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

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30 Jun 2025, 23:17

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1. Finding x

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

x = 260

2. Finding y

y = 200/1 = 200

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30 Jun 2025, 23:21

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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.

Lets start with Carla first

a. 1m/s relative to the walkway --> total speed = 1 + 0.5 = 1.5m/s
Time spent = 40 seconds
distance = 1.5m/s x 40s = 60m

b. Speed = 0 relative to the walkway --> total speed = 0 + 0.5 = 0.5m/s
Time spent = x
distance = 0.5 * x = 0.5x m

c. Speed = 1.5m/s relative to the walkway --> total speed = 1.5 + 0.5 = 2m/s
time spent = 5s
distance = 2m/s * 5s = 10m

Total distance covered by Carla = 60m + 0.5x m + 10m = 200
=> 70 + 0.5x = 200
=> 0.5x = 130
=> x = 260 seconds

Lets take Daniel now
Speed = 0.5m/s relative to the walkway --> total speed = 0.5 + 0.5 = 1m/s
distance = 200m
=> time = 200m/1m/s
=> time = 200 seconds

x for Carla is 260 seconds
y for Daniel is 200 seconds

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30 Jun 2025, 23:24

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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.

Ans X= 260 , Y=200

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01 Jul 2025, 00:00

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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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Carla: First at total speed of = 1 + 0.5 = 1.5 m/s, covered 60 meters, then at last at relative speed of 2 m/s (1.5 + 0.5) for 5 seconds: covered 10 meters.

And covered middle 130 (200-60-10) meters at 0.5 m/s which took her 260 seconds

Daniel: Walks at 0.5 m/s + 0.5 m/s of walkway , so total = 1 m/s for 200 meters which will take 200 seconds.

Hence, x=260,y=200

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01 Jul 2025, 00:42

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Carla

Total Distance = 200m
Speed of the walkway = 0.5 m/s
Carla's Speed = 1m/s

Distance covered by Carla while running at 1m/s = speed * time
= 1.5 m/s (1+0.5) * 40 = 60m
Remaining distance = 200-60 = 140 m
Let the distance covered by Carla in x seconds = d
d = speed * time
d = 0.5*x (as Carla is standing still, but walkway is moving, so speed of walkway is considered)

Now,

Remaining distance for Carla = 140-d, which she covered @ 2m/s (1.5+0.5) in 5 secs

140-d = 2*5
140-d = 10
d=130 m
Also d = 0.5 x
130 = 0.5 x
x = 260 secs

Daniel

Total Distance = 200m
Speed of the walkway = 0.5 m/s
Daniel's Speed = 0.5 m/s
Total Speed = 0.5+0.5=1m/s
Time taken to cover 200m = y
Time, y = Distance/Speed
y = 200/1
y = 200 secs

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 has three different speeds walking through the walkaway
- running in the first portion = 1+0.5
- standing = 0.5
- running in the last portion = 1.5 + 0.5

equating to total distance for Carla
1.5*40 + x*0.5 + 2*5 = 200
x = 260 sec

equating to total distance for Daniel
(0.5+0.5)*y=200
y = 200 sec

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Carla:
1. For 40 seconds, her speed is (0.5+1) or 1.5 m/s. she travels 1.5*40 = 60 m.
2. when she stops, she travels 0.5*x metres.
3. She runs at 1.5+0.5 or 2m/s for 5 seconds, covering 2*5 = 10 m and reaches the end.

thus, 200 - (60+10) = 130 m was covered during x seconds. x = 130/0.5, so x = 260 seconds.

Daniel:
1. He walks at (0.5+0.5) or 1m/s for y seconds, covering 200m. Thus y = 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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01 Jul 2025, 03:10

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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.

Length of walkway = 200 m, Sw= 0.5 m/sec
Carlas :
| -----1.5*40=60 meters-------0.5x meters---------2*5=10 meters----|
60+0.5x+10=200
0.5x=130-> x=260 secs

Daniel :
| ------------------------------------1*y =200----------------------------- ----|
Y=200secs

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Updated on: 01 Jul 2025, 10:21

Originally posted by Semshov on 01 Jul 2025, 03:15.
Last edited by Semshov on 01 Jul 2025, 10:21, edited 2 times in total.

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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.

The easiest is Daniel : Daniel's speed = 0.5 m for 1 second. He is also drived by the walkway at a constant speed of 0.5 m for one second. The combined rate for Daniel is 1 m for 1 second. So he will reach the 200 m in 200 second (value of y).

By analogy :
- Combined rate of Carla's first move : 1.5 m for 1 second. That gives 60 m distance covered in 40 seconds.
- The distance when carried only by the walkway is : 0.5x (0.5 m for 1 second ---> for x seconds we have 0.5méter * x seconds /1 second = 0.5x)
- The combined rate for the last move : 2 m for 1 second. That gives 10 m distance covered in 5 seconds.

