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# PS - Train meets

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Director
Joined: 25 Oct 2006
Posts: 635

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14 Mar 2009, 06:45
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If You're Not Living On The Edge, You're Taking Up Too Much Space

Kudos [?]: 637 [0], given: 6

SVP
Joined: 29 Aug 2007
Posts: 2472

Kudos [?]: 843 [0], given: 19

Re: PS - Train meets [#permalink]

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14 Mar 2009, 07:45
they each take z/(z/x + z/y) or xy/(x+y) hours to finish z miles.

miles covered by high speed train = (z/x) ((xy/(x+y)) = yz/(x+y)
miles covered by high regular train = (z/y) ((xy/(x+y)) = xz/(x+y)

diff = miles covered by high speed train - miles covered by regular train = yz/(x+y) - xz/(x+y) = z (y-x)/(x+y)
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Kudos [?]: 843 [0], given: 19

Intern
Joined: 25 Dec 2008
Posts: 18

Kudos [?]: 4 [0], given: 2

Schools: HBS, Stanford
Re: PS - Train meets [#permalink]

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14 Mar 2009, 08:37
I don't quite understand - could someone please post a bit more detailed explanation.

thanks

Kudos [?]: 4 [0], given: 2

Director
Joined: 25 Oct 2006
Posts: 635

Kudos [?]: 637 [0], given: 6

Re: PS - Train meets [#permalink]

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17 Mar 2009, 21:51
Ok, but question is asking the distance between first meet and second meet....!!
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If You're Not Living On The Edge, You're Taking Up Too Much Space

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Intern
Joined: 16 Jun 2008
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Location: Kiev, Ukraine
Re: PS - Train meets [#permalink]

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19 Mar 2009, 01:37
More detailed explanation for DaveGG

If two trains are travelling from the opposite directions towards each other, then time which they travel before the meeting is calculated by a formula:

t=S/(V1+V2), where S - total distance from A to B, V1 and V2 are the speeds of the trains respectively.

In our example, S=z, V1=z/x, V2=z/y. So time of meeting is:
t=z/(z/x+z/y). After some simplifications, we get t=xy/(x+y).
Before the meeting 1-st train has travelled such distance: S1=t*V1=xy/(x+y)*z/x
Before the meeting 2-nd train has travelled such distance: S2=t*V2=xy/(x+y)*z/y

The difference between the 1-st and the 2-nd distance is equal to:
S1-S2=xy/(x+y)*z/x - xy/(x+y)*z/y = z (y-x)/(x+y).

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Re: PS - Train meets   [#permalink] 19 Mar 2009, 01:37
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