<p>A man needs to go through a train tunnel.</p>

12.

A man needs to go through a train tunnel.

A man needs to go through a train tunnel. He starts through the tunnel and when he gets 1/4 the way through the tunnel, he hears the train whistle behind him. You don't know how far away the train is, or how fast it is going, (or how fast he is going). All you know is that
If the man turns around and runs back the way he came, he will just barely make it out of the tunnel alive before the train hits him.
If the man keeps running through the tunnel, he will also just barely make it out of the tunnel alive before the train hits him.
Assume the man runs the same speed whether he goes back to the start or continues on through the tunnel. Also assume that he accelerates to his top speed instantaneously. Assume the train misses him by an infinitesimal amount and all those other reasonable assumptions that go along with puzzles like this so that some wanker doesn't say the problem isn't well defined.
How fast is the train going compared to the man?

Submitted by tartle · Added 7 December 2007 · Updated 5 July 2026

Show Solution

Solution:

The train is moving twice the speed of the man. When the man runs back to the entrance of the tunnel, he covers 1/4 of the tunnel's length, while the train covers the entire length of the tunnel in the same time. Therefore, the train must be traveling at a speed that is double the man's speed to ensure both reach their respective exits at the same moment.

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Comments (7)

aatifahmed 7 January 2008

The train is moving at 2 times the speed of man.

Use www.rot-13.com to decipher the solution if need be.

Yrg gur genva or ng n qvfgnapr f sebz gur ghaary. Yrg gur ghaary or bs yratgu q naq gur fcrrq bs zna i, juvyr bs gur genva ki.
Gurersber:
Gvzr gnxra ol gur zna naq gur znpuvar vf fnzr ng gur ragenapr => (f/ki)=(q/4i)=((f+q)/(ki+4i)) ---- (v) (hfvat onfvp cebcregvrf bs engvb cebcbegvbaf).

Gvzr gnxra ol gur zna naq gur znpuvar vf fnzr ng gur rkvg =>
((f+q)/ki)=(3q/4i) ---- (vv)

hfr (v) & (vv):
(ki+4i)*q/4i=3q*ki/4i
fbyir gb trg k=2.

Axomomma 4 November 2008

The train is moving at 1.5 times the speed of the man.

Assume that the tunnel is 400 yards long.
The man hears the train when he is 1/4 of the way inside, or 100 yards in.
If the man travels back 100 yards the train will just miss him.
If he goes forward 300 yards the train will just miss him.
The train was 150 yards away from the entrance to the tunnel when the man first heard it.
If the man goes forward 300 yards the train will travel 450 yards. 300 x 1.5 = 450
If the man goes back 100 yards the train will travel 150 yards. 100 x 1.5 = 150

alexonfyre ★ Solved 22 December 2008

Actually the answer is that the train is moving twice the speed of the man, this forum doesn't have spoiler tags, so I will go ahead and post my work, hope that is okay.
First to Axomomma, for your hypothetical to be true, then the train would have to travel 550 yards while the man traveled 300 (since it has to go the distance to the tunnel, then all the way through it, 400 + 150 = 550) and then your two ratios wouldn't match, so that cannot be the answer. It was a decent guess, it took me a while to realise the error, but I think the solution to this problem is a system of equations).

There are 4 unknowns here: X (the length of the Tunnel), Y(the distance between the train and the tunnel), M (the speed of the man) and T (the speed of the train).
The ultimate goal here is to eliminate x and y and leave a relationship between m and t.
We have these two conditions given to us:

The time it takes the man to move 1/4 of x is equal to the time it takes the train to go all of y which gives us:

x/(4m) = y/t

also, the time it takes the man to go 3/4 x is equal to the time it takes the train to go all of y and x or:

3x/(4m) = (y+x)/t

For simplicity lets call the time it takes for the man to run through all of x, z or:

z = x/m

so

z/4 = y/t

and

3z/4 = (x+y)/t

now we can solve for z in terms of y and t:

z = 4y/t

make another substitution and we will eliminate t (don't worry it will be back):

(4y/t)*3/4 = (x+y)/t

3y/t=(x+y)/t

3y = x+y

x = 2y

(right here you know that the tunnel is twice the length of the distance between the train and tunnel, which is enough to solve it, but let's finish the elegant, mathematical way)

now with x in terms of y we can eliminate them both from our initial equation:
x/(4m) = y/t

becomes

2y/4m = y/t

y/2m = y/t

(cross multiply)

t = 2m

or the speed of the train is twice the speed of the man.

I couldn't get aatifahmed's decipher link to get his solution, it is broken now, so I don't know how similar his method was to mine, but I do know that his answer is correct.

einsteinager 17 June 2011

Yup thats the right way.

Zuko95 28 December 2014

Yes the answer is two times.

An easier way to solve it:
The time the train takes to reach the tunnel is equal to the time it takes the man to exit the tunnel going backwards , the time the man takes to walk 1/4 of the distance of the tunnel.

So when the train reaches the entrance of the tunnel (if the man is going forward) the man will be in the middle of the tunnel. So for both to reach the end of the tunnel at the same time the train must travel double the speed of the man.

qwertyuiopasdfg 26 September 2016

This is a very nice puzzle. My way probably doesn't make any mathematical sense :).

So the man is 1/4 through the tunnel when he hears the train. He would just make it alive whether he took the 1/4 tunnel way back or the rest of the 3/4 tunnel forward. You subtract these 3/4 - 1/4 which equals 2/4 or 1/2. The man is going 1/2 the speed of the train.

Breys 14 February 2017

Much faster than the ram, since a train moves at a normally quicker speed (I am being very literal and am giving a very literal answer.)

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