This is an old challenge from a puzzle forum that now ceased to exist. No luck searching for the solution there, but it can probably be found elsewhere in various forms.
A railway track runs parallel to a road before a railway crossing.
A cyclist rides along the road everyday at a constant speed of 10 km/h and he always meets a train at the crossing.
One day he was late by 25 min and met the train 5 km before the crossing.
At what speed and in what direction does the train travel?â
Youâre welcome to post your answer or any thoughts about it in this thread.
Please post both the answer and the formula for it, or how else you figured it out, so we can discuss eventual other solutions.
Edit)
The âmeetsâ and âmetâ doesânt suggest a direction, just that he arrives to the same place as the train.
The direction of the train versus the cyclist I think is determined by the relation of the time the cyclist is from the crossing and how late he is when he meets the train.
Wouldnât the pos/neg value of the time difference show if the train has passed the crossing or not?
If he usually meets the train exactly at the crossing, but being late they meet 5km before the crossing, it means the train moves in the opposite direction. Otherwise they wouldnât have met before the crossing.
They would never have met if the train moved faster in the same direction. Or after the crossing if it moved slower in the same direction
When they meet 5km before the crossing, the cyclist is 30min from the crossing. That means the train also is 5min before it reaches the crossing.
So the time cyclistâs travel-time minus late-time difference I thought about would be positive, indicating they both are 5 min before the crossing. Or am I cofusing things here?
Iâm not sure I agree with Guidoâs answer⌠at the moment my brainâs telling me thereâs two answers to the problem; the train could be moving in either direction, and the numbers change. The assumption that Guido seems to be making is that the train is moving faster than the cyclist, but thatâs not a given.
also, note that the problem does not say they met 25 minutes late, it says that the cyclist was 25 minutes late in starting.
Thereâs a lot missing from this puzzle. Off the top of my head:
Impossible, since the definition of parallel means the two will never meet.
You mean he passed the train. If he met the train, thereâd be a mess.
Depends on what rail system youâre in. If itâs the british rail system, the train isnât travelling, and has stopped because thereâs a problem down the line.
Okay, enough jokes, down to the actual problem.
Letâs make a couple of assumptions that are NOT in the puzzle:
The train is also moving at a constant speed at all times.
The train is on time on the day in question.
Both the cyclist and train are travelling in the same direction to reach the intersection point in question at all points in this question. (Otherwise âwhat directionâ is difficult to describe at best, and ambiguous at worst; if either direction is variable, then there are multiple answers to the question.)
The cyclist actually meets an intersecting road that crosses the train track. (otherwise, see definition of parallel, and we have to start adding angular calculations.)
Met, while not suggesting a direction, is an instantaneous event. (Otherwise, If the cyclist and train are going the same speed in the same direction, they are constantly âmeetingâ each other, and the whole thing becomes infinite.)
Iâm still muddling through this in my head, but i donât think its as simple as youâre thinking.
What direction the train moves is relative the cyclist, of course.
Finding both the speed and the direction the train travels in is not in the original puzzle, thatâs my own variant as I donât agree with the original version stating the direction as an requirement for solving the puzzle.
I search the net for this puzzle and thereâs many versions with different times and speeds, but all of them think the direction of the train is required to know to be able to solve the speed of the train.
I look forward to your next post, or what anyone else would say about the direction requirement.
The train and the cyclist have a start point that is theoretically infinite. (Otherwise, the answer may be âimpossible, because the train couldnât pass the cyclistâ)
Iâm going to let my brain spill out onto the forum here, and someone can tell me where iâm wrong.
First, let me translate the cyclistâs speed into Minutes, because my sanity and because weâre mixing minutes and hours.
10kph = 1/6 kpm.
If the cyclist starts 25 minutes late, the cyclist is simply 1/6*25, or 25/6, or 4 1/6 km âshortâ of where he should be at any given point in time on a regular day.
So, lets say they regularly meet at 8 AM. (Random time. Feel tree to call it T.)
at 8 AM, the train is at the intersection; the cyclist is 4 1/6 km away from it.
They met 5 km before the intersection.
at 8 AM, the meeting had already taken place. By how much?
The cyclist moves at 1/6 kpm, and is 5/6km past the meeting point.
So the meeting happened 5 minutes ago, at 7:55 AM.
Where was the train at that point?
The train was at the meeting point at 7:55, and will be at the intersection at 8.
Therefore, the train is moving towards the intersection.
Therefore, the train is moving in the same direction as the cyclist.
The train would cover 5 km in 5 minutes, so itâs moving at 1 kpm, or 60kph.
My answer is that the train and cyclist are moving in the same direction, and the train is travelling at 60kph.
Your logic seems coherent, do you think it would be viable to put together a formula or algorithm to solve any set of time and speed and other requirement for this puzzle?
