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

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New post 22 Mar 2017, 01:52
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At 1:00 PM, Train X departed from Station A on the road to [#permalink]

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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)


We have a converging rate problem in which we can use the following formula:

distance of train X + distance of train Y = total distance

Train X is traveling at a rate of r mph and train Y is traveling at a rate of s mph. Since train X left at 1:00 PM and train Y left at 1:30 PM, we can let the time of train Y = t and the time of train X = t + 0.5.
Thus:

r(t + 0.5) + st = p

rt + 0.5r + st = p

t(r + s) = p - 0.5r

t = (p - 0.5r)/(r + s)

However, t is relative to train Y, which left at 1:30 PM. Since we want the time relative to 1:00 PM, we need to add 0.5 hour to t; thus, the time needed for the two trains to pass each other is 0.5 + (p - 0.5r)/(r + s).

Answer: 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 25 Mar 2017, 07:49
ScottTargetTestPrep wrote:
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)


We have a converging rate problem in which we can use the following formula:

distance of train X + distance of train Y = total distance

Train X is traveling at a rate of r mph and train Y is traveling at a rate of s mph. Since train X left at 1:00 PM and train Y left at 1:30 PM, we can let the time of train Y = t and the time of train X = t + 0.5.
Thus:

r(t + 0.5) + st = p

rt + 0.5r + st = p

t(r + s) = p - 0.5r

t = (p - 0.5r)/(r + s)

However, t is relative to train Y, which left at 1:30 PM. Since we want the time relative to 1:00 PM, we need to add 0.5 hour to t; thus, the time needed for the two trains to pass each other is 0.5 + (p - 0.5r)/(r + s).

Answer: E


ScottTargetTestPrep

Sir,

I approached on your method.But I solved the equation in the following way:

p=rt+s(t-0.5)---->I subtracted 1/2 hour from the time taken by the train Y. Is that wrong??
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Re: At 1:00 PM, Train X departed from Station A on the road to [#permalink]

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the total distance between the stations be P

SPeed of X is r

SPeed of Y is S

when the trains are moving in the opposite direction then we add the speeds

Therefore total time taken should be p/r+s.

But Train X has already covered distance equal to 0.5r

hence the answer is option 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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techiesam wrote:
ScottTargetTestPrep wrote:
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)


We have a converging rate problem in which we can use the following formula:

distance of train X + distance of train Y = total distance

Train X is traveling at a rate of r mph and train Y is traveling at a rate of s mph. Since train X left at 1:00 PM and train Y left at 1:30 PM, we can let the time of train Y = t and the time of train X = t + 0.5.
Thus:

r(t + 0.5) + st = p

rt + 0.5r + st = p

t(r + s) = p - 0.5r

t = (p - 0.5r)/(r + s)

However, t is relative to train Y, which left at 1:30 PM. Since we want the time relative to 1:00 PM, we need to add 0.5 hour to t; thus, the time needed for the two trains to pass each other is 0.5 + (p - 0.5r)/(r + s).

Answer: E


ScottTargetTestPrep

Sir,

I approached on your method.But I solved the equation in the following way:

p=rt+s(t-0.5)---->I subtracted 1/2 hour from the time taken by the train Y. Is that wrong??


Mathematically, you can solve it the way you did. You let t represent the time counting from 1 p.m., and thus t - ½ was the time counting from 1:30 p.m.

However, after you solve for t, in terms of p, r, and s, you see that the equivalent expression of t is not in any of the answer choices. That is why in my solution, I made t relative to 1:30 p.m. rather than 1 p.m.
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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 29 Mar 2017, 19:24
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)


distance X travels alone in 0.5 hours=.5r miles
time X and Y travel together from 1:30 to passing=(p-.5r)/(r+s) hours
total time from 1PM to passing=0.5+(p-.5r)/(r+s) hours
E
At 1:00 PM, Train X departed from Station A on the road to   [#permalink] 29 Mar 2017, 19:24
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