kudos:

[1]

Given Kudos: 1

GMAT Focus 1: 645 Q83 V84 DI84 Verified score

GPA: 3.24

Products:

GMAT Focus 1: 645 Q83 V84 DI84 Verified score

Posts: 12

Kudos: 6

[1]

30 Jun 2025, 08:48

Kudos

Bookmarks

Since the speeds of Carla and Daniel are RELATIVE to the walkway, they get added to the walkway's speed.

So, getting the total distance covered in Carla's case-

(1+0.5) m/s x 40 sec + (o.5) m/s * X sec + (1.5+0.5) m/s * 5 sec = 200 m
So x= 260 secs

Similarly speed of Daniel is (0.5+0.5) m/s and distance is 200 m. so time is 200/1 = 200 seconds.

muuss

Joined: 10 Aug 2024

Last visit: 06 May 2026

Posts: 108

Own Kudos:

[1]

Given Kudos: 38

GMAT Focus 1: 615 Q84 V81 DI76 Verified score

Posts: 108

Kudos: 83

[1]

30 Jun 2025, 08:49

Kudos

Bookmarks

Speed of walkway=0.5m/s
Total Distance=200m
Speed of C for t(40s)=1.5m/s
D=40*1.5=60 m
Speed of C for t(x sec)=0.5 m/s
D= 0.5 x m
Speed of C for t(5 s)=2m/s
D=5*2= 10 m
200=60 +0.5x+10->x=260;

Speed of D for t(y sec)=1m/s
T(y)=200/1=200 sec

crimson_king

Joined: 21 Dec 2023

Last visit: 15 May 2026

Posts: 152

Own Kudos:

156

[1]

Given Kudos: 114

GRE 1: Q170 V170 Verified score

Posts: 152

Kudos: 156

[1]

30 Jun 2025, 08:49

Kudos

Bookmarks

Lets break it down for each one of them individually,

Daniel

Relative speed to walkway=0.5m/s
Walkway speed=0.5m/s
Total speed = 1m/s (Throughout)
Time Taken = Distance/Speed
=200/1= 200 seconds (y)

Carla

First 40 seconds
Relative speed to walkway=1 m/s
Walkway speed=0.5m/s
Total speed = 1.5 m/s (For first 40s)
Distance= Speed*Time
=40*1.5=60 m

Last 5 seconds
Relative speed to walkway=1.5 m/s
Walkway speed=0.5m/s
Total speed = 2 m/s (For last 5s)
Distance= Speed*Time
=2*5=10 m

Distance left = 200-(60+10)=200-70=130 m

These 130 meters she spent idly so the time taken if we just refer to the walkway speed (0.5 m/s) would be:
Distance/Time = 130/0.5 = 260 seconds (x)

Hence the final answers are x=260 & y=200.

Heix

Joined: 21 Feb 2024

Last visit: 05 May 2026

Posts: 473

Own Kudos:

168

[1]

Given Kudos: 63

Location: India

Concentration: Finance, Entrepreneurship

Schools: Ross MM "26 NUS Columbia '27

GMAT Focus 1: 485 Q76 V74 DI77 Verified score

GPA: 3.4

WE:Accounting (Finance)

Products:

Schools: Ross MM "26 NUS Columbia '27

GMAT Focus 1: 485 Q76 V74 DI77 Verified score

Posts: 473

Kudos: 168

[1]

30 Jun 2025, 08:52

Kudos

Bookmarks

Show Spoiler

Attachment:

GMAT-Club-Forum-arztzq3s.jpeg [ 240.59 KiB | Viewed 3351 times ]

Emkicheru

Joined: 12 Sep 2023

Last visit: 12 Sep 2025

Posts: 119

Own Kudos:

[1]

Given Kudos: 11

Location: Kenya

Schools: Anderson (UCLA) Sloan (MIT) Columbia '27

GMAT 1: 780 Q50 V48 Verified score

GRE 1: Q167 V164 Verified score

GPA: 3.7

Schools: Anderson (UCLA) Sloan (MIT) Columbia '27

GMAT 1: 780 Q50 V48 Verified score

GRE 1: Q167 V164 Verified score

Posts: 119

Kudos: 22

[1]

30 Jun 2025, 08:59

Kudos

Bookmarks

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

Carla starts running at a constant rate of 1 meter per second relative to the walkway, continues running for exactly 40 seconds, and then stops running for x seconds. While not running, Carla stands still and is carried only by the walkway.
Afterward, she resumes running at 1.5 meters per second relative to the walkway for exactly 5 seconds, at which point she reaches the end of the walkway.
Daniel walks at a constant rate of 0.5 meter per second relative to the walkway, from start to finish without stopping, and reaches the end in y seconds.

Select for x and y the value of of x and y, that are consistent with the information provided. Make only two selections, one in each column.

Carla runs at 1.9m/s for 40 secs implies total speed is 1.5 m/s hence total distance is 60m
Then runs at 1.5 m/s for 5 secs implies total speed is 2m/s for 5 secs ie 10m.But total distance is 200m implies remaining distance for X secs is 200-60-10=130m .at speed of 0.5 m/s implies x=260 secs
Daniel speed through out is 0.5 m/s ie total speed of 1m/s implies y= 200 secs

APram

Joined: 23 Jun 2024

Last visit: 17 Mar 2026

Posts: 708

Own Kudos:

280

[1]

Given Kudos: 248

Location: India

Schools: ISB '26 HEC INSEAD Jan '26

GMAT Focus 1: 605 Q86 V78 DI76 Verified score

GPA: 3.608

Schools: ISB '26 HEC INSEAD Jan '26

GMAT Focus 1: 605 Q86 V78 DI76 Verified score

Posts: 708

Kudos: 280

[1]

30 Jun 2025, 09:01

Kudos

Bookmarks

Speed of Carla for 40s = 1+0.5 m/s = 1.5m/s (Carla walks in same direction as walkway)
Speed of Carla for next x sec = 0.5m/s
Speed of Carla for last 5 sec = 1.5+0.5 = 2m/s

Total distance by Carla = 200m (As we are told that at last she reaches the end of walkway)
=>40*1.5 + 0.5*x + 2*5 = 200
=>60 + 0.5x + 10 = 200
=>x = 260 sec

Speed of Daniel = 0.5+0.5 = 1m/s (Daniel walk in same direction as walkway)
=> y*1 = 200 (Daniel reaches the end of walkway in y sec)
=> y=200 sec

So answer is x=260 and y=200

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

Carla starts running at a constant rate of 1 meter per second relative to the walkway, continues running for exactly 40 seconds, and then stops running for x seconds. While not running, Carla stands still and is carried only by the walkway.
Afterward, she resumes running at 1.5 meters per second relative to the walkway for exactly 5 seconds, at which point she reaches the end of the walkway.
Daniel walks at a constant rate of 0.5 meter per second relative to the walkway, from start to finish without stopping, and reaches the end in y seconds.

Select for x and y the value of of x and y, that are consistent with the information provided. Make only two selections, one in each column.