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Two trains cross each other in time T, when they are

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Two trains cross each other in time T, when they are [#permalink] New post 05 Jan 2006, 03:25
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Two trains cross each other in time T, when they are travelling in opposite directions. The faster train completely overtakes the slower train in time t if they are travelling in the same direction. If the trains are equal length, how long will the faster train take to completely overtake the second train, when the first one is in rest?

a) Tt/2(T+t)
b) (T-t)/(T+t)
c) (T+t)
d) 2Tt/(T+t)

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 [#permalink] New post 05 Jan 2006, 04:50
D)

s1 = speed of faster
s2 = speed of slower
L = length of the train

When trains are crossing in opp directions:
It is exactly same as a situation where, the faster train were travelling at at a speed s1+s2 and the slower would be stopped. (Please ignore grammar)

The distance it travels is length of the train (L)
L = (s1+s2)*T ---(1)

When the trains are travelling in the same direction:
It is exactly same as a situation where, the faster train were travelling at at a speed s1-s2 and the slower would be stopped. (Same distance L)
L = (s1-s2)*t --- (2)

(1)&(2) give:

(s1+s2)*T = (s1-s2)*t

or s1/s2 = (T+t)/(t-T) --(3)

Finally, when the slower train is stopped, the faster train has to travel the same distance L.

Time = L/s1
Using L from (1)
= (s1+s2)*T/s1
= (1+s2/s1)*T
Using the inverse of (3) here gives
= 2*t*T/(T+t)


Hope this is correct and these kind of problems dont appear in GMAT :( :?: :twisted:
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 [#permalink] New post 05 Jan 2006, 17:45
Good Question , keep them coming. Let us know the source.

Ans D

Let
L be the length of the Train
S be the speed of the slow Train
F be the speed of the fast Train


We have :

1. 2L/(F+S) = T

2. 2L/(F-S) = t

solving 1 & 2

F = L(t+T)/tT

we need to find 2L/F

= 2tT/T+t
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 [#permalink] New post 06 Jan 2006, 03:45
hkm_gmat wrote:
Good Question , keep them coming. Let us know the source.

Ans D

Let
L be the length of the Train
S be the speed of the slow Train
F be the speed of the fast Train


We have :

1. 2L/(F+S) = T

2. 2L/(F-S) = t

solving 1 & 2

F = L(t+T)/tT

we need to find 2L/F

= 2tT/T+t


It aren't two "lenghts" but one length that is passed by a train with speed s1+s2. In case this is unclear: Imagine two trains cross each other; they actually don't pass two lengths but one in a faster time.

Maybe this hint is useless or idle, but maybe it is helpful for other questions.
However in this case it doesn't matter since you use two lenghts in both statement 1 and 2.
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 [#permalink] New post 06 Jan 2006, 14:05
I think the two equations are not right. please see the red part...

giddi77 wrote:
D)

s1 = speed of faster
s2 = speed of slower
L = length of the train

When trains are crossing in opp directions:
It is exactly same as a situation where, the faster train were travelling at at a speed s1+s2 and the slower would be stopped. (Please ignore grammar)

The distance it travels is length of the train (L)
L = (s1+s2)*T ---(1) it should be 2L
When the trains are travelling in the same direction:
It is exactly same as a situation where, the faster train were travelling at at a speed s1-s2 and the slower would be stopped. (Same distance L)
L = (s1-s2)*t --- (2)

(1)&(2) give:

(s1+s2)*T = (s1-s2)*t

or s1/s2 = (T+t)/(t-T) --(3)

Finally, when the slower train is stopped, the faster train has to travel the same distance L.

Time = L/s1
Using L from (1)
= (s1+s2)*T/s1
= (1+s2/s1)*T
Using the inverse of (3) here gives
= 2*t*T/(T+t)


Hope this is correct and these kind of problems dont appear in GMAT :( :?: :twisted:
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 [#permalink] New post 06 Jan 2006, 14:08
The two equations are not right in my opinion. please see the red part...

hkm_gmat wrote:
Good Question , keep them coming. Let us know the source.

