# Train and man escaping from tunnel riddle

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A man walking home takes a shortcut through a train tunnel. A quarter of the way in, he hears a train whistle behind him.
The tunnel is not wide enough for the man to escape being hit by the train, so he must either turn back, or go forward, at his top speed of 20 mph. Either way he will escape, but by the slimmest of margins.

How fast is the train moving?

Solving it mathematically..

I              x              I   y/4    I               3y/4                  I
I———————I———–I——————————I
A                           B             C                                          D

A = is the train’s current position,
B = Start of tunnel
C = man’s current position
D = end of tunnel

AB = x miles
BD = y miles; so BC = y/4 and CD = 3y/4

Man’s speed is 20mph. Let S be the train’s speed.

Let the man take Tb time to get back to the start of the tunnel
So, 20mph = y/4 miles / Tb or Tb = y/4 miles / 20 mph
In this time, the train will also just make it to the entrance of the tunnel;
So Tb = x / S

So we can say that Tb = y/80 = x/S or Sy = 80x—Eq1

If the man runs in the other direction, toward the end of the tunnel, then, let him take time Td
Td = 3y/4 miles / 20
By this time the train would have reached the end of the tunnel as well.
So Td = (x+y)/S
So Td = 3y/80 = (x+y)/S or S3y = 80x + 80y—Eq2

Substituting for 80x from Eq1, Eq2 now becomes S3y = Sy + 80y or S3y = y(S+80) or S3 = S + 80 or S2 = 80 or S = 40.

So the speed of the train is 40mph.

It can be solve mathematically as Srihari Bharatthi did, but we can solve it by using common sense. We know that if he turns back he passes a quarter of the way and he’ll meet the train, so if he runs the opposite direction after a quarter of the way the train is now an the beginning of the tunnel.  So at this point he has to run half of the tunnel and the train travels the whole tunnel at the same time, which means that the train speed is twice his speed, so its 40mph