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's Journey: Phase Time (s) Total Speed (m/s) Distance covered (m) 1) Running at 1 m/s + walkway speed 0.5 m/s 40 1.5 60 2) Standing (no run) carried by walkway 0.5 m/s x 0.5 0.5x 3) Running at 1.5 m/s + walkway speed 0.5 m/s 5 2.0 10 Total distance = 60 + 0.5x + 10 = 200 → 0.5x = 130 → x = 260 seconds Daniel's Journey: He walks at 0.5 m/s relative + walkway 0.5 m/s = 1 m/s total To cover 200 m: y = 200 seconds

Carla: A = 40 * (0,5 + 1 ) = 60 B = 200 - A - B = 130 = x * 0.5 -> x = 260 C = 5 * (0,5 + 1,5 ) = 10 Daniel: y = 200 / (0.5 + 0.5 ) = 200

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

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Correct Answer: X= 260, Y= 200
Carla speed= (Speed of Carla + Walkway speed)
Daniel Speed= (Speed of Daniel + Walkway speed)
So Carla travelling distance,
Case (1) 40seconds*1.5 mps= 60 mtr
Case (2) 5seconds*2mps= 10mtr
Remaining by walkway speed= 130mtr./0.5mps= 260 seconds
Daniela travelled
200mtr./1mps= 200 seconds

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Clara net speed= 0.5+1.5
Daniel speed =0.5+0.5=1

1.5*40 + 0.5x + 2*5 =200
x=260

1y=200
y=200

Ans 260 & 200

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Since all speeds are relative to walkway, we shall add 0.5 m/s as a constant base speed for Carla and Daniel

Case 1 - Carla, Solving for 'x',

40(1+0.5) +x(0.5)+ 5(1.5+0.5) = 200, solving for x = 260 sec

Case 2 - Daniel, solving for 'y',

y(0.5+0.5) = 200, hence y = 200 sec

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Carla's Journey:

Phase	Time (s)	Total Speed (m/s)	Distance covered (m)
1) Running at 1 m/s + walkway speed 0.5 m/s	40	1.5	60
2) Standing (no run) carried by walkway 0.5 m/s	x	0.5	0.5x
3) Running at 1.5 m/s + walkway speed 0.5 m/s	5	2.0	10

Total distance = 60 + 0.5x + 10 = 200
→ 0.5x = 130 → x = 260 seconds

Daniel's Journey:

He walks at 0.5 m/s relative + walkway 0.5 m/s = 1 m/s total
To cover 200 m: y = 200 seconds

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Carla runs for 40 seconds at 1.5 m/s, so she covers 60 meters. Then she stands still and lets the walkway carry her. At the end, she runs again for 5 seconds at 2 m/s, covering 10 more meters. So she runs a total of 70 meters, which means the walkway carries her the remaining 130 meters.
Since the walkway moves at 0.5 m/s , it takes 130 ÷ 0.5 = 260 seconds. So x is 260.
Daniel walks the whole way at 0.5 meters per second relative to the walkway. So with the walkway also moving at 0.5, his total speed is 1 m/s. He covers 200 meters in 200 seconds. So y is 200.

Answer: x = 260, y = 200

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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.

The walk way is 200 meter long and it’s moving at a constant speed of 0.5 meters per second.

Speed of walkway W = 0.5 m/s

Carl running at a constant speed of 1 m/s.

For 40 seconds, relative speed = (1+0.5) = 1.5 * 40 = 60 metres.

While stopped running, the speed of Carl is zero, only walkway speed (0.5 m/s) acts on for those x seconds.

She again runs for 5 seconds at 1.5 m/s. So, the relative speed = (1.5+ 0.5) = 2 m/s * 5 = 10 metres.

Total distance = 200 metres.

Distance now covered running = 60+10 = 70 metres. Remaining distance = 200 - 70 = 130 metres.

For covering 130 metres at speed 0.5 m/s takes, 130/0.5 = 260 seconds.

X = 260 seconds.

Daniel walks at rate 0.5 m/s, then the relative speed = (0.5+0.5) = 1 m/s

1 m/s = 200 m / ( time in seconds )

time = 200 seconds.

y = 200 seconds

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Carla’s total time:
60 + (0.5 × x) + 10 = 200
→ 0.5x = 130
→ x = 260 seconds.

