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805+ Level|   Math Related|         
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 09:59
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Bob takes more time than Alice. 40 seconds
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 10:03
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Did not understand the question t the time in seconds after which Alice and Bob meet.
We can use relative speed formula
Alice walks upward at 2 steps/sec relative to the escalator, so her net speed = s+2
Bob walks downward at 3 steps/sec relative to the escalator, so his net speed = s−3
then we know relation of bobs and anns time t=4t
and then calculate s that is 14/3.

Alice time=80/s+2=12 sec

so steps taken by alice=12*2=24 steps





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Alice and Bob are on an escalator with 80 steps from bottom to top. The escalator moves upward at a constant rate of s steps per second.

Alice starts from the bottom and walks upward at a constant rate of 2 steps per second relative to the escalator. Bob starts from the top and walks downward at a constant rate of 3 steps per second relative to the escalator.

Bob takes four times as long to reach the bottom as Alice takes to reach the top.

Select for t the time in seconds after which Alice and Bob meet, and select for n the number of steps Alice would have walked on the escalator by the time she reaches the top. Make only two selections, one in each column.
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 10:05
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Alice and Bob are on an escalator with 80 steps from bottom to top. The escalator moves upward at a constant rate of s steps per second.

Alice starts from the bottom and walks upward at a constant rate of 2 steps per second relative to the escalator.

Distance covered in Steps by Alice in a sec: (2 + S)


Bob starts from the top and walks downward at a constant rate of 3 steps per second relative to the escalator.

Distance converted in Steps by Bob in a sec: (3 - S) for Bob to go downstairs S must be lower than 3;

Bob takes four times (4t) as long to reach the bottom as Alice takes (t) to reach the top.

(3-S)*4t = 80;

(2+S)*t = 80;

(3-S)*4=2+S;

12 - 4S = 2 + S;

5S = 10; S = 2;

Select for t the time in seconds after which Alice and Bob meet, and select for n the number of steps Alice would have walked on the escalator by the time she reaches the top. Make only two selections, one in each column.

To meet them, they must have cover 80 steps cumulative: (3-S) T + (2+S) T = 80;

T + 4T = 80; T = 16;

Alice Walk 2 Steps in a sec and Escalator also moves 2 steps in a sec, hence alice would have cover only 40 steps and 40 would be cover by escalator.
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 10:06
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For t, we can use the relative speed concept to quickly solve: 80/ (2+3) = 16 seconds when Alice & Bob met.
- We ignore the constant rate of escalator speed (s) since it will cancel out in the denominator: 2+s+3-s

For n, we need to find Alice speed x total time from Total Rate x Total Time = (2+s)*total time = 80 - Don't forget to take into the account of escalator speed!
- Since Bob takes four times as long to reach the bottom as Alice takes to reach the top, we can find (s): 2+s = 4(3-s) --> s = 2
- Total time = 80/4 = 20 secs for Alice to reach the top
- Her number of steps (n) = 2*20 = 40 steps
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 10:07
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  • escalator moves up at speed s.
  • Alice climbs up at 2 steps/sec relative to escalator her overall speed is 2 + s.
  • Bob climbs down at 3 steps/sec relative to his overall downward speed is 3 − s.
  • Escalator has 80 steps.
  • Bob takes 4× as long to go down as Alice takes to go up.

Set times:
  • Alice: 80⁄(2 + s)
  • Bob: 80⁄(3 − s)
And 80⁄(3 − s) = 4 × 80⁄(2 + s)
solving gives s = 2 steps/sec

Thus:
  • Alice speed = 4 steps/sec => time = 80⁄4 = 20 s
  • Bob speed = 1 step/sec => time = 80⁄1 = 80 s (4 times)


Meeting time t:
  • Alice covers 4t steps.
  • Bob covers t steps.
    Their paths sum to 80:
    4t + t = 80
    t = 16 seconds

Alice walks at 2 steps/sec (on the escalator) for her full 20 s:
n = 2 × 20 = 40 steps


t=16
n=40
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 10:15
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Bunuel
 


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Alice and Bob are on an escalator with 80 steps from bottom to top. The escalator moves upward at a constant rate of s steps per second.

Alice starts from the bottom and walks upward at a constant rate of 2 steps per second relative to the escalator. Bob starts from the top and walks downward at a constant rate of 3 steps per second relative to the escalator.

Bob takes four times as long to reach the bottom as Alice takes to reach the top.