(In my experience a problem is tackled in different ways depending the method/environment used. Like the Four Colour Theorem thatâs simple to solve in a graph even for any genus bodies, but extremely complex if you need to transform it to a math formula.)
(Again, spewing my brain out⌠this needs cleaning up/correcting.)
I will take for granted that the conversion from km/h to km/m is moot; that we can assume the first step is âconvert all values to the same units (minutes, hours, km, miles, whatever)â.
Definitions:
Dc = Delay of the Cyclist
Dm = Distance to the Meeting Point from the Intersection Point. (Thus, â5 km before the intersectionâ is -5)
Vc = Velocity of the Cyclist.
Ti = Time at Intersection.
Tm = Time at Meeting Point.
Vt = Velocity of the Train (target)
The cyclistâs position is a line such that Pc(Ti) = -(Dc*Vc), and the slope of the line is Vc.
The trainâs position is a line such that Pt(Ti) = 0, and the slope of the line is Vt.
Given a fixed point (Tm,Dm), find Vt.
It may be convenient to take Ti = 0.
Unknown Quantities: Tm, Vt. (Tm,Dm) must be on the line Pc(T) = Vc*T - (Dc*Vc), and the line Pt(T) = Vt*T, and Dm is known,
so Dm = Vc*Tm - (Dc * Vc) Dm + (Dc * Vc) = Vc*Tm (Dm + (Dc * Vc))/Vc = Tm Tm = (Dm + (Dc * Vc))/Vc
and into the next line⌠Pt(Tm) = Vt * Tm = Dm Vt = Dm / Tm Vt = Dm / (Dm + (Dc * Vc))/Vc) Vt = (Dm*Vc)/(Dm + Dc*Vc)
If Tm*Dm is positive, the cyclist is moving in the same direction as the train, if itâs negative theyâre moving opposite;
the train is travelling at Vt?
Mmmh⌠iâve done something wrong there. But my brain hurts. I will revist soon if noone beats me to it.
EDIT: No, i think what iâve done wrong there is that the direction of the train is indicated by the sign of Vt - if itâs negative, the train is moving towards the cyclist.
If the time for the cyclist to reach the crossing is less than the delay then the train has already passed the crossing and is traveling from the crossing while the cyclist is traveling towards it.
A pos value of the difference gives the same direction as the cyclist.
A neg value gives the opposite direction.
Congrats, I think you can solve all variants and find both the speed and the direction. Maybe it could be cleaner but it seems OK.
But, the speed and the directions are independent calculations even if they share details like the meeting place and the travel times for the distance to the crossing.
The speed is calculated by the time it takes the train and the cyclist to travel the distance of the meeting place from the crossing.
The direction relative each other is found by the time they are at the meeting place relative the crossing. E.g. if the cyclist is later than the time he needs to reach the crossing, the train has already passed the crossing.
Examples of combining the details:
Calculating the speed:
When they met they were both 10 km before the crossing. The time they needed to reach the crossing was 30 min for the cyclist and 5 min for the train (30 min - 25 min = 5 min).
â The trainâs speed is 5km in 5min = 60km/h.
Now, in case the cyclist was 35 min late:
â The trainâs speed is 5km in 5 min (30 min - 35 min = -5 min) = 60km/h
Finding the direction:
At the time both the cyclist and the train is moving towards the crossing.
â The trainâs direction is the same as the cyclist,
Now, in case the cyclist was 35 min late:
The train has passed the crossing 5 min ago (indicated by the neg 5 min value), while the cyclist is still heading for the crossing.
â The trainâs direction is opposite the cyclistâs.
You are welcome to elaborate the puzzle or add other variants you could find. Maybe one that actually needs a given direction to be able to find the speed of the train.
Still worth pointing out that this working out is based on all the assumptions not in the puzzle. As written, the puzzleâs answer is âNot enough Informationâ.
The train is also moving at a constant speed at all times.
The train is on time on the day in question.
Both the cyclist and train are travelling in the same direction to reach the intersection point in question at all points in this question. (IE: The train didnât decide to drive in the other direction today.)
4: The train and the cyclist have a start point that is theoretically infinite. (or at least, include the intersection and meeting point, and enough distance that the cyclist has already started his journey before meeting time.)
If the train isnât guaranteed to be moving at a constant speed, you canât answer the question, because the answer doesnt ask you what the trainâs speed is at a specific time, it just says âat what speed does it travelâ, to which the answer is âIt varies.â
If the train has suddenly decided to change directions today, you canât use the information about the intersection to determine anything, and the problem becomes unsolvable because you donât have a second fixed point on the trainâs line.
If the cyclist lives in a house 5km from the intersection, and is still at home when the train passes, heâs currently moving 0 kpm, and the train could be doing anywhere from 60 to Infinity kph. The false assumption here is that the cyclist has begun his journey at the meeting time. Gotta think laterally sometimes!
