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# Train T leaves from town A for town B and travels at a

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Senior Manager
Joined: 22 Sep 2005
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Train T leaves from town A for town B and travels at a [#permalink]  15 Oct 2006, 11:59
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Train T leaves from town A for town B and travels at a constant speed of x mph. Train S leaves town B for town A, later than the train T, and also travels at a constant speed of y mph. Will they meet closer to A or to B?

1) y>x
2) They finished their trips at the same time
SVP
Joined: 05 Jul 2006
Posts: 1519
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Kudos [?]: 111 [0], given: 39

Train T leaves from town A for town B and travels at a constant speed of x mph. Train S leaves town B for town A, later than the train T, and also travels at a constant speed of y mph. Will they meet closer to A or to B?

1) y>x
2) They finished their trips at the same time

ONE

INSUFF ...WE NEED TO KNOW THE TIME LAPSED BETWEEN BOTH TRAINS LEAVING THE STATIONS
2) IF THEY REACHED THE SAME DITINATION AT THE SAME TIME

THEN

THEY MET CLOSER TO A

Senior Manager
Joined: 05 Oct 2006
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Re: DS: Train [#permalink]  15 Oct 2006, 22:17
u have to know how much time later train s started in order to know where they meet...

choice 1...insufficient.
choice 2... if they finished at the same time ..means y>x...nothing can be said when they met..???..insufficient..

choice E as per me..

what's the OA???
Manager
Joined: 03 Jul 2005
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Train T leaves from town A for town B and travels at a constant speed of x mph. Train S leaves town B for town A, later than the train T, and also travels at a constant speed of y mph. Will they meet closer to A or to B?

1) y>x
2) They finished their trips at the same time

I would say E as well

Since we don't know how much later train S leaves than train T.

What is the OA?
Senior Manager
Joined: 22 Sep 2005
Posts: 279
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I haven't got OA, but in my opinion is E too.

1) NS, because although y>x, and train S leaves after train T, we can't know where they are going to meet.

2) NS. If they finished their trips at the same time, we deduce that y>x, and we knew that is NS.

1) and 2) are NS, because both give the same information.

So in my opinion the answer is E.
Manager
Joined: 12 Sep 2006
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(1) is insufficient for obvious reasons

(2) is sufficient, best way to see it is by drawing it on paper. We know that both trains finish their trips at the same time, so we know y>x

If we cut the distance in half, train S will be in the middle and, regardless of how much slower train T is, it will always be a little bit closer to B, therefore they meet closer to town B
Intern
Joined: 28 Jun 2006
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I agree with E.

By using both statements, I can think of situations which produce both answers.

Since we don't know how long train S waits to leave or how fast either of them go, train S could leave either before S gets to the halfway point, or after train S gets to the halfway point.

If train S waits to leave until train T has reached the halfway point, then they meet closer to B. If train S leaves only shortly after train T and travels much faster, then train S could pass the midway point of the track before train T does.

Sorry if the post is a little convoluted.
Director
Joined: 23 Jun 2005
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Kudos [?]: 29 [0], given: 1

B

If train S leaves later than T, but finishes at the same time, then T is traveling faster than S.
So, when they meet, the distance between the meeting point and B should be smaller than the distance between the meeting point and A. So, statement 2 is SUFF
Manager
Joined: 31 Aug 2006
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Kudos [?]: 2 [0], given: 0

anandsebastin wrote:
B

If train S leaves later than T, but finishes at the same time, then T is traveling faster than S.
So, when they meet, the distance between the meeting point and B should be smaller than the distance between the meeting point and A. So, statement 2 is SUFF

IF If train S leaves later than T, but finishes at the same time, then S is traveling faster than T and not vice versa..Can you pls. explain in little detail that how is it (B) ..and what is the OA ?
Manager
Joined: 07 Jun 2006
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Kudos [?]: 7 [0], given: 0

Code:
If train S leaves later than T, but finishes at the same time, then T is traveling faster than S.
So, when they meet, the distance between the meeting point and B should be smaller than the distance between the meeting point and A. So, statement 2 is SUFF

Anand, What if the difference in rates is negligible and the difference in time is huge???

Without actual numbers, its hard to decide on the meeting point. It can fall in either of the halves, depending on the difference in rates and difference in time.

Correct me if this line of thinking is wrong.
Director
Joined: 01 Oct 2006
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I guess B2 for the same reason as Anand
Manager
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two extreme examples
d=100
trains S =10, then T=10h
Train T :y=100 then T=1h

they meet closer to B
or
d=120
x=10, then T=12h
y=10,1 then T=11,88h

they meet closer to B
since right after T=6h
S is closer to B
and
T has not crossed the mid distance yet since it left 0,12h late... and it will cross the mid point at 0.12h+ 11,88/2=5,94+,12=6,16h
hope it helps
_________________

time is not on my side

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