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At 1:00 PM, Train X departed from Station A on the road to S

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At 1:00 PM, Train X departed from Station A on the road to S [#permalink] New post 27 Aug 2018, 09:09
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Question Stats:

37% (02:45) correct 62% (02:43) wrong based on 24 sessions
At 1:00 PM, Train X departed from Station A on the road to Station B. At 1:30 PM, Train Y departed Station B on the same road for Station A. If Station A and Station B are p miles apart, Train X’s speed is r miles per hour, and Train Y’s speed is s miles per hour, how many hours after 1:00 PM, in terms of p, r, and s, do the two trains pass each other?

A. \(0.5\) + \(\frac{(p - 0.5s)}{(r + s)}\)

B. \(\frac{(p - 0.5s)}{(r + s)}\)

C. \(0.5\) + \(\frac{(p - 0.5r)}{r}\)

D. \(\frac{(p - 0.5r)}{(r + s)}\)

E. \(0.5\) + \(\frac{(p - 0.5r)}{(r + s)}\)
[Reveal] Spoiler: OA

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Re: At 1:00 PM, Train X departed from Station A on the road to S [#permalink] New post 10 Sep 2018, 12:42
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say t hours is the time X takes.

then Y takes t-0.5 hours.

X goes tr miles and Y goes (t-0.5)s miles.

so tr+(t-0.5)s=p => t=(p+0.5s)/(r+s) which is written in another form in option E.


p.s.:one can very well start by taking t hours to be the time taken by Y. then X would take t+0.5 hours and in the end, this would be the answer.
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Re: At 1:00 PM, Train X departed from Station A on the road to S [#permalink] New post 13 Sep 2018, 11:31
Any detailed answers?
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Re: At 1:00 PM, Train X departed from Station A on the road to S [#permalink] New post 13 Sep 2018, 11:47
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I do think the explanation above is well perfect.

Simple and straight.

:?:
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Re: At 1:00 PM, Train X departed from Station A on the road to S   [#permalink] 13 Sep 2018, 11:47
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At 1:00 PM, Train X departed from Station A on the road to S

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