For rate-time-distance questions, try to reduce the question to as few unknown quantities as possible. Instead of a separate r, t, and d for each train, look for commonalities:

Both trains travel the same distance, so there's only one "d"

We don't know the time, but we know that Train Y took less time, since it left an hour and half later. So if Train X took t hours to travel some distance, Train Y only took (t-1.5) hours.

The rates aren't given, but we know that train Y was 4/3 the rate of train X.

Let's set it up. Since rate*time= distance, and the distances are equal, we can set Train X's rate * time equal to Train Y's rate * time

\(xt = \frac{4}{3}x(t-1.5)
\)

Working through the problem, we get

\(xt = \frac{4}{3}xt-4/3*1.5\) here it can be helpful to recognize that 1.5 is just \(\frac{3}{2}\), and cancel out the 3's: \(\frac{4}{3}*\frac{3}{2}=2\)
\(xt=\frac{4}{3}xt-2\)

Subtract xt from both sides, and add 2 to both sides, and we get:

\(2=\frac{1}{3}t\)

6 = t

So, adding 6 hours to Train X's starting time, we get 4:00 pm as the time that train Y will catch train X.

Answer: D