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.

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

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HansikaSachdeva

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Total distance = 200m
Walkway speed = 0.5m/s

For Carla:
First run:
Speed = 1 + 0.5 = 1.5m/s
Time = 40s
Distance = 1.5*40 = 60m

Second:
Speed = 0.5m/s
Time = x s
Distance = 0.5*x m

Third run:
Speed = 1.5 + 0.5 = 2m/s
Time = 5s
Distance = 2*5 = 10m

All will sum up to 200 m
=> 60 + 0.5x + 10 = 200
=> 0.5x = 130
x = 260

For Daniel:
Speed = 0.5 + 0.5 = 1m/s
Distance = 200m

Time = 200/1 = 200s
y = 200

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Total distance 200m
Walkway moves at 0.5 m/s
Carla first starts with 1 m/s so total relative speed is 1.5 m/s for 40 sec so distance covered is 60m
Then stops for x seconds and walkway moves at 0.5 m/s
And then walks at 1.5 m/s relative to walkway so total speed would be 2 m/s for 5 secs
So travel for 10 m
Total it travels by walking for 70 m
And remaining distance ,200-70 =130 m at 0.5 m/s
So x = 260
Daniel walks at constant speed of 0.5 m/s
Total relative speed 1 m/s
Distance 200 m
Time 200 s

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From the given information, we can deduce that,

Carla's combined speed was 1.5 m/s and 2 m/s at different points of time (after adding the speed of walkway)

In 40 seconds, she travelled 40*1.5 = 60m

In the last 5 seconds, she travelled 5*2 = 10.

In the intermediate period, she covered 130 m at 0.5 m/s and therefore took 260 seconds

Similarly, Daniel took 200/1 = 200 seconds

Therefore x = 260 and y = 200

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For Carla:
For t1 = 40s, Carla ran at 1 m/s RELATIVE to the walkway. This implies Carla's speed was 1 + 0.5 = 1.5 m/s
For t2 = x seconds, Carla moves at the speed of the walkway i.e. 0.5 m/s.
For t3 = 5s, Carla moves at 1.5 m/s RELATIVE to the walkway. Hence, Carla's speed was 1.5 + 0.5 = 2 m/s.

The walkway is 200m long. Thus, from the above three relations we have:
1.5*40 + 0.5*x + 2*5 = 200. Thus x = 260 seconds.

For Daniel:
He walks at constant speed of 0.5 m/s RELATIVE to the walkway. Hence, his speed is 0.5 + 0.5 i.e. 1 m/s.
y = 200/y i.e. y = 200.

Answer is:
X = 260
Y = 200

Bunuel

This question was provided by GMAT Club
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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.

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The moving walkway is 200 meters long and travels at a constant speed of 0.5 meters per second in the direction of motion.

Carla's Journey:

Carla runs at 1 meter per second relative to the walkway for 40 seconds.
Absolute speed = 1 + 0.5 = 1.5 meters per second.
Distance covered = 1.5 m/s × 40 s = 60 meters.

She stands still relative to the walkway for x seconds (carried only by the walkway).
Absolute speed = 0.5 meters per second.
Distance covered = 0.5 × x meters.

She resumes running at 1.5 meters per second relative to the walkway for 5 seconds.
Absolute speed = 1.5 + 0.5 = 2.0 meters per second.
Distance covered = 2.0 m/s × 5 s = 10 meters.

The total distance covered is 200 meters:
60 + 0.5x + 10 = 200
70 + 0.5x = 200
0.5x = 130
x = 260 seconds.

Daniel's Journey:

Daniel walks at a constant rate of 0.5 meters per second relative to the walkway from start to finish.
Absolute speed = 0.5 + 0.5 = 1.0 meters per second.
Distance covered = 200 meters.
Time y = distance / speed = 200 / 1.0 = 200 seconds.

Thus, x = 260 and y = 200.

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X=260, Y=200. Attached picture contains detailed solution

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Information given:
- Walkway: 200 meters long, moves at 0.5 m/s
- Carla: runs at 1 m/s relative to walkway for 40s, stops running for x seconds (just carried by walkway), runs at 1.5 m/s relative to walkway for 5s, finishing at end
- Daniel: walks at 0.5 m/s relative to walkway, non-stop to the end, total time = y seconds

Question:
- What are the values of x and y that are consistent with this setup?

