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Answer C...

Steve travels at a rate of 57 floors/60s~1floor/1s

Joyce travels at a rate of 63 floors/60s~1floor/1s

Between them there are approximatelly 40 floors, so they will both have to travel ~ 20 floors to cross each other.... 20 floors is ~ 20 s

We need to use this formula.... Rate*Time=Distance

Steve: 57/60*20=19

Joyce: 63/60*20=21....

Adding things up we get 30......
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yeap its 30 i kinda figured it out ina different way little later because it was bothering me...

but its pretty much the same :)
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Re: Steve gets on the elevator at the 11th floor of a building a [#permalink]
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Distance between Steve & Joyce = 40 floors

Lets say they meet on x floor

Time required for Joyce to reach x floor = Time required for Steve

So the equation would be

\(\frac{x}{57} = \frac{40-x}{63}\)

x = 19

Start is at 11th floor, so 11 + 19 = 30 = Answer = C
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Re: Steve gets on the elevator at the 11th floor of a building a [#permalink]
Steve gets on the elevator at the 11th floor of a building and rides up at a rate of 57 floors per minute. At the same time Joyce gets on an elevator on the 51st floor of the same building and rides down at a rate of 63 floors per minute. If they continue traveling at these rates, at which floor will their paths cross?

40 floors/120 floors per minute=1/3 minutes
11+57/3=30
51-63/3=30
C. 30
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Re: Steve gets on the elevator at the 11th floor of a building a [#permalink]
Here is an equation to solve the problem.


Let time it will take for that to happen be x.

Riding up will be.


11 + 57x


Riding down.

51 - 63x


Therefore

11 + 57x = 51 - 63x

57x + 63x = 51 - 11

120x = 40

x = 1/3.

Sub for x in any of the Expressions gives 30.

Posted from my mobile device
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Re: Steve gets on the elevator at the 11th floor of a building a [#permalink]
KarishmaB wrote:
AshikaP wrote:
Steve gets on the elevator at the 11th floor of a building and rides up at a rate of 57 floors per minute. At the same time Joyce gets on an elevator on the 51st floor of the same building and rides down at a rate of 63 floors per minute. If they continue traveling at these rates, at which floor will their paths cross?

A. 19
B. 28
C. 30
D. 32
E. 44


Even though this question uses elevators (which usually means a tougher question), it involves very simple relative speed calculation.

Steve is on 11th floor and Joyce on 51st floor so they have a distance of 40 floors between them.
Steve's speed is 57 floors/min and Joyce's speed is 63 floors/min in opposite direction (towards Steve).
Their relative speed will be the sum of their individual speeds = 57 + 63 = 120 floors/min

Time taken to meet = 40/120 = 1/3 min = 20 secs

Distance covered by Steve in 20 secs = Speed*Time = 57 * (1/3) = 19 floors.
So he is at 11+19 = 30th floor when they meet.

Answer (C)

Check this post on relative speed:
https://www.gmatclub.com/forum/veritas- ... elatively/
https://www.gmatclub.com/forum/veritas- ... -concepts/


Dear KarishmaB

Why the distance between them is 40 floors? Shouldn't the 11th and the 51st floor be included in that count and in that case the number of floors would be (51-11)+1= 41 floors, or maybe they should be excluded and in that case the number of floors between them would be (51-11)-1= 39?

Thank you in advance for your feedback :)
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Re: Steve gets on the elevator at the 11th floor of a building a [#permalink]
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GianKR wrote:
KarishmaB wrote:
AshikaP wrote:
Steve gets on the elevator at the 11th floor of a building and rides up at a rate of 57 floors per minute. At the same time Joyce gets on an elevator on the 51st floor of the same building and rides down at a rate of 63 floors per minute. If they continue traveling at these rates, at which floor will their paths cross?

A. 19
B. 28
C. 30
D. 32
E. 44


Even though this question uses elevators (which usually means a tougher question), it involves very simple relative speed calculation.

Steve is on 11th floor and Joyce on 51st floor so they have a distance of 40 floors between them.
Steve's speed is 57 floors/min and Joyce's speed is 63 floors/min in opposite direction (towards Steve).
Their relative speed will be the sum of their individual speeds = 57 + 63 = 120 floors/min

Time taken to meet = 40/120 = 1/3 min = 20 secs

Distance covered by Steve in 20 secs = Speed*Time = 57 * (1/3) = 19 floors.
So he is at 11+19 = 30th floor when they meet.

Answer (C)

Check this post on relative speed:
https://www.gmatclub.com/forum/veritas- ... elatively/
https://www.gmatclub.com/forum/veritas- ... -concepts/


Dear KarishmaB

Why the distance between them is 40 floors? Shouldn't the 11th and the 51st floor be included in that count and in that case the number of floors would be (51-11)+1= 41 floors, or maybe they should be excluded and in that case the number of floors between them would be (51-11)-1= 39?

Thank you in advance for your feedback :)


Take a simpler example. Say you are on the 3rd floor of a building and have to go to the 5th floor. How many floors do you have to climb?
From 3rd to 4th and then from 4th to 5th i.e. 2 floors.
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Re: Steve gets on the elevator at the 11th floor of a building a [#permalink]
Steve gets on the elevator at the 11th floor of a building and rides up at a rate of 57 floors per minute. At the same time Joyce gets on an elevator on the 51st floor of the same building and rides down at a rate of 63 floors per minute. If they continue traveling at these rates, at which floor will their paths cross?

This can be solved by the relative velocity concept
The difference in the floors = 51-11=40
Relative velocity=57+63=120 floors per minute

Time taken to cross each other = Difference of floors/relative velocity = 40/120 = 1/3 minute

Therefore the floor they will cross their path can be given by
=> 11+57*(1/3) = 11+19=30

Hence C
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Re: Steve gets on the elevator at the 11th floor of a building a [#permalink]
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