Riding the Gravy Train

Let the railroad be a segment $AB$ .Let $C$ be a point in this segment.Let $T$ be a point outside this segment collinear with the other points that represents train coming from $B$ towards $A$ .We know that the first friend ran towards $B$ and one the other friend ran toward $A$ ,in the direction of the train.Let $BC=l$ , let $AC=l_1$ let $TB=x$ where $l+l_1=100$ .We know ftom the problem that the first friend reached $B$ in the exact same moment that the train reached $B$ .Let $v$ be the speed of friends and $5v$ the speed of train.Since their times are equal we have:

$\frac{x}{5v}=\frac{l}{v}$ $(1)$ Similarly for the second friend , since their times are also equal, we have:

$\frac{x+100}{5v}=\frac{l_1}{v}$ $(2)$

From $(1)$ we have $x=5l$ and from $(2)$ we have $x+100=5l_1$ .Adding these two up we get:

$2x+100=5(l+l_1)$ .By substituting $l+l_1=100$ we get a linear equation and by solving it we get $\boxed{x=200}$ .Since $x=5l$ we get $\boxed{l=40}$ and since $x+100=5l_1$ we get $\boxed{l_1=60}$ .The distance that we are looking for is $TC$ .Observe that $TC=x+l=\boxed{240 km}$ .

Let x = the distance the train is from the bridge and y= distance the boys are from the train's end of the bridge, For the boy running towards the train: x/y = 5/1 and for the boy running towards the other end: x + 100/100-y = 5/1. From the 1st equation, x = 5y. From the 2nd equation, x + 5y = 400. Substituting we get 10y = 400 or y = 40 x = 200. Therefore the distance from the train to the boys when the train whistled is 240 m.