Jogger and Train

I think your question is missing a number of important points:

You say " he sees a train coming", but don't specify whether it is coming behind him or towards him.

Also, since he completed 3/4 of his run(presumably 3/4 of the length of the bridge", you say:

"and if it is possible for the jogger to just escape being hit by the train by running at full speed to either end of the bridge, what is the minimum speed, in km/h" !!

Why would he run to "either end of the bridge", since he had already completed 3/4 of the span of the bridge. Surely, he would finish running the remaining 1/4 of the bridge, to get out of harm's way, since it would take him a lot LESS time to get off the bridge.

I think......Because it is possible for him to survive by running to either end, then he must be heading toward the train.  If he runs backward, he has the advantage running in the same direction as the train, but he has to run  3/4  of the bridge.  If he runs forward, he has the advantage of only having to run  1/4  of the bridge, but he and the train are heading toward each other.

Although this subtly phrased question appears to lack sufficient information for a solution. It does in fact have one. 

Part of the information is contained in this phrase.

if it is possible for the jogger to just escape being hit by the train by running at full speed to either end of the bridge

Here the use of the word “if” is not so much a conditional clause, as it is a presented condition or supposition.  Rephrasing makes this more clear:

"Suppose the jogger can just barely escape being hit by the train by running to either end of the bridge at his greatest speed."

This tells us the runner will meet the train at end of the bridge and just miss being hit, whether he runs toward the train or away from the train. Along with this information, there are two pieces of numerical data: the train’s 30kph speed and the man’s relative position of 3/4 on the bridge. 

Using this information, set up an equation where the train and the man meet at either end of the bridge.   

\(\dfrac{(30 + F) }{(30-F)} = \dfrac{3}{1} \tiny \text { F is the speed the Fool can run.}\\ (30+F) =3(30-F)\\ 30+F = 90-3F\\ 4F = 60\\ \,\;F= \dfrac{60}{4}\\ \,\\ F = 15kph\\ \)

Answer  (A). 

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 

Good thing you got that "Fool" across the bridge, GA.......these guys get a liitle perturbed at late arrivals....!!!!!

Very nice, GA and Melody.....!!!

BTW.....if a Troll collects fares....does he do so from a "Troll" booth ??