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)

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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.

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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

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

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

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30 Jun 2025, 08:29

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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

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30 Jun 2025, 08:37

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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

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30 Jun 2025, 08:38

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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

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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

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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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30 Jun 2025, 08:45

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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

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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.

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

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30 Jun 2025, 08:46

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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.

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30 Jun 2025, 08:48

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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.

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30 Jun 2025, 08:49

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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

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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.

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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.

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

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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

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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.