Sam gets on the elevator at the 11th floor of a building and

[imane81]

Please help on this one, thank you


Sam gets on the elevator at the 11th floor of a building and rides up at a rate of 57 floors per min. At the same time, Jim gets on another elevator on the 51st floor of the same building and rides down at a rate of 63 floors per min. If they continue traveling at these rates, ,at which floor will their paths cross?
A. 19
B. 28
C. 30
D. 32
E. 44

[shashank.ism]

Please help on this one, thank you


Sam gets on the elevator at the 11th floor of a building and rides up at a rate of 57 floors per min. At the same time, Jim gets on another elevator on the 51st floor of the same building and rides down at a rate of 63 floors per min. If they continue traveling at these rates, ,at which floor will their paths cross?
A. 19
B. 28
C. 30
D. 32
E. 44

Let at Xth floor their paths cross.
so time taken by Sam= (X-11)/57
time taken by Jim = (51-X)/63

so (X-11)/57 = (51-X)/63 --> 63X-693 = 2907-57X --> 120X= 3600 --> X = 30.
Hence they meet at 30th floor Ans C

[harsh.champ]

Please help on this one, thank you


Sam gets on the elevator at the 11th floor of a building and rides up at a rate of 57 floors per min. At the same time, Jim gets on another elevator on the 51st floor of the same building and rides down at a rate of 63 floors per min. If they continue traveling at these rates, ,at which floor will their paths cross?
A. 19
B. 28
C. 30
D. 32
E. 44

Over here,it would be beneficial if we use the concept of relative velocity.
Now,relative distance between Sam and Jim = 40 floors.
Their relative velocity = 120 floors/min.
Hence,the distance will be covered in 40/120 = 1/3min
In 1/3 min. Sam's elevator would have come to [spoiler]11 + 1/3*57 = 11 + 19 = 30 floors[/spoiler]
This way,it takes only 1 minute to solve the question.

Ans would be C definitely.

[becnil]

It may be easier by way of Relative Speed. They have (51-11) 40 floors between them and their relative speed is (63+57) 120 floors per minute.

So, they will cross after 40/120 minutes, i.e. 20 seconds. In 20 sec, Jim rides down 63/3=21 floors, meaning they will cross at the (51-21) 30th floor.

[harsh.champ]

It may be easier by way of Relative Speed. They have (51-11) 40 floors between them and their relative speed is (63+57) 120 floors per minute.

So, they will cross after 40/120 minutes, i.e. 20 seconds. In 20 sec, Jim rides down 63/3=21 floors, meaning they will cross at the (51-21) 30th floor.

Hey becnil,
You posted the same solution as mine above.
I was doubtful whether such a question can be asked on GMAT because they are not testing your physics skills in the exams.
Or is relative velocity a very common concept??

Coming from an engg background I don't know how a layman thinks??

Can someone throw some light in this regard !!

[shashank.ism]

Please help on this one, thank you


Sam gets on the elevator at the 11th floor of a building and rides up at a rate of 57 floors per min. At the same time, Jim gets on another elevator on the 51st floor of the same building and rides down at a rate of 63 floors per min. If they continue traveling at these rates, ,at which floor will their paths cross?
A. 19
B. 28
C. 30
D. 32
E. 44

Over here,it would be beneficial if we use the concept of relative velocity.
Now,relative distance between Sam and Jim = 40 floors.
Their relative velocity = 120 floors/min.
Hence,the distance will be covered in 40/120 = 1/3min
In 1/3 min. Sam's elevator would have come to [spoiler]11 + 1/3*57 = 11 + 19 = 30 floors[/spoiler]
This way,it takes only 1 minute to solve the question.

Ans would be C definitely.

Harsh I think your have taken a right approach to solve the problem. But I think the way I have solved the problem will also take less than a minute to solve this problem.

[Stuart@KaplanGMAT]

It may be easier by way of Relative Speed. They have (51-11) 40 floors between them and their relative speed is (63+57) 120 floors per minute.

So, they will cross after 40/120 minutes, i.e. 20 seconds. In 20 sec, Jim rides down 63/3=21 floors, meaning they will cross at the (51-21) 30th floor.

Hey becnil,
You posted the same solution as mine above.
I was doubtful whether such a question can be asked on GMAT because they are not testing your physics skills in the exams.
Or is relative velocity a very common concept??

Coming from an engg background I don't know how a layman thinks??

Can someone throw some light in this regard !!

We're not talking relative velocity using relativity theory - it's simply a combined rate problem.

Here's the general rule: when two objects are travelling in opposite directions (i.e. directly toward each other or directly away from each other), you ADD their rates to get the combined rate.

In this question, one person is moving up and the other down, so the rule applies and we simply apply the distance/rate/time formula:

Total distance = total rate * total time

40 = 120 * t
1/3 = t

We can now use either person to answer the question.

If we choose Sam, in 1/3 of a minute he travels 1/3 * 57 = 19 floors. 11th + 19 = 30th floor.

If we choose Jim, in 1/3 of a minute he travels 1/3 * 63 = 21 floors. 51st - 21 = 30th floor.

* * *

We can also solve this question with weighted averages.

The combined rate is 120 fpm.

Sam covers 57/120 of the distance and Jim covers 63/120 of the distance.

If we solve with Sam, we get:

(57/120) * 40 floors = 57/3 = 19 floors. 11th + 19 = 30th floor.

If we solve with Jim, we get:

(63/120) * 40 floors = 63/3 = 21 floors. 51st - 21 = 30th floor.

[shashank.ism]

We can also solve this question with weighted averages.

The combined rate is 120 fpm.

Sam covers 57/120 of the distance and Jim covers 63/120 of the distance.

If we solve with Sam, we get:

(57/120) * 40 floors = 57/3 = 19 floors. 11th + 19 = 30th floor.

If we solve with Jim, we get:

(63/120) * 40 floors = 63/3 = 21 floors. 51st - 21 = 30th floor.

Stuart thanks for posting in this post. The second method you have given is really appreciable and fast..It is just a matter of few second in which we can solve this problem by your 2nd method and it adds a new dimension to solve this type of problem by combined rate.