Ans D

Let
L be the length of the Train
S be the speed of the slow Train
F be the speed of the fast Train


We have :

1. 2L/(F+S) = T

2. 2L/(F-S) = t it should be L
solving 1 & 2

F = L(t+T)/tT

we need to find 2L/F

= 2tT/T+t
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Re: PS - Relative velocity [#permalink] New post 06 Jan 2006, 14:16
From my working, the time it takes to completely overtake the second train, when the first one is in rest =2LTt/(2Lt+LT)


THE ANOTATIONS ARE SAME AS GIDDI77 HKM_GMAT!

by the way what is OA/OE?


krisrini wrote:
Two trains cross each other in time T, when they are travelling in opposite directions. The faster train completely overtakes the slower train in time t if they are travelling in the same direction. If the trains are equal length, how long will the faster train take to completely overtake the second train, when the first one is in rest?

a) Tt/2(T+t)
b) (T-t)/(T+t)
c) (T+t)
d) 2Tt/(T+t)

Thanks
Sreeni
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 [#permalink] New post 06 Jan 2006, 14:25
Sunshine,

allabout is right.

Either you have to take all three variables as 2L or L.

I just assumed the crossing point as the front of the first train to end of the same train. In this case the length is L

If you assume the crossing point as complete length of 2 trains the it should be 2L in all of the equations.

HTH.
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 [#permalink] New post 06 Jan 2006, 14:33
Thanks, giddi77, Now it is clear.....

In general the lenght is always L+L either travelling in one direction or opposite.
But the speed is S1+S2 when travelling in opposite direction and S1-S2 0r S2-S1 when travelling in the same direction.......

please correct me if wrong!


giddi77 wrote:
Sunshine,

allabout is right.

Either you have to take all three variables as 2L or L.

I just assumed the crossing point as the front of the first train to end of the same train. In this case the length is L

If you assume the crossing point as complete length of 2 trains the it should be 2L in all of the equations.

HTH.
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 [#permalink] New post 06 Jan 2006, 17:41
allabout wrote:
hkm_gmat wrote:
Good Question , keep them coming. Let us know the source.

Ans D

Let
L be the length of the Train
S be the speed of the slow Train
F be the speed of the fast Train


We have :

1. 2L/(F+S) = T

2. 2L/(F-S) = t

solving 1 & 2

F = L(t+T)/tT

we need to find 2L/F

= 2tT/T+t


It aren't two "lenghts" but one length that is passed by a train with speed s1+s2. In case this is unclear: Imagine two trains cross each other; they actually don't pass two lengths but one in a faster time.

Maybe this hint is useless or idle, but maybe it is helpful for other questions.
However in this case it doesn't matter since you use two lenghts in both statement 1 and 2.


Since the Trains completely pass each other the distance ought to be two times length of each train or 2L .
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 [#permalink] New post 06 Jan 2006, 18:39
hkm_gmat wrote:
allabout wrote:
hkm_gmat wrote:
Good Question , keep them coming. Let us know the source.

Ans D

Let
L be the length of the Train
S be the speed of the slow Train
F be the speed of the fast Train


We have :

1. 2L/(F+S) = T

2. 2L/(F-S) = t

solving 1 & 2

F = L(t+T)/tT

we need to find 2L/F

= 2tT/T+t


It aren't two "lenghts" but one length that is passed by a train with speed s1+s2. In case this is unclear: Imagine two trains cross each other; they actually don't pass two lengths but one in a faster time.

Maybe this hint is useless or idle, but maybe it is helpful for other questions.
However in this case it doesn't matter since you use two lenghts in both statement 1 and 2.


Since the Trains completely pass each other the distance ought to be two times length of each train or 2L .


Sorry hkm_gmat, I completely missed your post. You are right, 2L can also be taken if we assume that the trian cross completely.

IMO, the problem doesn't depend on whether you consider the length as L or 2L. In all three cases the relative speed/velocity is what is varying.
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Re: PS - Relative velocity [#permalink] New post 08 Jan 2006, 20:49
Very good discussion on this post. The Oa is 2Tt/(T+t). NO OE was given.Thanks for all your replies.

regards
Re: PS - Relative velocity   [#permalink] 08 Jan 2006, 20:49
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