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

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y is easier to calculate so I start with y. D walks at a constant rate of 0.5 meter per second relative to the walkway --> his speed = 0.5 + 0.5 (the speed of the walkway) = 1. y = 200/1 = 200s.

For x, we have: 1.5 * 40 + 0.5 * x + 2 * 5 = 200m --> 0.5 * x = 130 --> x = 260s

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 40*(1+0.5) + x*0.5 + 5*(1.5+0.5) = 200
70 + 0.5x = 200
0.5x = 130
x = 260

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

Answers: x = 260 and y = 200

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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.

269=X AND 200 =Y

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Equating the distance travelled by both Carla and Daniel
$(.5+1)40+.5x+(.5+1.5)5 = (0.5+0.5)y$ (since it is the relative speed that is given, add the speed of the walkway to obtain the absolute speed)|

Simplifying this gives you
$140 + x = 2y$

Therefore, $x = 260‎ ‎ and‎ ‎ y = 200$.

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Carla 1m + speed of walkway = 1.5m/sec X 40 sec = 60 m covered; then she runs for 5 sec at a rate of 2m/sec = 10m covered

total distance = 200m- 70m = 130 left at a rate of .5m/sec = 260 seconds that Carla stopped running.

whereas, daniel's walk rate + the rate of the walkway = 1m/sec
200m/1 = 200 seconds for Daniel to reach the end.

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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Daniel
speed=0.5+0.5=1
200 = 1*y
y = 200

Carla
speed1=1+0.5=1.5
distance1=1.5*40=60
speed2=0.5
distance2=0.5x
speed3=1.5+0.5=2
distance3=2*5=10

distance=70+0.5x=200
x = 260

Correct answers are x = 260 and 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.

for Carla
1.5*40 +0.5x + 2*5 = 200 ==> x=260

for Daniel
1*y =200 ==> y=200

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30 Jun 2025, 20:57

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x=260 and 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.

Daniel:
Walkway speed = 0.5 m/s
Daniel walks at 0.5 m/s relative to walkway ⇒ Absolute speed = 1 m/s
Time to cover 200 m = 200/1=200200 / 1 = 200200/1=200 seconds ⇒ y = 200

Carla:

Runs 40 s at 1 m/s relative to walkway ⇒ total speed = 1 + 0.5 = 1.5 m/s
→ Distance = 1.5×40=601.5 × 40 = 601.5×40=60 m
Then stands for xxx seconds ⇒ moves only with walkway at 0.5 m/s
→ Distance = 0.5x0.5x0.5x
Then runs 5 s at 1.5 m/s relative ⇒ total speed = 2 m/s
→ Distance = 2×5=102 × 5 = 102×5=10 m

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

✅ x = 260, y = 200

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i Think the concept tested here is that of relative speeds.
Two speeds are added if they move in thesame direction and subtracted if they move in opposite directions.
it is easier to start with Daniel as he moves with a constant speed of 0.5 durint the whole journey which makes his combined speed 1 m/s ( +0.5 of walkway) meaning he moves 200 meters in 200 seconds.

for Carla she running at 1 m/s + 0.5 of the walkway making 1.5 m/s for 40 seconds. applying D = ST formula we see she moved 60 meters durant the first 40 seconds

similarly during the last 5 seconds she was running at 1.5 + 0.5 of the walkway making a combined speed of 2 m/s. D = ST meaning she covered 10 meters during those five seconds.

Distance left for Carla = 200 - (60 + 10) = 130

she covered this 130 standing still ( he speed was only that of the walkway which is 0.5) therefore she covered those 130 meters in 260 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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30 Jun 2025, 21:19

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Carla:
-Runs 40s at 1 m/s relative to walkway, so actual speed = 1 + 0.5 = 1.5 m/s → covers 60 m
-Then stands still for x seconds, so walkway moves her at 0.5 m/s → covers 0.5x
-Then runs 5s at 1.5 m/s relative to walkway → actual speed = 1.5 + 0.5 = 2 m/s → covers 10 m
Total: 60+0.5x+10=200 → solve: x=260 ✅

Daniel:
-Walks entire way at 0.5 m/s relative to walkway → actual speed = 1 m/s
-Time = 200 ÷ 1 = 200 seconds ✅
Answer:
x = 260, y = 200

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Carla: 200 = 1.5*40 + 0.5x + 2*5 = 70 + 0.5x -> x = 260
Daniel: 200 = 1*y -> y = 200

The right answers are x = 260 and y = 200

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Step 1: Carla's Movement
Total distance = 200 meters
First 40 sec:
Speed = 1 + 0.5 = 1.5 m/s → Distance = 1.5 × 40 = 60 m

