# Train station, speed and time puzzle

• K Sengupta
In summary, George took a train from station X to Y, located 8 kilometers away. When the train left X, the minute hand of George's watch was on a minute mark. When the train reached Y, the minute hand and hour hand were coincident, but the minute hand was not on a minute mark. The train left X between 6:00 a.m. and 6:00 p.m. at an odd speed greater than 60 kph. The question is to determine the precise time when the train left X.
K Sengupta
George took a train at station X with the intention of going to Station Y (which is located 8 kilometers from station X). At the precise instant that the train started leaving station X, the minute hand of his 12 hour analog wristwatch was located precisely on a minute mark.

George looked at his watch again at the instant that the train reached station Y, when he observed that the minute hand and the hour hand were exactly coincident, but the minute hand was not situated on a minute mark.

Given that the train left station X between 6:00 a.m. and 6:00 p.m. inclusively, and that the speed of the train, in kilometers per hour, is an odd integer greater than 60, determine the precise time when the train left station X.

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Ok, I just had some surgery done, and I want to make sure that I understand the question.

First of all, do hour marks count as minute marks?

Second, just to be clear:

At Station X - Minute Hand = a degrees from 12 o'clock and is on a minute mark (multiple of 6); Hour Hand = b degrees from 12 o'clock

At Station Y - Minute Hand = c degrees from 12 o'clock; Hour Hand = c degrees from 12 o'clock; where c is not a multiple of 6

And the train is going at an odd speed greater than sixty kph, meaning that the minute hand will not move more than 8 marks (48 degrees).

Assuming they went at a constant speed, I get:
they left at 14:08:00, and went at 165 kph (102.5mph), arriving at precisely 14:10:54 and 6/11 seconds

DaveE

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davee123 said:
Assuming they went at a constant speed, I get:
they left at 14:08:00, and went at 165 kph (102.5mph), arriving at precisely 14:10:54 and 6/11 seconds

DaveE

How did you get that answer?

ƒ(x) said:
How did you get that answer?

Effectively, I started with the fact that the hour and minute hands will be EXACTLY in synch every 1 hour, 5 minutes, and 27 and 3/11 seconds, starting at 12:00:00 or 0:00:00. That narrowed it down to 11 possible arrival times. Next, in Excel, I made a quickie chart that had those arrival times in rows, and calculated nearest whole-minute increments in columns (I had 8 columns, since it was known that the maximum travel time was under 8 minutes. Next, I made a second grid that calculated the speed the train would need to travel in order to take the corresponding times. Turns out, there were only 2 odd-integer values for possible speeds, and one was 55 (vetoed for being under 60). Hence, there was only one possible option for a speed, arrival time, and departure time.

DaveE

The times are given in the format
hour:minute, e.g. 7:34.5 means 7 hours and 34.5 minutes.

Strange, I get exact times only for even velocities:
v=66; Left at 5:20.0 arrived at 5:27.272727272727273 Deviation in seconds: 0.0
v=110; Left at 3:12.0 arrived at 3:16.363636363636363 Deviation in seconds: 0.0
v=132; Left at 8:40.0 arrived at 8:43.63636363636363 Deviation in seconds: 0.0
v=176; Left at 6:30.0 arrived at 6:32.72727272727273 Deviation in seconds: 0.0
v=220; Left at 7:36.0 arrived at 7:38.18181818181818 Deviation in seconds: 0.0
v=264; Left at 4:20.0 arrived at 4:21.818181818181817 Deviation in seconds: 0.0

For odd velocities, the best I can get is:
v=115; Left at 7:34.00790513833992 arrived at 7:38.18181818181818 Deviation in seconds: 0.47430830039530747
v=123; Left at 2:7.0066518847006645 arrived at 2:10.909090909090908 Deviation in seconds: 0.39911308203986806
v=129; Left at 6:29.006342494714588 arrived at 6:32.72727272727273 Deviation in seconds: 0.38054968287525526
v=139; Left at 1:2.0013080444735114 arrived at 1:5.454545454545454 Deviation in seconds: 0.0784826684106843
v=143; Left at 3:13.006993006993007 arrived at 3:16.363636363636363 Deviation in seconds: 0.4195804195803987
v=147; Left at 5:24.007421150278294 arrived at 5:27.272727272727273 Deviation in seconds: 0.44526901669762253
v=151; Left at 7:35.00301023479831 arrived at 7:38.18181818181818 Deviation in seconds: 0.18061408789861844
v=189; Left at 10:52.00577200577201 arrived at 10:54.54545454545455 Deviation in seconds: 0.3463203463205389
v=265; Left at 4:20.006861063464836 arrived at 4:21.818181818181817 Deviation in seconds: 0.4116638078901502
v=279; Left at 6:31.006842619745846 arrived at 6:32.72727272727273 Deviation in seconds: 0.41055718475078606

The deviation tells you how great the difference is between an exact "integer minute mark" and the
actual departure. For example for the odd velocity of v=115km/h George leaves at 7:34.0079
That yields a deviation of 0.0079 minutes which corresponds to 0.0079*60seconds = 0.474 seconds.

