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

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At 1:00 PM, Train X departed from Station A on the road to [#permalink]

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New post 04 Oct 2010, 16:14
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Question Stats:

47% (02:44) correct 53% (02:23) wrong based on 213 sessions

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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 + (p - 0.5s)/(r + s)
B. (p - 0.5s)/(r + s)
C. 0.5 + (p - 0.5r)/r
D. (p - 0.5r)/(r + s)
E. 0.5 + (p - 0.5r)/(r + s)
[Reveal] Spoiler: OA
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Re: Rates Question [#permalink]

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New post 04 Oct 2010, 16:41
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krazo wrote:
Here is a tough rate question that took me some time to figure out the algebra for... give it a shot!
Quote:
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?


Quote:
    A. 0.5 + (p - 0.5s)/(r + s)
    B. (p - 0.5s)/(r + s)
    C. 0.5 + (p - 0.5r)/r
    D. (p - 0.5r)/(r + s)
    E. 0.5 + (p - 0.5r)/(r + s)


Rate of X - \(r\) miles per hour;
Rate of Y - \(s\) miles per hour;
Combined rate \(s+r\) miles per hour;
Distance between the stations \(p\) miles;

In 1/2 hours that X traveled alone it covered \(\frac{1}{2}*r\) miles, so together trains should cover \(p-\frac{1}{2}r=\frac{2p-r}{2}\) miles, which they will cover in \(\frac{\frac{2p-r}{2}}{r+s}=\frac{2p-r}{2(r+s)}\) hours.

Total time after 1:00 PM till they meet would be \(\frac{1}{2}+\frac{2p-r}{2(r+s)}=\frac{1}{2}+\frac{p-0.5r}{r+s}\) hours.

Answer: E.
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Re: Rates Question [#permalink]

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New post 04 Oct 2010, 17:30
The distance A is going to cover between 1:00 and 1:30
= .5r
now the distance between the two trains = (p-.5r)
the relative velocity = (r-(-s)) = r+s

From 1:30, time is going to take when they meet = (p-.5r)/(r+s)

so the ans is .5+((p-.5r)/(r+s)) [.5 is added for the time from 1:00 to 1:30]

ans is E
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Re: At 1:00 PM, Train X departed from Station A on the road to [#permalink]

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New post 16 Sep 2014, 20:23
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Hello Bunnel,

This definitely is a good procedure to solve this problem. But, may I know if this can be solved using picking up numbers or any easier method? IF yes, could you please share.

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Re: At 1:00 PM, Train X departed from Station A on the road to [#permalink]

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New post 16 Sep 2014, 23:16
Expert's post
krazo wrote:
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 + (p - 0.5s)/(r + s)
B. (p - 0.5s)/(r + s)
C. 0.5 + (p - 0.5r)/r
D. (p - 0.5r)/(r + s)
E. 0.5 + (p - 0.5r)/(r + s)



You can do it by plugging in numbers though with so many variables, it is hard to keep track of values for each.

Preferable here would be algebra (use relative speed concepts):

Time taken to meet starting from 1:30 \(= \frac{Total Distance}{Total Speed} = \frac{p - r/2}{r + s}\)
Note that since X covers first half hour alone, it covers r*0.5 distance alone so distance between the two trains at 1:30 is (p - r/2).

But we need the time taken from 1 onwards so time taken \(= 0.5 + \frac{p - r/2}{r + s}\)

Here is a post that discusses relative speed (including a similar trains example): http://www.veritasprep.com/blog/2012/07 ... elatively/

Solving using Plugging in:

Say p = 100, r = 100, s = 50.
X runs for half an hour and covers 50 miles in that time. So now X and Y are 50 miles apart.
Total time taken to cover 50 miles = 50/(100+50) = 1/3 hr
Total time taken to cover 100 miles = 1/2 + 1/3 = 5/6

Now put these values in the options. Remember, there are symmetrical options in r and s so you need to take different values for r and s.
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Re: At 1:00 PM, Train X departed from Station A on the road to [#permalink]

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New post 18 Dec 2015, 08:03
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Re: At 1:00 PM, Train X departed from Station A on the road to   [#permalink] 18 Dec 2015, 08:03
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