Daniel’s total time:
Total speed = 0.5 + 0.5 = 1.0 m/s

Distance = 200 m
Time = 200 ÷ 1.0 = 200 seconds = y.

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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.

We need to convert relative into absolute speed. (Person speed + Machine speed)
Based on the given information Carla travelled 1.5(40s) + 0.5(x) + 2(5) = 200.
So from this x = 2(130) = 260.

In case of Daniel, His absolute speed is 1m/s. So y = 200/1

Henc,e IMO X=260, Y=200.

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Carla's distance= $ 1.5* 40+ x*0.5 + 2*5 = 200; x= 260$
Daniel's $1*y= 200; y= 200$

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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Carla:

A = 40 [s] * (0,5 [m/s] + 1 [m/s]) = 60 [m]
B = 200 [m] - A - B = 130 [m] = x [s] * 0.5 [m/s] -> x = 260 [s]
C = 5 [s] * (0,5 [m/s] + 1,5 [m/s]) = 10 [m]

Daniel:

y = 200 [m] / (0.5 [m/s] + 0.5 [m/s]) = 200 [s]

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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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.

Carla:
Phase 1:
Speed = 1(run) + 0.5 (walkway) = 1.5m/s
Distance = 1.5(40s) = 60 m

Phase 2:
Speed = 0.5m/s
Distance= 0.5x

Phase 3:
Speed = 1.5+0.5 = 2
Distance = 2 x 5 = 10m

Total:
60+0.5x+10 = 200
x=260s

Daniel:
speed = 0.5 + 0.5 = 1
Time= 200/1 = 200

x=260
y=200

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Carla's Trip
Carla's journey involves different stages. The distance covered in each part can be calculated.

Part 1: Running (40 seconds)
- Carla's speed on the walkway: 1 m/s
- Walkway speed: 0.5 m/s
- Her speed relative to the ground: 1 m/s + 0.5 m/s = 1.5 m/s
- Distance covered: 1.5 m/s * 40 seconds = 60 meters
Part 2: Standing still (x seconds)
- Carla's speed on the walkway: 0 m/s
- Walkway speed: 0.5 m/s
- Her speed relative to the ground: 0 m/s + 0.5 m/s = 0.5 m/s
- Distance covered: 0.5 m/s * xseconds = 0.5x meters
Part 3: Running again (5 seconds)
- Carla's speed on the walkway: 1.5 m/s
- Walkway speed: 0.5 m/s
- Her speed relative to the ground: 1.5 m/s + 0.5 m/s = 2 m/s
- Distance covered: 2 m/s * 5 seconds = 10 meters

Total Distance for Carla
The total distance Carla travels is the sum of the distances from each part, which equals the walkway's total length (200 meters):
60 meters + 0.5x meters + 10 meters = 200 meters
70 + 0.5x = 200
0.5x = 130
x = 130 / 0.5 = 260 seconds
Daniel's Trip
Daniel walks at a constant speed relative to the walkway and reaches the end in y seconds.

Daniel's speed on the walkway: 0.5 m/s
Walkway speed: 0.5 m/s
His speed relative to the ground: 0.5 m/s + 0.5 m/s = 1 m/s
Distance covered: 1 m/s * y seconds = ymeters

Since Daniel reaches the end of the 200-meter walkway, the distance he covers is 200 meters:
y meters = 200 meters
y = 200 seconds

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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The length of the walkway is 200m, and the speed of the walkway (w) = 0.5 m/s

Quick note: When they say relative to the speed of the walkway, they just mean the speed of th person when speed of the walkway is not accounted for, in other words, If Carla is running on a walkway at a rate of 1 m/s relative to the walkway, then that means that speed is for Carla only, not Carla+speed of walkway

Let's decode the statements by person:

For Carla:

