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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