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GMAT Club Olympics 2025 (Day 1): In a certain mega-shopping mall, Carl 01 Jul 2025, 06:08
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Note to add the 0.5 meters per second to each speed because that is the new baseline with the moving walkway.

Calculate the total time it takes to complete the 200 meter walkway by just standing and subtracting each of the distance covered by each segment for Carla and Daniel.
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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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Total Distance Covered = 200 meter =-> Time * (Speed Relative to walkway + Walkway Speed)

1. 40 * (1 + 0.5) + x * (0.0 + 0.5) + 5 * (1.5 + 0.5) = 60 + 10 + 0.5x = 200;

0.5 x = 130; x = 260;

2. y (0.5 + 0.5) = 200; y = 200;
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Carla, (1+0.5)*40+0.5x+(1.5+0.5)*5 = 200 which implies x = 260

Daniel, (0.5+0.5)y = 200 which implies y = 200
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GMAT Club Olympics 2025 (Day 1): In a certain mega-shopping mall, Carl 01 Jul 2025, 06:52
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We are discussing relative speed concept where the platform is moving as well.
In this case, if the object on the platform moves in the same direction as the platform, the speeds are added up, while moving in the opposite dorection of the platform, speeds are subtracted (Imagine rowing a boat agains the water current and along with water currents, while going against, it is tougher to row as the forces are opposing, while going along, it's easier as the current helps rowing).

Coming to question:
We use distance = speed * time,
Carla: Total distance covered = 200 meters = 1.5*40 + x*0.5 + 2*5, (where the first term represents the distance covered when Carla was running 1m/s + 0.5m/s(speed of platform) for 40 seconds, the second term is where she is still so total speed is just speed of platform which is 0.5m/s for x seconds, next third term is where she is running at 1.5 m/s + 0.5m/s for 5 seconds), gives us x as 260 seconds

Daniel: Total distance covered = 200 meters = 1*y, (Daniel walks at 0.5 m/s + 0.5 m/s(speed of the platform), for y seconds to cover a total distance of 200 meters), gives us y as 200 seconds

So, 260 and 200.
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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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Speed of Carla for the first 40 sec = 1.5m/s
Speed of Carla for the next x sec = 0.5,/s
Speed of Carla in last 5 sec = 2m/s
Total distance travelled by Carla = 40 X 1.5 + 0.5x + 2 X 5 = 200
x = 260sec

Speed of Daniel for y secs = 1m/s
Total distance travelled by Daniel = 1 X y = 200
y = 200sec

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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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GMAT Club Olympics 2025 (Day 1): In a certain mega-shopping mall, Carl 01 Jul 2025, 07:21
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Total distance = 200m
Walkway speed = 0.5m/s

Carla :

Runs at 1 m/s relative to the walkway for 40 seconds
Then stands still (only carried by the walkway) for x seconds
Then runs again at 1.5 m/s relative to the walkway for 5 seconds, reaching the end


D1 = 1.5* 40 = 60m
D2 = 0.5 * x

Remaining distance => 200 - (50 + 0.5x) = 2* 5
=> x = 260m

Daniel:
Walks at 0.5 m/s relative to the walkway without stopping until the end
His total time is y seconds

Time taken by daniel = 200/1 = 200

y = 200

x= 260
y = 200
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GMAT Club Olympics 2025 (Day 1): In a certain mega-shopping mall, Carl 01 Jul 2025, 07:27
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C & D travel times

D
Let's start with D since it is easier.
Speed of D + speed of walkway = total speed
.5m/s + .5m/s = 1m/s
200 meters will take 200 seconds

C
1st phase. speed of C 1m/s, + speed of Walkway .5m/s= 1.5m/s
1.5m/s * 40 sec= 60 meters traveled --------------------------------------{1}

2nd phase halt (will come back to this later)

3rd phase speed of C 1.5m/s, + speed of Walkway .5m/s= 2m/s
2m/s * 5 sec= 10 meters traveled --------------------------------------{2}

total distance covered 60+10=70
remaining distance 200-70= 130

2nd phase halt
speed of C 0m/s, + speed of Walkway .5m/s= .5m/s
time taken to cover 130 meters = 130 * 2 =260



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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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Carla : 1.5 m/s * 40 seconds = 60m distance covered
in x seconds when she is resting = 0.5 m/s * x seconds = 0.5x m covered
for the last 5 seconds = 2m/s * 5 seconds = 10m covered

60 + 0.5x + 10 = 200
x = 260 seconds.