Then x sec:
Speed = 0.5 m/s → Distance = 0.5 × x

Final 5 sec:
Speed = 1.5 + 0.5 = 2.0 m/s → Distance = 2 × 5 = 10 m

Total:
60+0.5x+10=200⇒0.5x=130⇒x=26060+0.5x+10=200⇒0.5x=130⇒x=260

Step 2: Daniel's Movement

Speed = 0.5 + 0.5 = 1.0 m/s
Time = 200 ÷ 1 = 200 seconds ⇒ y = 200

Therefore,
x = 260
y = 200

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Segment	Carla’s speed relative to walkway	Walkway speed	Ground speed
Run #1	1 m/s	+0.5	1.5 m/s
Stand	0	+0.5	0.5 m/s
Run #2	1.5 m/s	+0.5	2.0 m/s
Daniel	0.5 m/s	+0.5	1.0 m/s

Distances Carla covers
[*]First run (40 s): 1.5×40=60 m
[*]Pause (x s): 0.5x m
[*]Second run (5 s): 2.0×5=10 m
Total must equal the 200 m walkway:
60+0.5x+10=200 ⟹0.5x=130⟹x=260 s.

Daniel’s time
Ground speed =1.0 m/s
y= (200 m)/(1 m/s)=200 s

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Ans: x = 260 and 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.

for x = Carla 1.5 for 40 seconds + .5 fo x + 2 for 5 seconds (we added 0.5 m/sec to the speed for relative speed on the walaway; if carla runs at a speed of v, carla's total speed on walaway becomes v+0.5)

so from the above equation = 1.5 * 40 + x* .5 + 2 * 5 = 200 = from this = x = 260

for y = no change in the speed, only 0.5 m/sec walking so relative speed becomes 1m/sec = 200 seconds for y

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in short) on translating statement 1 and 2 we get >> 60+ .5x+ 10 =200 >>x=260
on translating statement 3 we get>> 1 x y = 200

detailed translation) Carla runs at 1.5 meters per second for 40 seconds, covering 60 meters. Then she stands still for x seconds while the walkway moves her at 0.5 meters per second, covering 0.5x meters. Then she runs at 2 meters per second for 5 seconds, covering 10 meters. So total distance is 60 plus 0.5x plus 10 equals 200. That gives 0.5x equals 130, so x equals 260.

Daniel walks at 1 meter per second relative to the ground, since he walks at 0.5 meters per second relative to the walkway and the walkway moves at 0.5 meters per second. To cover 200 meters at 1 meter per second, he takes 200 seconds. So y equals 200.

x is 260
y is 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.

The walkway moves at $0.5$ m/s.
Carla’s trip:

First leg:
Runs at $1$ m/s relative to walkway
→ Ground speed = $1 + 0.5 = 1.5$ m/s
Time = 40 s
Distance = $1.5 \times 40 = 60$ m
Second leg:
Stands still for $x$ seconds
→ Carried only by walkway at $0.5$ m/s
Distance = $0.5x$
Third leg:
Runs at $1.5$ m/s relative to walkway for 5 seconds
→ Ground speed = $1.5 + 0.5 = 2$ m/s
Distance = $2 \times 5 = 10$ m

Total walkway length = 200 m:
$60 + 0.5x + 10 = 200 \Rightarrow 0.5x = 130 \Rightarrow x = 260$

Daniel’s trip:
Walks at $0.5$ m/s relative to walkway
→ Ground speed = $0.5 + 0.5 = 1$ m/s
$y = \dfrac{200}{1} = 200$ seconds

Final Answer: x = 260, y = 200

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Quote:

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.

[color=#000000]x[/color]	[color=#000000]y[/color]
		[color=#000000]100[/color]
		[color=#000000]130[/color]
		[color=#000000]200[/color]
		[color=#000000]250[/color]
		[color=#000000]260[/color]
		[color=#000000]300[/color]

Distance = speed * time
Hence both will have covered 200 mts (moving walkway)
Carla' s distance = (1+0.5)40 +x(0.5)+ (1.5+0.5)5
=70+0.5x
This will be equal to 200, hence x = 260
Daniel's distance = (0.5+0.5)y
This will be equal to 200
hence Y = 200

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Quote:

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.
x y
100
130
200
250
260
300

A)Distance = speed * time
Hence both will have covered 200 mts (moving walkway)
Carla' s distance = (1+0.5)40 +x(0.5)+ (1.5+0.5)5
=70+0.5x
This will be equal to 200, hence x = 260
Daniel's distance = (0.5+0.5)y
This will be equal to 200
hence Y = 200

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Speed of Carla in Case 1= 1+0.5=1.5
Distance travelled by carla in case 1= 1.5*40=60m
in second case carla travels with the speed of the runway so distance travelled= 0.5x
third case carla travelled a distance of 2*5=10m
Total DIstance=200=60+0.5x+10
from here x= 1300/5=260

Speed of Daniel=1m/s
Distance travelled=200m
time taken= 200/1=200 sec

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