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# Trains

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VP
Joined: 18 May 2008
Posts: 1300
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Trains [#permalink]  02 Apr 2009, 22:14
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Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?

A. z(y – x)/x + y

B. z(x – y)/x + y

C. z(x + y)/y – x

D. xy(x – y)/x + y

E. xy(y – x)/x + y

I followed this approach: Since trains are approaching each other, relative speed =x+y
Let t be the time taken then t= z/(x+y)

Now distance travelled by high speed train = t*x=zx/(x+y)

distance travelled by average speed train = t*y=zy/(x+y)

Hence difference between both = z(x-y)/(x+y)

But the OA is z(y-x)/(x+y)

What is wrong in my approach?
Intern
Joined: 03 Apr 2009
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Re: Trains [#permalink]  03 Apr 2009, 03:25
Your t isn't correct. x and y are not the speeds of the trains. Hint: The speed of the high-speed train is z/x.
VP
Joined: 18 May 2008
Posts: 1300
Followers: 13

Kudos [?]: 171 [0], given: 0

Re: Trains [#permalink]  03 Apr 2009, 05:20
Oh my god! In a hurry, i misread vthe time as speed. Bad day!
Raison wrote:
Your t isn't correct. x and y are not the speeds of the trains. Hint: The speed of the high-speed train is z/x.
Intern
Joined: 03 Apr 2009
Posts: 11
Followers: 0

Kudos [?]: 8 [0], given: 0

Re: Trains [#permalink]  03 Apr 2009, 15:20
I figured that's what happened. It happens to all of us at times.
Re: Trains   [#permalink] 03 Apr 2009, 15:20
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