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

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04 Oct 2010, 15:14

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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)

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.

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

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16 Sep 2014, 19:23

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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.

Pretz
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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)

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 \(= 0.5 + \frac{p - r/2}{r + s}\)

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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18 Dec 2015, 07:03

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

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22 Mar 2017, 00:52

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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)

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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25 Mar 2017, 06:49

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).

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).

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.
_________________

Scott Woodbury-Stewart Founder and CEO

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

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

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29 Mar 2017, 18:24

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