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A building with 600 floors has two elevators. The floors

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A building with 600 floors has two elevators. The floors [#permalink] New post 12 Dec 2006, 00:02
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A building with 600 floors has two elevators. The floors are evenly spaced, such that an elevator must travel 40 feet to reach an adjacent floor. Elvator A travels at a rate of 10 feet per second, whereas elevator B travels at a rate of 20 feet per second. If elevator A begins going down from the 450th floor at the same time that elevator B begins traveling up from the 5th floor, at what floor will the two elevators pass each other, assuming neither makes any stops?

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 [#permalink] New post 12 Dec 2006, 01:02
A travels 1 floor in 4 seconds. = 1/4
B travels 1 floor in 2 seconds. = 1/2

Common rate=1/4+1/2=3/4

There are 450-5+1=446 floors between them
The travel time of this distance assuming the common rate:
3/4*t=446
t=592 seconds (approximately)

So A will travel:
1/4*592=d
d=148 floors

B will travel:
1/2*592=d
d=296 floors

They will meet at:
A: 450-148=302
B: 5+296=301

They will pass each other at approximately 302nd floor....
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 [#permalink] New post 12 Dec 2006, 13:52
SimaQ wrote:

There are 450-5+1=446 floors between them
The travel time of this distance assuming the common rate:
3/4*t=446
t=592 seconds (approximately)



I think the part highlighted is wrong because if A was on the 10th floor and B was on the 5th floor then the amount of distance between them is 5*40
If you draw it:

10th---40ft----9th---40ft---8th---40ft---7th---40ft---6th---40ft---5th

Now you can see there are five lots of 40ft inbetween the 10th and 5th floor so if we want to calculate the distance in feet it should be (10-5)*40.

Furthermore, if you used the original rationale then the no. of floors worth of distance between the 10th and 9th floor would be (10-9+1)*40 = 80 ---- when the actual distance is 40
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 [#permalink] New post 12 Dec 2006, 15:44
Both traveling opp direction
relative speed = (10/40)+(20/40) floor per sec= 3/4 floor per sec

Distance to be traveled 450-5 = 445 floor

3/4 in 1 sec
445 in 445/(3/4)= 593 sec

Floor by B= 593*1/2=296

B was already at 5

Now its at 296+5= 301 floor

B 593/4 = 148.25 450-148.25 = 301.5

b/w 301 and 302 answer
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 [#permalink] New post 12 Dec 2006, 23:51
MBAlad wrote:
SimaQ wrote:

There are 450-5+1=446 floors between them
The travel time of this distance assuming the common rate:
3/4*t=446
t=592 seconds (approximately)



I think the part highlighted is wrong because if A was on the 10th floor and B was on the 5th floor then the amount of distance between them is 5*40
If you draw it:

10th---40ft----9th---40ft---8th---40ft---7th---40ft---6th---40ft---5th

Now you can see there are five lots of 40ft inbetween the 10th and 5th floor so if we want to calculate the distance in feet it should be (10-5)*40.

Furthermore, if you used the original rationale then the no. of floors worth of distance between the 10th and 9th floor would be (10-9+1)*40 = 80 ---- when the actual distance is 40


I converted everything into floors in order not to mess with feet.... That way it is easier to calculate....

You would have been right if i used the rate of A and B in feet rather than floors.....
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 [#permalink] New post 13 Dec 2006, 07:47
Distance between 450th and 5th floor = (445 -1) * 40ft = 444 * 40
Both the lifts together will cover this distance in 444 * 40 / (20 + 10) sec = 148 * 4 sec
B traveled = 148 * 4 * 20 ft
Number of floors B traveled = 148 * 4 * 20 /40 = 296
They will meet = 5 + 296 + 1 = 302

(Consider the 1st floor that is not at 40ft, the distance between 1st and 2nd flr is 40ft. If a lift travels 40ft, it will travel from 1st flr to 2nd flr)
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 [#permalink] New post 13 Dec 2006, 10:41
SimaQ wrote:
MBAlad wrote:
SimaQ wrote:

There are 450-5+1=446 floors between them
The travel time of this distance assuming the common rate:
3/4*t=446
t=592 seconds (approximately)



I think the part highlighted is wrong because if A was on the 10th floor and B was on the 5th floor then the amount of distance between them is 5*40
If you draw it:

Even if you're talking about floors then there are
10th---40ft----9th---40ft---8th---40ft---7th---40ft---6th---40ft---5th

Now you can see there are five lots of 40ft inbetween the 10th and 5th floor so if we want to calculate the distance in feet it should be (10-5)*40.

Furthermore, if you used the original rationale then the no. of floors worth of distance between the 10th and 9th floor would be (10-9+1)*40 = 80 ---- when the actual distance is 40


I converted everything into floors in order not to mess with feet.... That way it is easier to calculate....

You would have been right if i used the rate of A and B in feet rather than floors.....


I'm still not sure - between 10th & 5th there are 4 floors between them (9,8,7,6) or 5 floors worth of space (200ft). Using your calc it would give (10-5)+1 = 6 floors?

10th---40ft----9th---40ft---8th---40ft---7th---40ft---6th---40ft---5th
  [#permalink] 13 Dec 2006, 10:41
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