Thanks Ginger, I like your logic
I have done this but I have effectively said that the train has no length and the man has no width and they both arrive at the end of the bridge at the same time. So effectively the speed would have to be greater.
Since he can only just get to either end in time then the train must be coming in the opposite direction to the man.
Let the length of the bridge be 4b, so the man has crossed a distance of 3b and has 1b left to traverse.
Let the man be jogging v km/h
Let the distance that the train is from the bridge be k km
The train is traveling at 30km/h
\(\boxed{speed=\frac{distance}{time}\\ time=\frac{distance}{speed}}\)
If the man continues forward then it will take him b/v to get to the end of the bridge.
In this time the train will travel for x/30 hours
so
\(\frac{b}{v}=\frac{x}{30}\\ x=\frac{30b}{v}\)
If the man runs in the opposite direction then it will take him 3b/v hours to get to the end of the bridge.
In this time the train will travel for (x+4b)/30 hours
so
\(\frac{3b}{v}=\frac{x+4b}{30}\\ \frac{90b}{v}=x+4b\\ \frac{90b}{v}-4b=x\\\)
so
\(\frac{90b}{v}-4b=\frac{30b}{v}\\ \frac{90}{v}-4=\frac{30}{v}\\ 90-4v=30\\ 60=4v\\ v=15km/hour \)
The speed has to be minimally greater than 15km/hour
So I think A is the answer sought
I can see that i made my answer a little more complicated than necessary