She runs at a speed of 1 m/s. So her combined speed when running on the walkway is 1 + w = 1 + 0.5 = 1.5 m/s (we add the speeds since they are moving in the same direction)
She runs at this speed for 40 seconds. Thus, the distance covered is d1 = 40* (1.5) = 60m
She stops for x seconds. Thus, she is moving with the walkway at the speed of w=0.5 m/s for x seconds. So, the distance covered is d2 = x*(0.5)m
She resumes to run, now at 1.5 m/s; so her combined speed is 1.5+w = 1.5+0.5 = 2 m/s; she does that for 5 seconds. Thus, the distance covered is d3 = 5*(2) = 10m
Note that now she has reached the end of the walkway

The total distance covered by Carla = Length of the walkway
d1 + d2 + d3 = 200
60 + 0.5x + 10 = 200

Solving for x, we get x = 260

For Daniel:

He walks at 0.5 m/s. So his combined speed is 0.5+w=0.5+0.5= 1 m/s
He maintains that for y seconds and reaches the end of the walkway. So distance covered by Daniel is y*(1) = y meters

Now, the total distance covered by Daniel = Length of the walkway

so, y=200

Final answer:

x = 260
y = 200

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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.

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30 Jun 2025, 10:43

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walkway speed= 0.5m/s
For X
Carla actual speed for 40 sec= 1+0.5= 1.5 m/s

carla speed for x sec=0.5 m/s
carla speed for 5 sec= 1.5+0.5 m/s

Total Distance= 200= 1.5*40+0.5x+10, x=260 sec

For Y
Daniel speed= 0.5 +0.5 m/s

1*y=200
y=200 sec

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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.

Speeds add up in this case as Carla is running with relative to walkway

Speed of Carla when Carla is running at 1 meter per second = 1 + 0.5 = 1.5 m/sec
Speed of Carla when Carla is running at 1.5 meter per second = 1.5 + 0.5 = 2 m/sec

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

0.5x = 200 - (60 + 10) = 130

x = 260

As Daniel walks at a constant speed, his speed = 0.5 + 0.5 = 1m/sec

Time taken = 200/1 = 200

x = 260 seconds
y = 200 seconds

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carla - 40 second she will travel 1+0.5 = 1.5, then 60m
final with 1.5+0.5 =2, then 10 m, total 70 and remaining 130 with 0.5, X = 130/0.5 = 260
For Y = 200/(0.5+0.5 ) = 200

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Walkway = 200 m long

Carla runs (1+0.5) 1.5m/s for 40 seconds, 0.5m/s for x seconds and 2m/s for 5 seconds to cover 200 m length.
1.5*40 + 2x + 2*5 = 200 => x= 260 ... A

Daniel 1m/s for y seconds => 1*y= 200 => y = 200 ... B

Concept: Relative speed. Since they are moving in same direction as walkway, for an observer not on walkway, walkway speed will add to their speed, to cover the 200m distance.

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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Distance, D = 200 m
Speed of walkway, w = 0.5 m/s (meters = m, second = s)
Speed of Carla = c m/s
Speed of Daniel = d m/s

Relative velocity of Carla, (Vc') = c - w
Relative velocity of Daniel, (Vd') = d - w

Motion for Carla has 3 parts,
1.) (Vc') = c - w; c = 1 + 0.5 = 1.5 m/s
2.)c = 0 m/s, as she stops
3.)c = 1.5 + .5 = 2 m/s

Adding the 3 parts; 1.5*40 + 0.5*x + 2*5 = D = 200

∴x = 260 s

Equation for Daniel,
(Vd') = d - w
d = 0.5 + 0.5 = 1 m/s
Given, 1y = D = 200
∴y = 200 s

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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.

For Carla:

She ran at constant rate for 1m/s for 40s => 40 x 1 = 40m. + walkway: 0.5m/s x 40 = 20m. Total for Phase 1: 60m
And she stop for x second but walkway still move for 0.5m/s => 0.5 x x = (1)
Finally, she ran at constant rate for 1.5m/s for 5 second => 1.5 x 5 = 7.5m + walkway 0.5 x 5 = 2.5. Total for Phase 3: 10m
=> (1) = 200 - 60 - 10 = 130 = 0.5x => x=260 seconds

For Daniel:

He walks nonstop at a constant rate of 0.5m/s + walkway's speed 0.5m/s => Speed 1m/s => 200m : 1m/s = 200 seconds = y

Answer: x=260, y=200

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