Solution:
- Carla's total distance must add up to 200 meters
- Segment 1: Carla runs for 40s. Her speed relative to the ground is 1 + 0.5 = 1.5 m/s. Distance: 40 x 1.5 = 60 meters
- Segment 2: Carla stops running for x seconds. Her speed relative to the ground is 0.5 m/s. Distance: 0.5x = ? meters
- Segment 3: Carla runs for 5s. Her speed relative to ground is 1.5 + 0.5 = 2 m/s. Distance: 5 x 2 = 10 meters
- Carla's total time: 60 + 0.5x + 10 = 200
- 70 + 0.5x = 200
- 0.5x = 130, x = 260

- Daniel's speed relative to ground: 0.5 + 0.5 = 1 m/s
- Distance: 200 meters
- Daniel's total time: 200/1 = 200, y = 200

Answer: x = 260, y = 200

Bunuel

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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.

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distance is 200 and speed is 0.5

for Carla
200= 1.5*40+x*0.5+ 2*5
solve for x = 260
for Daniel
1*y= 200
y= 200
x is 260 & y is 200

Bunuel

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

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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.

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Carla=
Distance.........Speed...............Time
(60 m)........... 1+0.5 mps........ 40 seconds
—(1x m) ............ 1 mps ............... x seconds
(10 m) ............ 1.5+0.5 mps ....... 5 seconds
Total Distance= 200 m
60+x+10=200
x=200-60-10=130

Daniel time (y)= Distance/Speed= 200/(0.5+0.5)=200/1=200

x=130 and y=200

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Total distance = 200
Walkway speed = 0.5m/s

Carla Details
First Leg: Relative speed is 1m/s. Actual speed = 1.5m/s
D=s*t= 1.5*40=60m

Second Leg: Actual speed = 0.5m/s
D=s*t= 0.5*x=0.5x

Third Leg: Relative speed is 1.5m/s. Actual speed = 2m/s
D=s*t= 2*5=10m

60+0.5x+10=200
x=130/0.5=260

Daniel's details
Relative speed is 0.5m/s. Actual speed = 1m/s
D=s*t= 1*y=y
y=200

Bunuel

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

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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.

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30 Jun 2025, 09:31

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For Carla:
absolute speed for 40 seconds = 1 + 0.5 (walkway speed) = 1.5 m/s
distance travelled in 40 seconds = 1.5 * 40 = 60 m

similarly distance travelled in 5 seconds is 10 m

Total distance travelled by carla = 70m

distance travelled by walkway is x seconds = 130 m
so 130/0.5 = 260 seconds

for Daniel absolute speed is 0.5 + 0.5 = 1m/s
so time needed for 200m is 200 seconds

so x is 260 and y is 200

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30 Jun 2025, 09:31

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Answer

x = 260
y = 200

Explanation

First, let's find y (Daniel's time), because it's easier.

Daniel's walking speed (0.5 m/s) and the walkway's speed (0.5 m/s) combine.
His total speed is 1 m/s.
To travel 200 meters at 1 m/s, it takes 200 seconds.
So, y = 200.

Next, let's find x (Carla's standing time) by calculating the distance she covers in each part of her journey.

First Run (40 seconds):
- Her speed: 1 m/s (running) + 0.5 m/s (walkway) = 1.5 m/s.
- Distance covered: 1.5 m/s × 40 s = 60 meters.
Second Run (5 seconds):
- Her speed: 1.5 m/s (running) + 0.5 m/s (walkway) = 2.0 m/s.
- Distance covered: 2.0 m/s × 5 s = 10 meters.
Total Distance Covered Running: 60 m + 10 m = 70 meters.
- This leaves 200 m - 70 m = 130 meters to cover while standing still.
Standing Still (for x seconds):
- During this time, she only moves at the walkway's speed of 0.5 m/s.
- Time = Distance / Speed
- x = 130 meters / 0.5 m/s = 260 seconds.
- So, x = 260.

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30 Jun 2025, 09:32

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constant rate of 1 meter per second relative to the walkway= 1.5m/s
1.5 meters per second relative to the walkway for exactly 5 seconds = 2m/s
1.5*40+0.5x+2*5=200 x=260

0.5 meter per second relative to the walkway=1ms
1m/s Y=200
Y=200

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Here x is 260 and y is 200.
For Daniel.
Daniels speed and walkway speed are given.So total speed is 1m/s.
Distance of speedway is given 200 m.
So time is 200s.

