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Train station, speed and time puzzle

  1. Jul 30, 2009 #1
    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.
     
    Last edited: Jul 30, 2009
  2. jcsd
  3. Jul 30, 2009 #2
    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).
     
  4. Jul 30, 2009 #3
    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
     
    Last edited: Jul 30, 2009
  5. Jul 30, 2009 #4
    How did you get that answer?
     
  6. Jul 30, 2009 #5
    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
     
  7. Aug 6, 2009 #6
    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.
     
  8. Aug 6, 2009 #7
    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.
     
    Last edited: Aug 6, 2009
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