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.

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. For Carla: First 40 seconds, speed is 1+0.5=1.5m/s Total distance covered= 40*1.5=60 meters Last 5 seconds speed is 1.5+0.5= 2m/s Total distance covered=5*2=10 meters Balance distance covered in middle is 200-60-10=130 meters Speed is 0.5m/s Let x be the time 0.5x=130 x=260 For Daniel Speed=0.5+0.5= 1 m/s Time=y 1y=200 y=200

The speed of the moving walkway is 0.5 meters per second. For Carla: For 40 sec, she runs at a constant rate of 1 meter per second relative to the walkway, this means that her total speed will be (1+0.5 m) since she and the walkway are moving in the same direction, thus increasing her overall speed. Therefore, distance travelled by her in the first 40s: 40*1.5= 60 metre. Towards the end, her speed= (1.5+0.5)= 2 meter/ sec. This means that the distance she travels in the last 5 sec is 10 meters (2*5). Thus, the distance remaining by Carla= 200- (60+10)= 130. this is the distance that carla just stood still and was carried by walkway. Therefore time taken to cover 130 meters is 260 sec (130/0.5)x= 260sec. For Daniel: Daniel walks at a constant rate of 0.5 meter per second relative to the walkway for the entire distance of 200 meters. Therefore time take by Daniel (y)= 200/ (0.5+05)= 200 sec.

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

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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.

For Carla:
First 40 seconds, speed is 1+0.5=1.5m/s
Total distance covered= 40*1.5=60 meters

Last 5 seconds speed is 1.5+0.5= 2m/s
Total distance covered=5*2=10 meters

Balance distance covered in middle is 200-60-10=130 meters
Speed is 0.5m/s
Let x be the time
0.5x=130
x=260

For Daniel
Speed=0.5+0.5= 1 m/s
Time=y
1y=200
y=200

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x= 260 seconds, y= 200 seconds

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Total distance is 200 m
speed of walkway is 0.5m/s

Carla's speed for first 40 seconds: (1+0.5) m/s => dist. covered 40*1.5=60m
The distance for next "x" second is 0.5xm
rest of the journey (1.5+.5) *5 = 10 m

Total distance he had to cover 200
total he covered 60+10+0.5x=200=>x= (200-70) *2=260seconds

For Daniel, speed constant =0.5m/s
speed of walkway=0.5m/s , Total speed = 1m/s

time = d/speed=200/1=200 seconds (y)

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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You need to add the speed of the walkway to the speed of the person.

Carla:
First part: Distance = 1.5m/sec * 40 sec =60 m
Second part: Distance = 0.5m/sec * x sec
Third part: Distance = 2m/sec * 5 sec =10 m
For the second part you can substract the distance of part 1 & 3 to get the distance of part 2 = 200m - 60m - 10m =130 m
To get x: 130m =0.5m/sec * x sec
X = 260m

For Dan:
200m =1m/sec * y
Y= 200

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30 Jun 2025, 12:22

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ANS
x= 260
y=200

Carla

distance she ran

1 part
Speed of Carla running in the moving walkway = Speed of the moving walkways+ Speed of her running = .5+1 =1.5 m/sec
Distance she runs at speed of 1.5 Mts/Sec
D= V*T = (1.5 mts/seg)*(40 seg)= 60 meters

2 part
Speed of Carla running in the moving walkway= Speed of the moving walkways+ Speed of her running = .5+1.5 =2 m/sec
Distance she runs at speed of 2 Mts/Sec
D= V*T = (2 mts/seg)*(5 seg)= 10 meters

Total distance she ran= 60+10 = 70mts

Distance she kept still is 200mts-70 Mts = 130mts
Time she spent still on the moving walkways= Remaining distance/Speed of the moving walkaways = 140*/(.5mts/sec)= 260 sec

Because she was still her speed of walking was 0 Mts/sec that's why we just divided by the Speed of the moving walkaway.

Daniel
He walks at a rate of 0.5 Mts/seg
so
Speed of Daniel walking in the moving walkway = Speed of the moving walkways+ Speed of his walking = .5+.5 =1 m/sec
Time he took to reach the end is
Time= Distance/Speed of Daniel walking in the moving walkway= 200mts/1 m/sec= 200 sec

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The speed of the moving walkway is 0.5 meters per second.

For Carla:

For 40 sec, she runs at a constant rate of 1 meter per second relative to the walkway, this means that her total speed will be (1+0.5 m) since she and the walkway are moving in the same direction, thus increasing her overall speed. Therefore, distance travelled by her in the first 40s: 40*1.5= 60 metre.
Towards the end, her speed= (1.5+0.5)= 2 meter/ sec. This means that the distance she travels in the last 5 sec is 10 meters (2*5).
Thus, the distance remaining by Carla= 200- (60+10)= 130. this is the distance that carla just stood still and was carried by walkway.
Therefore time taken to cover 130 meters is 260 sec (130/0.5)x= 260sec.

For Daniel:

Daniel walks at a constant rate of 0.5 meter per second relative to the walkway for the entire distance of 200 meters.
Therefore time take by Daniel (y)= 200/ (0.5+05)= 200 sec.

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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Carla speed with respect to the walkway - Indicates that the someone on the walkway seeing Carla whereas when someone sees Carla from the ground the speed of escalator and Carla will be added.

Distance = Speed * Time

Carla distance covered in 1st phase = 1+0.5 = 1.5*40 = 60 m
Carla distance covered in 2nd phase = 0.5*x
Carla distance covered in 3rd phase = (1.5+0.5) * 5 = 10 m
Total length of walkway = 200 m = 60+10+(0.5x)
x = 130/0.5 = 260 m

Daniel
(0.5+0.5)*y = 200
y = 200 m

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The total distance of the moving walkway is $200$ meters. Carla and Daniel both cover the same distance.
The walkway is moving at a speed of $0.5$ m/s .
If a person is moving $x$ m/s relative to the walkway, then their speed, relative to the earth is $x + 0.5$ m/s .