Daniel : 1m/s * y seconds = 200m
y = 200 seconds.
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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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D speed is 1 (0.5 +0.5 as in same direction) = 200 sec
C known speed and time leads to 70m remaining 130m is covered at 0.5m/s(stand still) =260 sec
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GMAT Club Olympics 2025 (Day 1): In a certain mega-shopping mall, Carl 01 Jul 2025, 07:46
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Total Distance = 200m
Walkway adds to your speed is the key idea behinds this.

So for Carla, (1+0.5)x40 + (0+0.5)xX + (1.5+0.5)x5 = 200, i.e
200 - 10 - 60 = 0.5xX,
130=0.5x,
x = 260.

For Daniel, (0.5+0.5)y = 200,
i.e y = 200.
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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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Bunuel
 


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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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Well first solve x. 1 meter for 40 seconds = 40 meters + the 20 meters that the walkway moved, so 60 meters. Standstill period is x, then 1.5 meters plus .5 of the walkway is 2 meters for 5 seconds so 10 meters. That is 70 meters plus .5x = 200 or .5x=130 or 260

For Y it is just 1 meter times Y = 200 or 200
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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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Understanding the Scenario
We have a moving walkway that's 200 meters long, moving at a constant speed of 0.5 meters per second (m/s).
Carla's Journey:
  1. [size=100]First Running Phase:
    [/size]
    • Runs at 1 m/s relative to the walkway for 40 seconds.
    • Effective speed: Her speed + walkway's speed = 1 + 0.5 = 1.5 m/s.
    • Distance covered: 1.5 m/s * 40 s = 60 meters.
  2. [size=100]Standing Still Phase:
    [/size]
    • Stands still for x seconds (only moved by the walkway).
    • Effective speed: 0 + 0.5 = 0.5 m/s.
    • Distance covered: 0.5 * x.
  3. [size=100]Second Running Phase:
    [/size]
    • Runs at 1.5 m/s relative to the walkway for 5 seconds.
    • Effective speed: 1.5 + 0.5 = 2 m/s.
    • Distance covered: 2 * 5 = 10 meters.
Total distance covered by Carla: 60 (first run) + 0.5x (standing) + 10 (second run) = 200 meters.
So, 60 + 0.5x + 10 = 200 → 70 + 0.5x = 200 → 0.5x = 130 → x = 260 seconds.
Daniel's Journey:
  • Walks at 0.5 m/s relative to the walkway constantly.
  • Effective speed: 0.5 (his walk) + 0.5 (walkway) = 1 m/s.
  • Total distance: 200 meters.
  • Time taken (y): Distance / Speed = 200 / 1 = 200 seconds.
Verifying the Calculations
Let me double-check Carla's total distance:
  1. First run: 1.5 m/s * 40 s = 60 m.
  2. Standing: 0.5 m/s * x = 0.5 * 260 = 130 m.
  3. Second run: 2 m/s * 5 s = 10 m.
    Total: 60 + 130 + 10 = 200 m. ✔️
Daniel:
  • Speed: 0.5 (his) + 0.5 (walkway) = 1 m/s.
  • Time: 200 m / 1 m/s = 200 s. ✔️
Final Answer
  • x (Carla's standing time): 260 seconds
  • y (Daniel's total time): 200 seconds
Selecting from Given Options
Assuming the options provided are numerical values, you would select:
  • x = 260
  • y = 200
These are the values consistent with the information given.
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Answer: x = 260, y = 200.

Let's do y first.

Notice that a speed that's relative to the walkway means that we're excluding the speed of the walkway itself.
If Daniel "walks at a constant rate of 0.5 meter per second relative to the walkway", that means that the speed of Daniel without the walkway is 0.5 m/s.
However, we have to adopt the standpoint of a static person. That is why, for someone who's not moving, the total speed of Daniel would be:
\([m]s_{Daniel}\) = \(s_{speed relative to the walkway}\) + \(s_{walkway}\)[/m]
That means \([m]s_{Daniel}\) = 0.5 + 0.5 [/m]
Therefore, \(s_{Daniel}\) = 1 m/s.

