Can you solve it? - Cyclist Meets Train

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#1

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?​ :thinking:

You’re 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 “meets” and “met” does’nt suggest a direction, just that he arrives to the same place as the train.

#2

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

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#3

Sounds plausible…

:thinking: Now, he’s 30 min from the crossing while 25 min late. Is the time difference positive or negative?

#4

I don’t understand the question

#5

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.

Wouldn’t the pos/neg value of the time difference show if the train has passed the crossing or not?

#6

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 wouldn’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

#7

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 cyclist’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?

#8

I’m not sure I agree with Guido’s answer… at the moment my brain’s telling me there’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 that’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.

There’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. :stuck_out_tongue:

You mean he passed the train. If he met the train, there’d be a mess.

Depends on what rail system you’re in. If it’s the british rail system, the train isn’t travelling, and has stopped because there’s a problem down the line.

Okay, enough jokes, down to the actual problem.

Let’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 ‘what direction’ 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 ‘meeting’ each other, and the whole thing becomes infinite.)

I’m still muddling through this in my head, but i don’t think its as simple as you’re thinking.

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#9
  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, that’s my own variant as I don’t agree with the original version stating the direction as an requirement for solving the puzzle.

I search the net for this puzzle and there’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. :thinking:

#10
  1. The train and the cyclist have a start point that is theoretically infinite. (Otherwise, the answer may be ‘impossible, because the train couldn’t pass the cyclist’)

I’m going to let my brain spill out onto the forum here, and someone can tell me where i’m wrong.

First, let me translate the cyclist’s speed into Minutes, because my sanity and because we’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 ‘short’ 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 it’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.

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#11

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 that’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.)

#12

(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 ‘convert all values to the same units (minutes, hours, km, miles, whatever)’.

Definitions:
Dc = Delay of the Cyclist
Dm = Distance to the Meeting Point from the Intersection Point. (Thus, “5 km before the intersection” is -5)
Vc = Velocity of the Cyclist.
Ti = Time at Intersection.
Tm = Time at Meeting Point.
Vt = Velocity of the Train (target)

The cyclist’s position is a line such that Pc(Ti) = -(Dc*Vc), and the slope of the line is Vc.
The train’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 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 it’s negative they’re moving opposite;
the train is travelling at Vt?

Mmmh… i’ve done something wrong there. But my brain hurts. I will revist soon if noone beats me to it.

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

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#13

Exactly my thought too. :slightly_smiling_face:

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.

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#14

To wrap this up. :slight_smile:

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 you’ll find that that’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).
– The train’s speed is 5km in 5min = 60km/h.

Now, in case the cyclist was 35 min late:
– The train’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.
– The train’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.
– The train’s direction is opposite the cyclist’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. :slight_smile:

#15

Still worth pointing out that this working out is based on all the assumptions not in the puzzle. As written, the puzzle’s answer is “Not enough Information”.

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#16

Please mention what’s missing. :crazy_face:

#17

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 didn’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.)
#18

I’m trying to not break any law of nature or logic, I’m thinking:

  1. Not necessarily
  2. Guilty.
  3. Not necessarily
  4. The meeting points are sufficient
#19

If the train isn’t guaranteed to be moving at a constant speed, you can’t answer the question, because the answer doesnt ask you what the train’s speed is at a specific time, it just says “at what speed does it travel”, to which the answer is “It varies.”

If the train has suddenly decided to change directions today, you can’t use the information about the intersection to determine anything, and the problem becomes unsolvable because you don’t have a second fixed point on the train’s line.

If the cyclist lives in a house 5km from the intersection, and is still at home when the train passes, he’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! :stuck_out_tongue_winking_eye:

#20

It’s the speed at the meeting place that’s asked for, but your’e right, a constant speed is assumed.

It doesn’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?