The total distance is 60 + 0.5x + 10 = 200. By solving this equation we find x=260 seconds

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01 Jul 2025, 03:40

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X=260, Y=200

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Carla runs 1.5 m/s for 40 seconds: 1.5×40=60 meters
Then Carla stands still for x seconds. The walkway carries her at 0.5 m/s, so she covers 0.5x meters
Then Carla runs again at 2 m/s for 5 seconds: 2×5=10 meters
Total distance Carla covers: 60+0.5x+10
Since the walkway is 200 meters long:70+0.5x=200⇒x=260

Daniel walks the entire 200 meters at 1 m/s ⇒ Time = 200 seconds

x = 260 seconds, y = 200 seconds

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01 Jul 2025, 04:12

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Carla S = (1+0.5)X40 = 60M
0.5X
(1.5 + 0.5)X5 = 10
60+10+0.5X= 200
0.5X=130
X = 260
(0.5+0.5)Y=200
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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01 Jul 2025, 04:36

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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.

First, Carla
1m/s relative to walkway => 1.5m/s x 40 s = 60 m
stand still => 0.5m/s x x = 0.5x m
1.5m/s relative to walkway = 2 m/s x 5 s = 10 m

=> 60 + 0.5x + 10 = 200m => x = 260

Second, Daniel
0.5m/s relative to walkway => 1m/s x y = 200 m => y = 200

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Total walkway length = 200 meters
The walkway moves at a constant speed of 0.5 meters per second.
[hr]
Carla:
Part 1: Running for 40 seconds at 1 m/s relative to the walkway
Since the walkway is also moving at 0.5 m/s, Carla's actual ground speed is:
1 + 0.5 = 1.5 m/s
Distance covered in 40 seconds:
1.5 × 40 = 60 meters
[hr]
Part 2: She stops running for x seconds and stands still
During this time, she moves only with the walkway at 0.5 m/s.
Distance covered in this phase: 0.5x
Total distance so far = 60 + 0.5x
[hr]
Part 3: She resumes running at 1.5 m/s relative to the walkway for 5 seconds
So ground speed = 1.5 + 0.5 = 2 m/s
Distance = 2 × 5 = 10 meters
Now the total distance Carla covers in all three parts is:
60 (first part) + 0.5x (second part) + 10 (third part) = 200 meters
That gives us the equation:
60 + 0.5x + 10 = 200
=> 0.5x = 130
=> x = 260
So the correct value for x is 260
[hr]
Daniel:
He walks from start to finish without stopping at a constant rate of 0.5 m/s relative to the walkway.
So his actual ground speed = 0.5 + 0.5 = 1 m/s
To travel 200 meters at 1 m/s, time = 200 / 1 = 200 seconds
So the correct value for y is 200
[hr]
Final Answers:
x = 260
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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Speed of walkway: 0.5 m/s

<Carla>
* For the first 40 seconds:
Speed = 1 + 0.5 = 1.5 m/s
Distance = T x S = 40 x 1.5 = 60 m

* For the last 5 seconds:
Speed = 1.5 + 0.5 = 2 m/s
Distance = T x S = 5 x 2 = 10 m

Total distance covered by running: 70 m
Distance on walkway: 130 m
Time (to cover 130m on walkway): 130 / 0.5 = 260 seconds (labeled as 'x')

<Daniel>
His speed: 0.5 + 0.5 = 1 m/s
Time to cover 200m: 200 / 1 = 200 seconds (labeled as 'y')

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

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.

Given:
Speed of the walkway: 0.5m/s
Length of the walkway: 200m

Carla's Case:

Running for 40s:
Running speed= 1m/s
Total speed relative to walkway: 1+0.5=1.5m/s.
Total time=40s.
Distance covered (d1)= 40*1.5=60m.

Break for x seconds:
Speed= 0.5m/s (only the walkway's speed is considered; otherwise, she is at rest)
Distance covered=d2

Starts running again:
Running speed= 1.5m/s
Total time=5s
Total Speed= 1.5+0.5= 2m/s
Distance covered= 2*5=10m.

Remaining distance=d2= 200-(60+10)=130m
This distance was covered during the x seconds at the speed of 0.5m/s.
=> x= 130/0.5=260s.

Daniel's Case:
Walking speed= 0.5m/s
Total Speed= (0.5+0.5)=1m/s
Total distance=200m
Time= y= 200/1=200s

Ans: x=260
y=200

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01 Jul 2025, 05:57

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Total distance = 200m
moving walkway =. 0.5 m/s

Daniel = 0.5m/s relative to the moving walkway -> 1m/s relative to ground.
d=rt -> 200 = 1m/s * y -> y = 200 meters.

Carla = 1m/s relative to the moving walkway -> 1.5m/s relative to ground for 40 seconds -> d = 1.5m/s * 40 seconds = 60 meters.

Then she runs 1.5m/s relative to the moving walkway -> 2m/s relative to ground for 5 seconds -> d = 2m/s * 5 seconds = 10 meters.

x = 200 - 60 - 10 = 130 meter

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01 Jul 2025, 06:04

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x=260 y=200
r t w
walkway .5m/s 200m

c1 1+.5 40s 60m
c2 1.5+.5 5s 10m

total distance walked by C is 70 m which means that walkway helped her cover 130 [email protected] m/s.
thus, 130/.5 which is 260

D .5+.5 y 200m

y = 200/1, thus, y = 200

hope this solution is correct.

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.