School-Safe Puzzle Games

It is Time….

Starting today, we have a series of 4 clock puzzles.  The first three are easy to medium in difficulty, and can be considered a warm up for the fourth one. This last is extremely difficult and should satisfy any of our math wizards.

Here is the 1st:

At 12 o’clock, the hour and minute hands of the clock are superimposed. At exactly what time will they next be superimposed?

The 2nd will go up early Thursday am. Big Thanks go out to Bilbao for helping prepare this challenge!

Also, don’t forget to check out the recently posted Click-A-Flick, as well as the Find the Fault.

31 Comments to “It is Time….”


  1. Migrated | Profile

    At 1:05 AM/PM depending on whether 12 o’clock refers to midnight or midday.


  2. cheddarmelt | Profile

    1:05:27.27


  3. Benson | Profile

    im pretty sure its about 1:05 or 1:06


  4. joe | Profile

    The hands will overlap 11 times as they travel round the full face of the clock (12 including the 12 oclock position again).
    A full turn of the face of the closk is 360 degrees so they will superimpose every ( 360 / 11 =) 32.727272degrees as they travel round.
    We can calculate how many minutes this equates to on our face in order toknow the position of the hands. So:
    min 32.727272
    —- = ———
    60 360


    Solving for min = 5.454545
    So at every 5.4545 minute mark of the clock.


    After 12 oclock this will be at 1.054545 minutes
    So 1:05 and 27.27secs


  5. BillyPilgrim | Profile

    Sorry, disregard my last comment, my formula was wrong. I believe the answer is 1:05.45, with the 45 in the decimal repeating. Or if you want seconds too, 1:05:27.27 with the 27 repeating.


  6. Shawn | PUZZLE GRANDMASTER | Profile

    1:05:27 plus 27/100 of a second.


    It is obvious that the next occurrence is sometime after 1:05, at which point the minuted hand will be at the “1″ and the hour hand will be slightly past the “1.” To make this less confusing, I structured the math beginning at 1:05, so that all calculations are relative to the number “1″ on the clock face.


    The minute hand travels from one hash-mark on the watch to the next in 1 minute or 60 seconds. The hour hand travels the same distance 5 times per 60 minute interval, or once every 12 minutes.


    At 1:05, the minute hand is at the “zero” point relative to the number “1″, and the hour hand is at 5/12 (5 minutes times 1/12) the distance from “zero” to the first hash-line between the numbers 1 and 2.


    The formula is:


    [starting point + (speed per minute)*(minutes passed)] minute hand
    = equals =
    [starting point + (speed per minute)*(minutes passed)] hour hand


    minute hand starting point = 0
    minute hand speed per minute = 1 hash-mark/minute
    hour hand starting point = 5/12
    hour hand speed per minute = 1/12 hash-mark/minute
    number of minutes elapsed = X


    0 + (1)*(X) = 5/12 + (1/12)*(X)
    X = 5/11 = .454545 minutes = 27.27 seconds


  7. michaelc | Profile

    Minute hand will be 60/11 past or 5 5/11 and the hour hand will be 1/11 past.


    Time will be about 1:05::27.


    Can’t wait for the other 3! :)


  8. Oneiric | Profile

    At five past one. (Or five past one, 25 seconds and some milliseconds, counting the movement of the hour hand according to the minutes and seconds passed.)


  9. alderney girl | Profile

    the next time the two clock arms will be superimposed is 5 minutes past one 1:05


  10. doar823 | Profile

    Just a little after 1:05…if the minute hand is rotating once every minute and the hour hand is rotating 1/60 every minute, I think it will be at 1:05:01


  11. Obiwan | Profile

    1:05 1 hr=1/12 of the dial, 1/12 of 60 minutes=5


  12. aaronlau | Profile

    Guess it is how precise should the answer be.
    1 o’clock and 5 minutes.
    And 45.45 seconds?


  13. bach | Profile

    just registered, my first time answering here… not sure if i’m right.


    a “tick” is 1/60th of the circumference of the circle.
    the minute hand moves 1 tick every minute.
    the hour hand moves 1 tick every 12 minutes.


    then, we can say the pace of the minute hand is x and the pace of the hour hand is 1/12x when x is the number of minutes elapsed.


    using common sense we know the two hands will meet sometime after 1:00, so we can represent the time 1:00 in terms of ticks as 0 for the minute hand and 5 for the hour hand. since we want to find the time at which the ticks on both hands are equal, we factor in the rates: x + 0= 1/12x + 5. once solved, x= 60/11, where x is the time elapsed after 1:00. thus, the answer is 01:05:27:45, hh:mm:ss:ms.


  14. bach | Profile

    oooops i meant 01:05:27:27!


