Can you solve it? - Cyclist Meets Train

Fancy a pastime?

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

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 âmeetsâ and âmetâ doesânt suggest a direction, just that he arrives to the same place as the train.

If it takes the train 25 minutes to do 5 km, itâs going at a speed of 12km/h towards the cyclist.

1 Like

Sounds plausibleâŚ

Now, heâs 30 min from the crossing while 25 min late. Is the time difference positive or negative?

I donât understand the question

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.

Wouldnât 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 wouldnât 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 cyclistâs 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?

Iâm not sure I agree with Guidoâs answerâŚ at the moment my brainâs telling me thereâs 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 thatâs 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.

Thereâs 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, thereâd be a mess.

Depends on what rail system youâre in. If itâs the british rail system, the train isnât travelling, and has stopped because thereâs a problem down the line.

Okay, enough jokes, down to the actual problem.

Letâs make a couple of assumptions that are NOT in the puzzle:

1. The train is also moving at a constant speed at all times.
2. The train is on time on the day in question.
3. 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 âwhat directionâ is difficult to describe at best, and ambiguous at worst; if either direction is variable, then there are multiple answers to the question.)
4. 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.)
5. 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 âmeetingâ each other, and the whole thing becomes infinite.)

Iâm still muddling through this in my head, but i donât think its as simple as youâre thinking.

1 Like
1. 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, thatâs my own variant as I donât agree with the original version stating the direction as an requirement for solving the puzzle.

I search the net for this puzzle and thereâs 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.

1. The train and the cyclist have a start point that is theoretically infinite. (Otherwise, the answer may be âimpossible, because the train couldnât pass the cyclistâ)

Iâm going to let my brain spill out onto the forum here, and someone can tell me where iâm wrong.

First, let me translate the cyclistâs speed into Minutes, because my sanity and because weâre 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 âshortâ 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 itâs 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.

1 Like

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 thatâs 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 outâŚ 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 âconvert 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, â5 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 cyclistâs position is a line such that `Pc(Ti) = -(Dc*Vc)`, and the slope of the line is `Vc`.
The trainâs 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 lineâŚ
`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 itâs negative theyâre moving opposite;
the train is travelling at `Vt`?

MmmhâŚ iâve done something wrong there. But my brain hurts. I will revist soon if noone beats me to it.

EDIT: No, i think what iâve done wrong there is that the direction of the train is indicated by the sign of `Vt` - if itâs negative, the train is moving towards the cyclist.

1 Like

Exactly my thought too.

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.

1 Like

To wrap this up.

The original puzzle had the directions of the train and the cyclist given as a reqiuirement for finding the speed of the train. If you search the puzzle youâll find that thatâs how all variants are setup:
https://duckduckgo.com/?q=+Every+day+a+cyclist+meets+a+train+at+a+particular+crossing

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 trainâs speed is 5km in 5min = 60km/h.

Now, in case the cyclist was 35 min late:
â The trainâs 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 trainâs 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 trainâs direction is opposite the cyclistâs.

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 puzzleâs answer is âNot enough Informationâ.

1 Like

We assume:

1. The train is also moving at a constant speed at all times.
2. The train is on time on the day in question.
3. 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 didnât 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.)

Iâm trying to not break any law of nature or logic, Iâm thinking:

1. Not necessarily
2. Guilty.
3. Not necessarily
4. The meeting points are sufficient

If the train isnât guaranteed to be moving at a constant speed, you canât answer the question, because the answer doesnt ask you what the trainâs speed is at a specific time, it just says âat what speed does it travelâ, to which the answer is âIt varies.â

If the train has suddenly decided to change directions today, you canât use the information about the intersection to determine anything, and the problem becomes unsolvable because you donât have a second fixed point on the trainâs line.

If the cyclist lives in a house 5km from the intersection, and is still at home when the train passes, heâs 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!

Itâs the speed at the meeting place thatâs asked for, but yourâe right, a constant speed is assumed.

It doesnât need to change, the direction could be the same every day and still fit the case at hand.

He could still travel the 5km in the stated speed.

OK, a quantum variant. Now the cyclist is 30 min late, where do you find the train?