Select for t the time in seconds after which Alice and Bob meet, and select for n the number of steps Alice would have walked on the escalator by the time she reaches the top. Make only two selections, one in each column.
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Let's say escalator moving at rate of s steps per second

So relative rate of alice 2+s steps per second (as she is also going up)
relative speed of bob 3-s steps per second (as he is coming down we subtract rate of escalator)

also given if time taken b y bob = 4* time taken by alice

tb = 4 * ta
80/(3-s) = 4*80/(2+s)
2+s = 4*(3-s)
2+s = 12-4s
5s = 10
s = 2
so rate of escalator = 2 steps per second

alice rate = 4
bob rate = 1

Now to calculate their meeting point time t
the distance each travelled will sum to 80

da+ db = 80

4*t + 1*t = 80
t=16

To calculate steps taken by Alice
first we need to calculate total time for her entire journey
which is
80/(2+s) = 80/4 = 20

steps she took will be her rate* time = 2*20 = 40

So
t=16
n=40
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 10:16
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4 A time = B time
A's speed is actually
2 + elevator speed X
B's speed is actually
3 - elevator speed X

4 80 / (2 + X) = 80 / (3 - X)

X = 2
(4 + 1) t = 80
t = 16

80 / (2 + 2) = 20
20* 2 = 40
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 10:27
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We know that:
vA = 2+s
vB = 3-s

4 x (80/(2+x)) = 80 / (3-s)
s = 2

So vA = 4, vB = 1
t = 80 / (vA + vB) = 16
Regarding n, for each second Alice walks 2 steps and the escalator moves 2 steps. So she must walk half the distance = 40 steps. n = 40
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 10:39
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Speed of Elevator (Se) = s/sec
Distance = 80 steps
Speed of Alice (Sa) = 2/sec + s/sec (as it is relative to the escalator), similarly,
Speed of Bob (Sb) = 3/sec - s/sec

Time taken by B is 4 times A i.e.,
80/3-s=4*80/2+s
solve for s and we will get s = 2/sec

Value of n (no. of steps by Alice) = 80/2+2 = 20
Value of t (Time when A&B meet) = 16, how?
Distance by A in 16 secs is 64 steps and Distance by B in 16 secs is 16 steps which totals to 80 steps.

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Alice and Bob are on an escalator with 80 steps from bottom to top. The escalator moves upward at a constant rate of s steps per second.

Alice starts from the bottom and walks upward at a constant rate of 2 steps per second relative to the escalator. Bob starts from the top and walks downward at a constant rate of 3 steps per second relative to the escalator.

Bob takes four times as long to reach the bottom as Alice takes to reach the top.

Select for t the time in seconds after which Alice and Bob meet, and select for n the number of steps Alice would have walked on the escalator by the time she reaches the top. Make only two selections, one in each column.
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 10:43
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Information given:
- Escalator: 80 steps from bottom to top, moves upward at s steps per sec
- Alice: walks upward at 2 steps/sec relative to escalator, so net speed: s+2
- Bob: walks downward at 3 steps/sec relative to escalator, so net speed: s-3
- Bob takes 4x as long to reach the bottom as Alice takes to reach the top
- They start at the same time from opposite ends

Question:
- Pick t: how many seconds after starting do Alice & Bob meet
- Pick n: how many steps does Alice walk on the escalator by the time she reaches the top?

Solution:
- First, find the escalator speed s
- Alice time to top: 80/(s+2)
- Bob time to bottom: 80/(3-s) (note that this is inverted because Bob's net speed is down)
- 80/(3-s) = 4 * (80/s+2), which gives 1/(3-s) = 4/(s+2), which gives s+ 2 = 12 - 4s, 5s = 10, s = 2
- The escalator speed is 2 steps/sec

- Alice's total time to reach the stop is therefore 80/(2+2) = 20 sec
- Her rate relative to escalator is 2 steps/sec, so n = 2 x 20 = 40
- n = 40

- Alice net up speed is 4 (2+2)
- Bob's net down speed is -1 (1-3 = -1) (1 step/sec downward)
- Closing speed is 4 + 1 = 5 steps/sec
- Initial distance = 80 steps
- So, t = 80/5 = 16

Answer: t = 16 sec, n = 40 steps
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This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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Alice and Bob are on an escalator with 80 steps from bottom to top. The escalator moves upward at a constant rate of s steps per second.

