A train flies round trip at a distance of X miles each away. The velocity with the head wind is 160mph, while the velocity with tail wind is 240mph, What is the average speed for the round trip?

To get the average speed for the round trip, recall that...

$$distance = (speed)(time)$$ $$d = vt$$

We also know that $$average \space speed = \frac{total \space distance}{total \space time}$$

With that in mind...

The time required by train (a flying train, I suppose, hehe) to reach its destination is $\frac{X}{160}$ hours. The time required by train to return from its destination is $\frac{X}{240}$ hours.

So...

The total distance for the round trip would be $X+X = 2X$ miles.
The total time spent flying would be $\frac{X}{160}+\frac{X}{240} = \frac{X}{96}$ hours.

Therefore, the average speed for the entire round trip would be $\frac{2X \space miles}{\frac{X}{96} \space hours}$ = $\color{green}{192 \space miles \space per \space hour}$

To get the average speed for the round trip, recall that...

$$distance = (speed)(time)$$ $$d = vt$$

We also know that $$average \space speed = \frac{total \space distance}{total \space time}$$

With that in mind...

The time required by train (a flying train, I suppose, hehe) to reach its destination is $\frac{X}{160}$ hours. The time required by train to return from its destination is $\frac{X}{240}$ hours.

So...

The total distance for the round trip would be $X+X = 2X$ miles.

The total time spent flying would be $\frac{X}{160}+\frac{X}{240} = \frac{X}{96}$ hours.

Therefore, the average speed for the entire round trip would be $\frac{2X \space miles}{\frac{X}{96} \space hours}$ = $\color{green}{192 \space miles \space per \space hour}$

Alternate solutions are encouraged...