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

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

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

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

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.

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.

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

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