Alice starts from the bottom and walks upward at a constant rate of 2 steps per second relative to the escalator. Bob starts from the top and walks downward at a constant rate of 3 steps per second relative to the escalator.

Bob takes four times as long to reach the bottom as Alice takes to reach the top.

Select for t the time in seconds after which Alice and Bob meet, and select for n the number of steps Alice would have walked on the escalator by the time she reaches the top. Make only two selections, one in each column.
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 10:46
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Let s be the escalator speed in steps per second (the escalator is moving from bottom to top)
Speed of Alice be A
Speed of Bob be B

Alice starts from the bottom and walks upward at a constant rate of 2 steps per second relative to the escalator.
This means that on the Escalator, Alice is walking at 2 steps / sec
Alice's speed relative to ground is then
A = s + 2

Bob starts from the top and walks downward at a constant rate of 3 steps per second relative to the escalator.
Similarly, for Bob, the relative speed is
B = - s + 3 (because the escalator is in reverse)

There are a total of 80 steps, and Time for Bob = 4 * Time for Alice
Time = Distance / Speed
tA = 80 / (s + 2)
tB = 80 / (3 – s)


tB = 4 * tA
80 / (3 – s) = 4 * 80 / (s + 2)
1 / (3 – s) = 4 / (s + 2)
s + 2 = 4(3 – s)
s = 2

n the number of steps Alice would have walked on the escalator by the time she reaches the top
From tA = 80 / (s + 2)
We can substitute value of s and determine time that Alice walks on escalator as
tA = 80 / 4 = 20 seconds

In those twenty seconds, total steps from bottom to top are 80
Escalator moving automatically at 2 steps per second -> 2 * 20 seconds = 40 steps
The remaining 80-40 = 40 steps Alice WALKS (at 2 steps per second for 20 seconds)

t the time in seconds after which Alice and Bob meet
Escalator speed is 2 steps per second
A = s + 2 = 4 steps per second upwards
B = -s + 3 = 1 steps per second downward

in t seconds, Alice goes 4t steps up and Bob goes 1t step down

When they meet, total steps they would have travelled would be 80
so 4t + t = 80
which means t =16

Correct answer:
n = 40 & t = 16
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 11:19
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Bunuel
 


This question was provided by GMAT Club
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Alice and Bob are on an escalator with 80 steps from bottom to top. The escalator moves upward at a constant rate of s steps per second.

Alice starts from the bottom and walks upward at a constant rate of 2 steps per second relative to the escalator. Bob starts from the top and walks downward at a constant rate of 3 steps per second relative to the escalator.

Bob takes four times as long to reach the bottom as Alice takes to reach the top.

Select for t the time in seconds after which Alice and Bob meet, and select for n the number of steps Alice would have walked on the escalator by the time she reaches the top. Make only two selections, one in each column.
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There are 80 steps in the escalator. The escalator moves upward ( bottom to up) at the constant rate of s steps per second.

Speed of Escalator = E = s steps/second.

There are two participants : Alice and Bob.

Alice moves from the bottom towards up. Speed of Alice (A) = 2 steps/second.

Bob moves from top to downward. Speed of Bob (B) = 3 steps/ second.

Time to reach bottom = 4 * time to reach top.

Tb = 4* ta

Case 1: Alice moving up along with the escalator.

Number of steps = 80 = s*ta + 2* ta

= 80 = (s+2) * ta

ta = 80 /(s+2)

Case 2: Bob moves down along with the escalator.

Number of steps = 80 = 3*tb - s*tb

tb = 80/(3-s)

Substitute ta and tb in the equation above

80/(3-s) = 4* [80/(s+2)]

s+2 = 12 -4s

5s = 10

Speed of escalator E = 2 steps/ second.

TIME OF MEET:

Relative speed of Alice = 2+2 = 4 m/s

Relative speed of Bob = 3-2 = 1 m/s

Time = 80 / (4+1) = 16 seconds

T = 16 seconds

Number of steps taken by Alice to reach top: (IF HE HAD WALKED)

Time taken to reach top = 80/4 = 20 seconds.

The number of steps taken by Alice in 20 seconds = 20* 2 = 40 steps.
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 11:52
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Answer: t=16 seconds and n=40 steps

Alice and Bob are on an escalator with 80 steps from bottom to top.

The escalator moves upward at a constant rate of s steps per second.

Rate of Alice going up the escalator=( 2+s)

Rate of Bob going down the escalator =( 3-s)

it is given that t(Bob)= 4 t (Alice)
so 80/(3-s)=4*(80/2+s)
therefore, s=2

t( Alice)= 20 seconds and t(Bob) = 80 seconds

the total distance= 80 steps
for Bob and Alice to meet, time is the same T

rate(Alice)*t+ rate (Bob)*t=80
4t+t=80
or 5t=80 so t=16
Hence Bob and Alice meet after 16 seconds.

II) for n- Alice walks up at the rate of 2 steps relative to the escalator and the escalator moves up at the rate of s=2

Hence for going up 80 steps on the escalator, Alice walks up 40 steps.
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 11:54
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Distance = 80
Alice:
Speed = s+2
So, time taken by Alice to reach the top, \( t_A = \frac{80}{(s+2)}\)

Bob:
Speed = 3-s
So, time taken by Bob to reach the bottom, \(t_B =4t_A = \frac{320}{(s+2)}\)
and, \(t_B = \frac{80}{(3-s)}\)

\(\implies \frac{320}{(s+2)} = \frac{80}{(3-s)}\)
\(\implies 80(s+2)=320(3-s)\)
\(\implies 80s+160=960−320s\)
\(\implies s=2\)

So, escalator speed, s = 2 steps/second

So, Alice's speed = 2+2 = 3 steps/second
\(t_A=20\) sec
Alice walks 2 steps/sec relative to the escalator for 20 seconds
So, \(n=2 \times 20=40\)

When Alice and Bob meet, they will have travelled for the same time t and total distance covered = 80 steps
\(\implies 80=(s+2)t+(3-s)t\)
\(\implies 80=4t+t\)
\(\implies t=\frac{80}{5}=16\)

Answer: t=16, n=40.
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 11:59
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Va= 2+s, Vb= 3-s => t=80/(2+s+3-s)=80/5=16
80/2+s = 80/3-s => s=0.5
=> n = t x (Va+s) = 16x(2+0.5)=40
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Alice and Bob are on an escalator with 80 steps from bottom to top. The escalator moves upward at a constant rate of s steps per second.

Alice starts from the bottom and walks upward at a constant rate of 2 steps per second relative to the escalator. Bob starts from the top and walks downward at a constant rate of 3 steps per second relative to the escalator.

Bob takes four times as long to reach the bottom as Alice takes to reach the top.

Select for t the time in seconds after which Alice and Bob meet, and select for n the number of steps Alice would have walked on the escalator by the time she reaches the top. Make only two selections, one in each column.
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 12:35
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Time taken by B is 4 times time taken by A
B=4*A

80/(3-s)=4*80/(s+2)

therefore, s= 2 steps/sec

To find the value of "n" = No. of total steps/Relative speed = 80/s+2
n=80/(2+2)= 20

n=20

To find the value of "t" = Total steps/ A & B's relative speed moving in opposite direction

A's speed= s+2=4
B's speed= 3-2=1

t= 80/(4+1)
t=16

I Hope everyone finds the solution useful :)
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 13:05
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the time in seconds after which Alice and Bob meet - since they are moving towards each other, the speeds will get added.. meeting time will be D/s1+s2 = 80/ 2+3 = 80/5 = 16 = t

The number of steps Alice would have walked on the escalator by the time she reaches the top = 80/2 = 40 steps = n
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 13:22
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t = 16 (Option B) and n = 40 (Option D) are the Correct answers.

First of all lets understand the question type. So this question is the same as the type of question in which we are told that a ship is sailing with and against the ocean currents.
And to find the speed of ship in these type of question we have to find two speeds: one when the ship is sailing with the current and another when the ship is sailing against the current.
Speed with the current speed: Speed of ship + Speed of the current
Speed again the current: Speed of ship - Speed of the current

Similarly in this question we are told that: "Alice and Bob are on an escalator with 80 steps from bottom to top. The escalator moves upward at a constant rate of s steps per second". From here we can identify the distance and the direction in which the escalator is moving.

Now lets look at the next piece of information: "Alice starts from the bottom and walks upward at a constant rate of 2 steps per second relative to the escalator. Bob starts from the top and walks downward at a constant rate of 3 steps per second relative to the escalator". From here we can conclude that Alice is walking with the speed of escalator and Bob is walking against the speed of escalator.
With this information we can calculate the speed of Alice & Bob while they are on escalator:
Alice's Speed: 2 + s (s is the speed of escalator)
Bob's Speed: 3 - s (s is the speed of escalator)

Now lets see the last piece of information: "Bob takes four times as long to reach the bottom as Alice takes to reach the top" which basically means that: Time taken by Bob = 4*Time taken by Alice.

Now after checking all the information lets start solving this question with the help of all the information mentioned in the question.

Time taken by Bob = 4*Time taken by Alice

We all know the formula to calculate the time taken to travel a distance: Time = Distance/Speed and in question the distance of the escalator is give as 80.

So, 80/(3-s) = 40*80/(2+s) from here we can cancel out 80 from each side after which we will get:
12-4s = 2+s, after further solving it we will get s = 2

So from this we can calculate the exact speed of Alice and Bob which will be: Alice - 4 (2+s), Bob - 1 (3-s).

Now the question asks us about the time after which Alice and Bob will meet on escalator.

Time = Distance/Total Speed (Alice's Speed + Bob's Speed)
Time = 80/5 (4+1)
Time = 16 seconds (Option B)

Next the question asks us the number of steps Alice would have walked on the escalator by the time she reaches the top.

So we know that Alice's Speed is 4 i.e. 2 steps per second for escalator and 2 steps per second which she is walking by her-self. Therefore from here we can conclude that half the distance was covered by escalator and other half she covered by walking:

Steps taken = 80/2
Steps taken = 40 steps (Option D)


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Alice and Bob are on an escalator with 80 steps from bottom to top. The escalator moves upward at a constant rate of s steps per second.

Alice starts from the bottom and walks upward at a constant rate of 2 steps per second relative to the escalator. Bob starts from the top and walks downward at a constant rate of 3 steps per second relative to the escalator.

Bob takes four times as long to reach the bottom as Alice takes to reach the top.

Select for t the time in seconds after which Alice and Bob meet, and select for n the number of steps Alice would have walked on the escalator by the time she reaches the top. Make only two selections, one in each column.
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Dav2000
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 13:22
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Alice and Bob are on an escalator with 80 steps from bottom to top. The escalator moves upward at a constant rate of s steps per second.

Alice starts from the bottom and walks upward at a constant rate of 2 steps per second relative to the escalator. Bob starts from the top and walks downward at a constant rate of 3 steps per second relative to the escalator.

Bob takes four times as long to reach the bottom as Alice takes to reach the top.

Select for t the time in seconds after which Alice and Bob meet, and select for n the number of steps Alice would have walked on the escalator by the time she reaches the top. Make only two selections, one in each column.
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According to the question

The speed of Alice is (2+s) step/sec
and speed of Bob is (3-s) steps/sec

According to information given Time of Bob for 80 steps is 4 times Time taken by Alice.
Therefore, (80/(2+s)) * 4 = 80/(3-s) which gives s = 2 steps/sec

So the total speed of Alice is 4 steps/sec. And since is speed of Alice is only half of the total speed of Alice combined with escalator the she covered only have the distance Hence n = 40.

And for t we have the total distance equation since the total distance covered by Alice and Bob will be same in t time then 80 = (Speed Bob)*t + (Speed Alice) * t
=> 80 = 1*t + 4*t which gives t= 16. Hence the answer.
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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator 02 Jul 2025, 14:10
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Alice and Bob are on an escalator with 80 steps from bottom to top. The escalator moves upward at a constant rate of s steps per second.

Alice starts from the bottom and walks upward at a constant rate of 2 steps per second relative to the escalator. Bob starts from the top and walks downward at a constant rate of 3 steps per second relative to the escalator.

Bob takes four times as long to reach the bottom as Alice takes to reach the top.

Select for t the time in seconds after which Alice and Bob meet, and select for n the number of steps Alice would have walked on the escalator by the time she reaches the top. Make only two selections, one in each column.
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Called T is Alice’s time to reach the top. => Bob’s time to reach the bottom is 4T.
We have: distance= speed*time
80=T*(2+s) and 80=4T*(3-s)
=> (2+s)*T=(3-s)*4T => 2+s=12-4s => s=2.
The time (t) in seconds after which Alice and Bob meet: t= 80/(combine rate) = 80 /(2+s)+(3-s)=80/5=16
The numbers (n) of steps Alice would have walked on the escalator:
T=80/(2+s)=80/4=20 seconds and Alice walks with 2steps/seconds => n=20*2=40 steps

Answer: t=16 and n=40
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