For the odd velocities I set the deviation to less than 0.5 seconds.

Update:
I read davee123's solution and wondered why it didn't occur in my list. Thanks to to the "round" function in java his solution came up too. Indeed his solution has the smallest deviation. Good job davee123!
v=91; Left at 5:21.998001998001996 arrived at 5:27.272727272727273; Deviation in seconds: 0.11988011988023572
v=115; Left at 7:34.00790513833992 arrived at 7:38.18181818181818; Deviation in seconds: 0.47430830039530747
v=123; Left at 2:7.0066518847006645 arrived at 2:10.909090909090908; Deviation in seconds: 0.39911308203986806
v=129; Left at 6:29.006342494714588 arrived at 6:32.72727272727273; Deviation in seconds: 0.38054968287525526
v=139; Left at 1:2.0013080444735114 arrived at 1:5.454545454545454; Deviation in seconds: 0.0784826684106843
v=143; Left at 3:13.006993006993007 arrived at 3:16.363636363636363; Deviation in seconds: 0.4195804195803987
v=147; Left at 5:24.007421150278294 arrived at 5:27.272727272727273; Deviation in seconds: 0.44526901669762253
v=151; Left at 7:35.00301023479831 arrived at 7:38.18181818181818; Deviation in seconds: 0.18061408789861844
v=155; Left at 9:45.99413489736071 arrived at 9:49.09090909090909; Deviation in seconds: 0.3519061583574512
v=165; Left at 2:7.999999999999999 arrived at 2:10.909090909090908; Deviation in seconds: 5.3290705182007514E-14
v=189; Left at 10:52.00577200577201 arrived at 10:54.54545454545455; Deviation in seconds: 0.3463203463205389
v=195; Left at 1:2.9930069930069925 arrived at 1:5.454545454545454; Deviation in seconds: 0.419580419580452
v=203; Left at 3:13.99910434393193 arrived at 3:16.363636363636363; Deviation in seconds: 0.05373936408425095
v=211; Left at 5:24.997845756139597 arrived at 5:27.272727272727273; Deviation in seconds: 0.12925463162417827
v=229; Left at 9:46.99483922191346 arrived at 9:49.09090909090909; Deviation in seconds: 0.30964668519246175
v=251; Left at 2:8.996740311481346 arrived at 2:10.909090909090908; Deviation in seconds: 0.19558131111921284
v=263; Left at 4:19.993086761147595 arrived at 4:21.818181818181817; Deviation in seconds: 0.4147943311443214
v=265; Left at 4:20.006861063464836 arrived at 4:21.818181818181817; Deviation in seconds: 0.4116638078901502
v=277; Left at 6:30.99442074171316 arrived at 6:32.72727272727273; Deviation in seconds: 0.334755497210395
v=279; Left at 6:31.006842619745846 arrived at 6:32.72727272727273; Deviation in seconds: 0.41055718475078606
v=293; Left at 8:41.998138380390934 arrived at 8:43.63636363636363; Deviation in seconds: 0.11169717654397004

Again: The deviation tells you how great the difference is between an exact "integer minute mark" and the actual departure. For example for the odd velocity of v=115km/h George leaves at 7:34.0079 .That yields a deviation of 0.0079 minutes which corresponds to 0.0079*60seconds = 0.474 seconds.
For the odd velocities I set the deviation to less than 0.5 seconds.

davee123's solution is indeed exact:
One can derive an equation for the minute marks when minute and hour hand overlap: minuteMarks = k*60/11, with k from 0 to 11.
The time that elapses during George's train ride (given in minutes) is: timeElapsed = 8/v*60.
The difference is for k=2 and v=165km/h: minuteMarks-timeElapsed = 2*60/11 - 8/165*60 = 8 which is an integer.

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## 1. What is the "train station, speed and time puzzle"?

The "train station, speed and time puzzle" is a classic mathematical problem that involves calculating the time it takes for two trains to meet at a train station when they are traveling towards each other at different speeds.

## 2. How do you solve the "train station, speed and time puzzle"?

To solve the puzzle, you need to use the formula: time = distance / speed. First, determine the distance between the two trains by adding their distances from the train station. Then, plug in the distances and speeds into the formula to calculate the time it takes for the trains to meet.

## 3. What are the key factors in the "train station, speed and time puzzle"?

The key factors in the puzzle are the distance between the two trains, the speeds at which they are traveling towards each other, and the time it takes for them to meet at the train station.

## 4. Are there any variations to the "train station, speed and time puzzle"?

Yes, there are variations to the puzzle that involve factors such as one train starting from a different location or changing speeds during the journey. These variations may require different formulas or additional steps to solve.

## 5. How can the "train station, speed and time puzzle" be applied in real life?

The puzzle can be applied in real life situations such as calculating the time it takes for two cars to meet on a road, or the time it takes for two people walking towards each other to meet at a certain point. It can also be applied in more complex scenarios, such as determining the time it takes for two planes flying towards each other to cross paths.

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