algebra precalculus - how far has bird flown when two trains cross each other

Here's the question:

A train leaves point A and heads to point B (which is 100 km away) at 20 km/hr (the track is a straight line between the two points).

Another train leaves point B and heads towards point A at 30 km/hr.

A bird sitting on the firs train takes off as soon as the train starts and flies back and forth between the two trains until the trains pass each other.

If the bird flies 40 km/hr, how far has the bird flown at the time the trains pass each other?

Here's what I've tried:

Distance = 100
Rate of train A = 20
Time it takes train A to travel the track = (100/20) = 5 hours
Time it takes train B to travel the track = (100/30) = 3.33 hours

So is the time that the two trains meet 5 - 3.33 = 1.67 hours

And then I would multiply 40 *1.67 = 66.8 km to get the distance the bird traveled.

Is my logic correct, and did I arrive at the correct answer? If not, please let me know where I have erred.

Thanks!

Answer

Well, your last idea to find the distance the bird flew is nice. However, the first part kind of isn't. For the first part, what you need to do is to find $x$ such that train $A$ travels $x$ in the same amount of time $B$ needs to go $100-x$ kilometers. This leads to solving:
$$ \frac{x}{20} = \frac{100 - x}{30} $$
This is a linear equation and it should not be too hard for you to solve by yourself.

To illustrate the next part, let me give away that the answer is given by $x=40$. This means that $A$ travels $40$ km in the same time as $B$ $60$ km. The time it took the two trains to meet is thus given by $\frac{40}{20} \left(=\frac{100-40}{30}\right) = 2$ hours. This then implies that the bird flew in total $80 ( = 2\cdot 40)$ km.