Carcass 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. \(0.5\) + \(\frac{(p - 0.5s)}{(r + s)}\)

B. \(\frac{(p - 0.5s)}{(r + s)}\)

C. \(0.5\) + \(\frac{(p - 0.5r)}{r}\)

D. \(\frac{(p - 0.5r)}{(r + s)}\)

E. \(0.5\) + \(\frac{(p - 0.5r)}{(r + s)}\)

This is very tricky!

First let's denote \(t\) as the time Train Y travelled to get to Station B.

Remember this, as we'll be using this at the end.When the two trains pass eachother what we're actually asking is: when do they meet? They meet when the sum of the distances of Train X and Train Y have covered the entire distance \(p\).

We're also told that Train X departs at 1:00 and Train Y departs at 1:30.

So Train X has been travelling for a half hour longer than Train Y. Since we let \(t\) be the time it takes Train Y to get to Station B, \(t + \frac{1}{2}\) must be the time Train X travels to get to Station A.

Given that the rate of Train X is \(r\) and the rate of Train Y is \(s\), putting all these pieces together:

\(d_A = r(t + \frac{1}{2})\)

\(d_B = st\)

And we know that \(d_A + d_B = p\), so we can add both equations to eachother:

\(r(t + \frac{1}{2}) + st = p\)

Now, let's isolate \(t\).

\(rt + 0.5r + st = p\)

\(rt + st = p - 0.5r\)

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

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

And we've found \(t\)

But we're not done

Recall (first line above in blue) that we let \(t\) be the time it takes train Y to get to Station B,

which left at 1:30. We're being asked how long it took them to meet after 1:00.So if it took \(t = \frac{(p - 0.5r)}{(r+s)}\) for them to meet when we started counting at 1:30, then we need to factor in the extra half an hour of travel. So:

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

Giving E as the answer.