For Carla
In the first part,the total speed is 1+0.5=1.5m/s.Time is 60 sec.So the distance is 60m.
In the third part,the total speed is 1.5+0.5=2m/s.Time is 5 sec. So distance travelled is 10m.
So the distance that needs to be travelled in the second part is 200-(60+10)=130m.
The speed in second part is speed of walkway that is 0.5m/s and the distance is 130.So the time taken is 260 s.

Bunuel

This question was provided by GMAT Club
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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.

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30 Jun 2025, 09:36

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Correct answer: x = 260, y = 200

- Carla & Daniel travel at same distance: S = 200 (m)
- Carla x Daniel move in the same direction with walkway, which is 0.5m/s => Their actual moving rate have to add 0.5 (m/s) for each person
=> Actual rate of Carla = Rate Carla + 0.5 (m/s)
Actual rate of Daniel = Rate Daniel + 0.5 (m/s)

	Actual Rate (m/s)	Time (s)	Distance Traveled	Find x/y
Carla	1 + 0.5 = 1.5 0 + 0.5 = 0.5 1.5 + 0.5 = 2	40 x 5	1.5 x 40=60 0.5 x X = [1][/2]X 2 x 5 = 10 Total S = 200	==> S = 200 = 60 + (1/2)x + 10 => (1/2)x = 130 => x= 260
Daniel	0.5 + 0.5 = 1	y	Total S = 1 x y = 200	==> S = y = 200

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30 Jun 2025, 09:39

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1.5 x 40 + 0.5x + 2 x 5 = (0.5 + 0.5) x y
y = 0.5x + 70

So x must be 260, and y must be 200

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Bunuel

This question was provided by GMAT Club
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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.

Distance = Speed * Time

Carla :

Speed when Carla was running @ 1m/s = 1 + 0.5 = 1.5 m/s
Speed when Carla was running @1.5m/s = 1.5 + 0.5 = 2 m/s

1.5*40 + 0.5 * x + 2 * 5 = 200

60 + 10 + 0.5x = 200

x = 130/0.5 = 260

Daniel :

Constant Speed = 0.5 + 0.5 = 1m/s

1*y = 200

y = 200

Therefore:

x = 260
y = 200

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30 Jun 2025, 09:43

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If relative speed is 1 m/s then total speed is 1.5 m/s and that walk was for 40 second so distance covered is 60 meters.
In the last phase carla travelled at relative speed of 1.5 m/s so total speed is 2 m/s for 5 seconds so distance covered is 5*2 which is 10

so total distance covered in phase 1 & 3 is 70 m so pending distance is 130 m which is travelled at a speed of 0.5 m/s so the time to cover the distance is 260 seconds

whereas for daniel the total speed is 1 m/s which is 200/ 1 and total seconds are 200

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Carla ran in three phases:

1. Her overall speed = Speed of the moving walkway + Her own Speed => 0.5 m/s + 1 m/s = 1.5 m/s
Distance travelled by Carla = Speed x Time = 1.5 x 40 = 60 meters

2. She stood still for x seconds - we will come back to this

3. Then she ran at the speed of 1.5 m/s relative to walkway speed (which is 0.5 m/s), this means her overall speed for the distance covered here was 2 m/s. Now, Distance covered in the third phase = Speed x time => 2 m/s x 5 seconds = 10 meters

So, in total, she ran for 70 meters. Rest (200-70) 130 meters, she covered in x seconds, with the walkway speed of 0.5 m/s.
Time = Distance/ Speed => x = 130/0.5 = 260 seconds

For Daniel:
Overall speed = 0.5 + 0.5 = 1 m/s
Time = Distance/Speed => y = 200/1 = 200 seconds

Bunuel

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

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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.

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30 Jun 2025, 09:48

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Answer :- x = 260 ; y = 200

explanation :-

For Carla ,
60 + x *(.5) + (2*5) = 200
x/2 = 130
x= 260 sec

For Daniel,
y= 200/ ( .5 + .5 ) = 200 sec

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From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards