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GMAT Club Olympics 2025 (Day 3): Alice and Bob are on an escalator

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Mardee

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03 Jul 2025, 06:13

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Let escalator speed be x

Alice's speed relative to the escalator = 2 steps/s (up)
=> Alice's speed relative to the ground = x + 2

Bob's speed relative to the escalator = 3 steps/s (down)
=> Bob's speed relative to the ground = 3 - x (up)

We know that Bob takes 4 times as long to reach the bottom as Alice takes to reach the top
=> If lets say Alice time to reach top is T, Bob's time to reach down is 4T

=> For Alice, Total length =>(x+2)*T = 80 => T = 80 / (x+2)
=> For Bob, Total length =>(3-x)*4T = 80
So,
(3-x)*4*(80/(x+2)) = 80
x = 2 (Speed of escalator)

Alice total speed while going up = 2 + 2 = 4
Bob total speed while going down = 3 - 2 = 1

Relative speed while approaching each other = 4 + 1 = 5
Time taken for them to meet each other t = 80 / 5 = 16 seconds

Time taken by Alice to reach the top = 80 / 4 = 20 seconds
Number of steps taken by Alice to reach the top on the escalator n = 20* (2 steps/sec) = 40 steps

t = 16
n = 40

SumnerSCB

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03 Jul 2025, 06:15

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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.

So make the equations for both Alice = 80/(2+s)=TimeA Bob = 80/(3-s)= Time B..... It tells us time b is equal to 4 times time A so 4(80/(2+s))=80/(3-s) this ends up giving s=2 which means the escalator is moving at a rate of 2. another more logical way and less math would just to realize that Bobs denominator must have been 4 times the size. So with 2 for s we then plug it in with 4t=80 or time for alice is 20 seconds so for n she walks 40 steps half the distance because the escalator covers the other half. For when they meet 4t=80-t which means t is 16 for when they meet to verify plug in. 4*16 = 64 and 80-1*16 = 64 so good

jkkamau

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03 Jul 2025, 06:18

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Let s be speed of elevator...Alice speed = 2+s and Bob 3-s
4(80/2+s) = (80/3-s)
s=2
t= 80/(4+1)= 16
n= 80/4 =20x2= 40

Bunuel

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RedYellow

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03 Jul 2025, 06:44

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timeBob=4*timeAlice
80/speedBob=4*80/speedAlice
4*speedBob=speedAlice
4(3-s)=2+s
12-4s=2+s
s=2

speedBob=1 -> finalTimeBob=80
speedAlice=4 -> finalTimeAlice=20

The easiest way of calculate the meet time is to add the speeds: 1+4=5 and dividing 80 by this sum: 80/5=16

Alice walks at 2 steps per second relative to the scalator, so she walks 2*20=40 steps.

Correct answers are t=16 and n=40

Manu1995

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03 Jul 2025, 06:46

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Given:
Escalator length = 80 steps

Escalator speed = s steps/sec (upward)

Alice walks upward at 2 steps/sec relative to the escalator
⇒ Her net speed = 2+s

Bob walks downward at 3 steps/sec relative to the escalator
⇒ His net speed = 3−s

Bob takes 4 times as long to reach the bottom as Alice takes to reach the top

Let:

t = time in seconds after which Alice and Bob meet

T = time Alice takes to reach the top ==> T = 80/(2+s)

Then Bob takes 4T = 320/(2+s) to go from top to bottom
==> his speed = 3−s

So, Bob also travels 80 steps in 320/(2+s) seconds

==> 80 = (3-s)* 320/(2+s)

On solving for s:
s=2

Now Alice speed ==> 2+s = 4
Time to reach to the top = 80/4= 20sec
So T= 20
Bob takes 4× longer = 80 sec, going down at 3−2 =1 step/sec

Alice goes up at 4 steps/sec
Bob comes down at 1 step/sec
Relative speed = 4+1=5 steps/sec
They meet when they’ve together covered 80 steps

Therefore,time when they meet = 80/5= t= 16sec
Alice walks at 2 steps/sec relative to escalator in 20 sec
==>2 × 20 = n= 40 steps

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03 Jul 2025, 07:13

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Because Bob’s ride takes four times as long as Alice’s, the escalator must be helping both by 2 steps per second. That means Alice actually goes up 4 steps each second and Bob comes down 1 step each second. Starting 80 steps apart, they close the gap at 5 steps per second, so they meet after 16 seconds. Alice reaches the top in 20 seconds and, walking 2 steps every second on the moving escalator, she physically climbs 40 steps.