Because everything is constant, we can adopt v = s / t. Or t = s / t.
In Daniel's case, we have that t = 200 m / 1 m/s = 200 s.
y = 200.

Now let's look at x.
Applying the same logic, we have:
  • First section: s = 40 s * 1.5 m / s = 60 m
  • Second section: s = 5 s * 2 m / s = 10 m
  • Third section: we have 200 - 60 - 10 = 130 m to cover, and a speed of walkway of 0.5 m / s. That means that t = 130 m / 0.5 m/s = 260 s.
x = 260.
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GMAT Club Olympics 2025 (Day 1): In a certain mega-shopping mall, Carl 01 Jul 2025, 12:31
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Total Distance of Walkway = 200 meters

Daniel's Speed:

0.5 m/s (Walkway) + 0.5 m/s (Self speed) = 1 m/s

Time Taken by Daniel (y) = 200 = 1m/s * y

Hence, y = 200


Carla's Journey is in 3 parts:

  • 1.5 m/s for 40 seconds
  • 0.5 m/s for x seconds
  • 2 m/s for 5 seconds=

Total journey by Carla (1.5 * 40) + (0.5 * x) + (2 * 5) = 200

Solving, x = 260

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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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Walkway speed: 0.5 m/s

Everything is relative to the walkway unless stated otherwise.

⸻

Carla’s movement breakdown:
1. First phase:
• She runs at 1 m/s relative to walkway, so total speed = 1 + 0.5 = 1.5 m/s
• Time = 40 seconds
• Distance covered = 1.5 \times 40 = 60 meters
2. Second phase:
• She stops running and stands still.
• So she is only moved by the walkway = 0.5 m/s
• Time = x seconds
• Distance covered = 0.5 \times x meters
3. Third phase:
• She resumes running at 1.5 m/s relative to walkway, so total speed = 1.5 + 0.5 = 2 m/s
• Time = 5 seconds
• Distance covered = 2 \times 5 = 10 meters

⸻

Total walkway length: 200 meters

So total distance Carla covers is 200 m.

From her 3 phases:

\text{Total distance} = 60 + 0.5x + 10 = 200
70 + 0.5x = 200
\Rightarrow 0.5x = 130
\Rightarrow x = 260

✅ So, x = 260 seconds

⸻

Daniel’s speed:
• Walks at 0.5 m/s relative to walkway
• Walkway is moving at 0.5 m/s
• So total speed = 0.5 + 0.5 = 1 m/s
• Walkway length = 200 m
• Time = 200 / 1 = 200 seconds

✅ So, y = 200 seconds


And E , C
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GMAT Club Olympics 2025 (Day 1): In a certain mega-shopping mall, Carl 01 Jul 2025, 13:17
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Elite097
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Walkway speed: 0.5 m/s

Everything is relative to the walkway unless stated otherwise.

⸻

Carla’s movement breakdown:
1. First phase:
• She runs at 1 m/s relative to walkway, so total speed = 1 + 0.5 = 1.5 m/s
• Time = 40 seconds
• Distance covered = 1.5 \times 40 = 60 meters
2. Second phase:
• She stops running and stands still.
• So she is only moved by the walkway = 0.5 m/s
• Time = x seconds
• Distance covered = 0.5 \times x meters
3. Third phase:
• She resumes running at 1.5 m/s relative to walkway, so total speed = 1.5 + 0.5 = 2 m/s
• Time = 5 seconds
• Distance covered = 2 \times 5 = 10 meters

⸻

Total walkway length: 200 meters

So total distance Carla covers is 200 m.

From her 3 phases:

\text{Total distance} = 60 + 0.5x + 10 = 200
70 + 0.5x = 200
\Rightarrow 0.5x = 130
\Rightarrow x = 260

✅ So, x = 260 seconds

⸻

Daniel’s speed:
• Walks at 0.5 m/s relative to walkway
• Walkway is moving at 0.5 m/s
• So total speed = 0.5 + 0.5 = 1 m/s
• Walkway length = 200 m
• Time = 200 / 1 = 200 seconds

✅ So, y = 200 seconds


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