Can you solve it? - Cyclist Meets Train

Community
21

The case ONLY fits if the direction is the same every day. That’s the point.

He could, or he could not. You don’t know, so the answer is indeterminate.

Both at the intersection and meeting place at the same time, and 5km apart.

1 Like
22

“The train travels 5km in 5 minutes, so it was doing 60kpm”… except, it doesnt have to be. Maybe it was doing 30kpm at the meeting point, did that for 2 1/2 km, and then sped up to 120kpm for the last 2 1/2km. Which is why the constant speed must be assumed.

1 Like
23

It only needs to fit the case as described. Other values have other results.

What matters is if he travels the 5km at the stated speed as it’s given.

Or a 5km long train. :wink:

Are we on the same page?

24

I’ve already pledge guilty in that point. (In number 2.) Point 1. :blush:

Brittish trains was out of my picture. :slight_smile:

25

(That was point 1).

We’re on the same page, it’s just we have to have those caveats, otherwise, lateral-thinking puzzlers will come along and say “According to the riddle, this scenario could prove you wrong.”

1 Like
26

I think you’ve covered all sides, now if you could mend your algorithm to reflect all possible cases. :stuck_out_tongue:

(I’m off to bed now.)

27

Algorithm holds given the assumptions listed.

Vt = (Dm*Vc)/(Dm + Dc*Vc)
Direction is the sign of the value.

If the assumptions are NOT given, the problem is either unsolvable, or has infinite solutions.

28

I’d reckon they were in my original setup? With a train in Germany.

29

I mean that, for example, if the train’s speed is not a constant, there are infinite solutions, because the train could have been doing any speed at the meeting point, and then compensated during the transit between meeting point and intersection to make the math work.

If all 4 of the assumptions in post 17 are true, the algorithm holds for all values of Dm, Vc, and Dc, such that Dm+Dc*Vc != 0. (cause dividing by 0 is bad, mkay?)

The assumptions are not given as true in the original problem.

1 Like
30

Can we agree on this puzzle as solvable?

Puzzle:

A railway track runs parallel to a road before a railway crossing.

A cyclist rides along the road everyday at a speed of 10 km/h and he always meets a train at the crossing. Their speed is always constant.

One day he was late by 25 min and met the train 5 km before the crossing.

At what speed does the train travel?​
What’s the train’s direction versus the cyclist?

:sleeping:

31
  1. The train travels at 60 km/h
  1. The train’s direction is identical to that of the cyclist.
  2. At the commencement of the cyclist’s journey the
    train is 50 km behind.
  3. After one hour the train will have traveled 60 km and
    will then catch the cyclist who has traveled 10 km.
  4. On the one day he was late by 25 minutes, it will
    take him 55 minutes - ( 25 + 30 ) - to cover 5km.
  5. In that 55 minutes the train will have traveled the
    55 km required to catch the cyclist.

coothead

1 Like
32
  1. The train’s direction is identical to that of the cyclist.

I agree. Glad you joined the party. :slight_smile:

Nice table list.

Would the train’s direction be different if instead the cyclist was 35 min late?

What would change in the list if, say, the cyclist was 25 min late but rode in the speed of 15 km/h?

:thinking:

33
  1. No.
  2. It would take the train 53 minutes to catch the
    cyclist who had traveled 3km in 18 minutes.
  3. If the cyclist was 25 min late but rode at the
    speed of 15 km/h, then it would take the train
    between 58 and 59 minutes to catch the cyclist
    who would have traveled between 8¼km and
    8½km in between 33 and 34 minutes.

I could not be bothered to work out that last bit precisely. :unhappy:

coothead

34

You sure about that? (It would, in fact, be different, if the meeting point did not move.)
Keep in mind there are two points of concern in the puzzle - the meeting point, and the intersection, which the train still has to reach at the same time as a normal day, and the cyclist will be 35 minutes late to.

When the train reaches the Intersection, the Cyclist is 35 minutes away from it. As such, the cyclist travelling 10 kph (1/6 kpm), must be 35/6 (5 and 5/6) km short of the intersection.
If he’s 5 5/6 km away from the intersection, he has not yet reached the meeting point (which is 5km short of the intersection) when the train passes through the intersection. Since the meeting point is in the direction of the cyclist at that point in time, and the meeting has yet to happen (because the cyclist hasn’t reached the meeting point), the train must be heading towards the meeting point, from the intersection point - meaning it’s heading towards the cyclist from the other direction.

If the cyclist is 25 minutes late but rode at the speed of 15km/h (1/4 km per minute), and the meeting point is still 5 km before (from the perspective of the cyclist) the intersection, the train must be moving towards the cyclist at 60kph.

1 Like
35

Well explained, @m_hutley. :slight_smile:

I think that’s an overly complex view, the question (assumed, I assume) was with what speed the train traveled the 5km distance.

Then the solution is quite simple, I think. :thinking:

36

This topic was automatically closed 91 days after the last reply. New replies are no longer allowed.