Bunuel

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Lemniscate

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03 Jul 2025, 07:30

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Speeds:
SB=3-s
SA=s+2

Comparing times:
TB=4*TA
80/SB=4*80/SA
4SB=SA
4*(3-s)=s+2
12-4*s=s+2
5*s=10
s=2

Bob reaches the botton in 80/1=80 seconds
Alice reaches the top in 80/4=20 seconds

Aggregate speed B y A = 1+4 = 5
80/5 = 16 -> time they meet

steps of Alice = time Alice reaches the top * relative speed of Alice = 20*2=40

Answers are t=16 and n=40

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Speed in steps/sec	Alice	Bob
Relative Speed	2	3
Effective Speed	2+s	3-s
Time taken	x	4x

They both covered the same distance, so
x(2+s)=4x(3-s)
2+s=12-4s
s=2

Effective speeds are 4 and 1step/sec for Alice and both
If they walk simultaneously their relative speed = 5
Time taken to meet = 80/5=16 secs

In 16secs, Alice covered = 16*4 = 64 steps

Bunuel

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03 Jul 2025, 07:36

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Alice's speed = 2 + S
Bob's speed = 3-s
Bob takes four times as long to reach the bottom as Alice takes to reach the top.
4(80/(2+s)) = 80/(3+s)
s = 2, Alice's speed = 4, Bob's speed = 1
t= 80/(4+1) = 16 seconds
to calculate n, time taken by Alice to reach the top= 80/4 = 20 seconds
20 * Alice's relative speed = 20 *2 = 40 steps

t = 16
n = 20

Bunuel

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UfuomaOh

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03 Jul 2025, 07:40

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Let Ta= time alice takes to reach the top
Tb= time bob takes to reach the bottom

Alice total speed = 2+s

Bobs total speed as he is moving against the escalator's constant speed=3-s

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

Hence Tb=4Ta

Meanwhile Ta= (80/2+s)

so Tb= 4*(80/2+s)

Solve for s using the expression Tb=4Ta

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

400s=800

s = 2

Ta= 20 seconds
Tb=80 seconds

The number of steps Alice would have walked on the escalator by the time she reaches the top.

Alice walks 2 steps per second and takes 20 seconds to complete the entire step in the escalator. Hence the total number of steps she would have worked to reach the top is

2 X 20= 40 steps.

n= 40

Select for t the time in seconds after which Alice and Bob meet

Alice net speed is 4

so she has climbed 4t steps from the bottom

Bob's net speed is 1

so he has descended 1 x t steps = t steps from the top, meaning that he is at 80-t steps from the bottom

so they meet where

4t=80 - t

t= 16

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03 Jul 2025, 07:46

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Let alice and bob be A and B respectively and s be speed of escalator. As per the statements provided, we can deduce that the speed of A is 2+s and B is 3-s

We are given that

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

Solving for s, we get s = 2

Therefore Speed of A = 4 and B = 1

Time taken for A and B to meet => 80/(4+1) = 16 seconds = t

Similarly, the number of steps Alice would have walked on the escalator by the time she reaches the top

time taken to reach the top by A => 80/4 = 20

In 20 seconds, A would have walked 20*2 = 40 steps.

Therefore, t = 16 and n = 40

GMAT1034

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03 Jul 2025, 07:52

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The time for meeting has to be calculated with s=D/T formula. speeds for Alice and Bob will be s+2, and 3-s respectively, it will come out to be 16. And total steps alice would have walked will be 80, as thats the total steps required to reach at the Top

Bunuel

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03 Jul 2025, 07:57

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Bunuel

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Total steps: 80 steps

Speed of escalator: s steps/sec.
For Alice:
Bottom to Top:
Rate: 2 steps/sec
Total rate: 2+s steps/sec

For Bob:
Top to Bottom:
Rate: 3 steps/sec
Total Rate: 3-s steps/sec

Bob is against the escalator and takes 4 times are long are Alice.
2+s= 4(3-s)
=> s=2 steps/sec

Alice takes 4 steps/sec and Bob takes 1 step/sec.

Since, Alice is moving from Bottom to Top,
After 16 seconds: She will cover 64 steps whereas Bob will cover 16 steps.
Hence, they'll meet after 16 seconds.

t=16 seconds

Number of steps Alice takes till she reaches the top: 80/4= 20steps.

ANSWER:
t=16
n=20