Apply the same logic to the information given in the question. Use $Distance = Speed*Time$ to find the total distance travelled in terms of $x$ and $y$.

Total distance covered by Carla -> $(1.5*40) + (0.5*x) + (2*5) = 60 + \frac{x}{2} + 10 = \frac{x+140}{2}$
Total distance covered by Daniel -> $1*y$

Because they covered the same distance, $\frac{x+140}{2} = 1*y$
$x = 2y-140$

This equation is satisfied when $y=200$ and $x=260$.

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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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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.

x:
Carla walks: 1x 40 Seconds with a speed of one second/Meter; 1x 5 Seconds with 1.5 seconds/Meter:
Therefore: 40*1 + 1.5*5 = 47,5; During that time (45 Seconds), the moving walkway moved 22.5 Meters. Together: 70 Meters.
As a result, the residual equals 130.
X = 130.

With Y its easier. Same logic, but we just assume, that the move as one "unit" -> 1 Meter/Second.
That results in 100 Seconds total.

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30 Jun 2025, 14:46

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Carla: (1+0.5)*40 + 0.5x + (1.5+0.5)*5 = 200
=> 0.5x = 130 => x = 260

Daniel: (0.5+0.5)*y = 200 => y=200

So, x=260 and y=200

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Carla's speed at first: (1m + 0.5m)/sec = 1.5m/sec
Carla's speed while stopping: 0.5m/sec
Carla's speed at last: (1.5m+0.5m)/sec = 2m/sec
Thus, 1.5m*40 sec + 0.5m*x + 2m*5 sec = 200m <=> x=260 seconds

Daniel's speed: (0.5m + 0.5m)/sec = 1m/sec
Thus, 1m*y = 200m <=> y= 200 seconds

Answer: x=260, y=200

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X= 260, Y=200.

Starting with Y.
200 meters. Daniel walking rate = .5m/s. Walkway Speed = .5m/s. Daniel+Walkway = 1m/s

200 meters takes Daniel 200 seconds to cover.

Carla.
[(run) 1m/s + (walkway) .5m/s ] [ 40seconds] = [1.5m/s][40 seconds] = 60meters
Stand still (x)(walkway .5m/s)
[(run) 1.5m/s + (walkway) .5m/s][5seconds] =[2m/s][5seconds] =10meters

200 meters - 70 meters = 130 meters
Stand still (130 meters) / (walkway .5m/s) = 260 seconds

Y= 200
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.

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, 16:06

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Carla
(1+0.5)*40 + 0.5x + (1.5+0.5)*5 = 200
60 + 0.5x + 10 = 200
0.5x = 130
x = 260

Daniel
(0.5+0.5)y = 200
y = 200

x = 260 and y = 200

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30 Jun 2025, 16:36

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Carla and Daniel travel the same distance, which is the the length of the moving walkway = 200m.

Distance travelled by Carla = (0.5+1)*40 + 0.5x + (0.5+1.5)*5
Therefore, (0.5+1)*40 + 0.5x + (0.5+1.5)*5 = 200, or x = 260.

Distance travelled by Daniel = (0.5+0.5)*y
Therefore, y = 200.

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X= 260 s - Carla stayed hold moving only by the walkway for 130 meters, if the walkway speed was 0,5m/s, she needed 260 seconds. Y= 200, Daniel walked at 0,5m/s relative to walkway, so his total speed was 1,0m/s. To cross 200 meters he needed 200 seconds.

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For carla
200= (1.5+0.5)5 + 0.5x + (1+0.5)40
x=160/.5 = 260

For daniel
200= (0.5+0.5)y
y=200

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They go the same distance, so we can say:
D[Carla]= D[Daniel]
Distance=Velocity x Time
since they are going the same direction as walkway, in each part, the speed is the sum of their relative speed and the walkway speed.
[(1+0.5)*40] + [(0+0.5)*x] + [(1.5+0.5)*5] = (0.5+0.5)*y
70 + x/2 = y
out of the options, x=260 and y=200 satisfy the equation.

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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30 Jun 2025, 18:47

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Carla runs for 40 s at an effective speed of 1.5 m/s, covering 60 m, then stands still for xxx seconds and is carried 0.5 m/s by the walkway, and finally runs for 5 s at an effective 2.0 m/s to cover the last 10 m; setting 60+0.5x+10=20060 + 0.5x + 10 = 20060+0.5x+10=200 gives x=260x = 260x=260 s. Daniel walks the entire 200 m at 1.0 m/s (his 0.5 m/s over ground plus the 0.5 m/s walkway), so y=200/1.0=200y = 200/1.0 = 200y=200/1.0=200 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, 19:02

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Carla:

Runs at $1 m/s$ relative to the walkway for $40$ seconds. Actual speed = $1.5 m/s$. Distance covered, $a= 40*1.5=60 m$

Then stops running, still moves at $0.5 m/s$ for x seconds. Distance covered, $b= 0.5x$

Resumes running at 1.5 m/s for 5 seconds. Actual speed = $2 m/s$. Distance covered, $c= 2*5 = 10m/s$

$a+b+c=200$

$60+10+0.5x=200$

$0.5x=130$

$x=260$

Daniel:

walks at a constant rate of $0.5m/s$, Actual speed is $1m/s$. Covers $200m$ in $y$ seconds.

$y=\frac{200}{1}=200 sec$

x: 260
y: 200