  15. deadskin | Profile

    Angle(Hour)=(H/12)*360 + (M/60)*30
    Angle(Minute) = (M/60)*360


    (H/12)*360 + (M/60)*30 = (M/60)*360


    30*H + (1/2)*M = 6*M
    60*H + M = 12*M
    60*H = 11*M
    (60/11)*H = M


    For each Hour (1, 2, 3, …;) calculate the minutes that need to pass for the minute hand to catch up to the hour hand.


    HH:MM::SS
    12:00::00am
    01:05::27+(3/11)am
    02:10::54+(6/11)am
    03:16::21+(9/11)am
    04:21::49+(1/11)am
    05:27::16+(4/11)am
    06:32::43+(7/11)am
    07:38::10+(10/11)am
    08:43::38+(2/11)am
    09:49::05+(5/11)am
    10:54::32+(8/11)am
    12:00::00pm
    01:05::27+(3/11)pm
    02:10::54+(6/11)pm
    03:16::21+(9/11)pm
    04:21::49+(1/11)pm
    05:27::16+(4/11)pm
    06:32::43+(7/11)pm
    07:38::10+(10/11)pm
    08:43::38+(2/11)pm
    09:49::05+(5/11)pm
    10:54::32+(8/11)pm


  16. deejam | Profile

    At 1:05 the hands superimposed after 12 0′clock…


  17. lograh | Profile

    Depends somewhat on the mechanics of the hands, if they move in a smooth motion or in steps — I’ll assume they have a smooth movement. Also, I’ll assume we are talking about a standard 12-hour clock face (I’ve seen some 24-hour clock faces where the hour hand takes 24 hours to go around).


    All that assumed, I’d guess they are superimposed next at somewhere around the 13:05-13:06 minute. Not precisely at 13:05, as the hour hand would have moved 1/12 of the distance between the “1″ and the “2″, but that’s a really small movement so it wouldn’t be so late as 13:07 when they line up. Probably roughly 13:05:10 when it happens. Maybe?


    Even more interesting, how much longer before this kind of puzzle becomes undo-able for the majority of people since they have been raised on digital clocks? :)


  18. hex | PUZZLE MASTER | Profile

    m: minutes
    h: hours
    The angular position of the hours hand = 2 x Pi / 12 x (h + m / 60)
    The angular position of the minutes hand = 2 x Pi x m / 60


    The hands superimpose when both angular positions are equal (actually separated by 2 x k x Pi where k is an integer, but the conditions on the hours and minutes dictate that k=0):
    m = 60 / 11 x h
    h=1 -> m=60/11 = ~5.45 minutes
    so ~1:05


  19. Hendy | Profile

    The actual answer is beyond the resolution of the standard dial-face clock. It would be sometime between 1:05 and 1:06. I am sure there is a formula that works, but I only look up formulas, I don’t generate them any more. I gave up that practice in my first attempt at college. However, I solved it via a form of interpolation and stopped at somewhere between 12567/138240 and
    12568/138240 where the units are fractions of a rotation about the circle of the face (clockwise of course). To convert these to times it is somewhere between 1:05.45442708333333
    and 1:05.45486111111111.



    Here’s the premise: for each minute about the clock-face, the hour hand has to turn 1/60th of an hour beyond the current hour. In other words when the clock shows 1:01 to human recognition, the actual time is in separate hour hand rotation and minute hand rotation. The hour hand will have moved 1/60th of an hour past one o’clock and the minute hand has moved 1 minute (also 1/5th or 12/60th of an hour mark) past 12 o’clock on the dial. If you take the greatest common factor of 720 (12 hours * 60 minutes in a trip around the clock with the hour hand) and 60 (minutes in a trip around the clock with the minute hand) you get 720. Therefore, 1:01 is now 61/720th rotations on the hour hand and 12/720th (or 1/60th) rotations on the minute hand. Additionally, 1:02 is now 62/720th hours and 24/720th (or 2/60th) minutes. To further illustrate, 1:30 is 90/720th hours (half way between 1 and 2 on the clock-face) and 360/720th (30/60th or one-half) minutes.


    Or more simply, the hour hand still moves while the minute hand moves. It does not stop exactly at 1 and wait until after 1:59 to click to 2 (unlike a digital clock).


    If the question is changed to include, “within the resolution of a minute mark on the face of the clock”, then the answer would be only 12:00 (twice a day). This is the only spot where the two superimposed clock hands land on an exact minute mark.


  20. Roel | Profile

    I went physical on this one, i.e. I got my physics book from the shelves.


    I assumed the hands of the clock have a constant angular velocity, say v_h for the hour hand and v_m for the minute hand. The formulas for the angles of the hands at a given time t are thus given by:


    h = h_0 + v_h * t
    m = m_0 + v_m * t


    Assuming that the starting position (t=0) is 12 o’clock, this means that


    h_0 = m_0 = PI/2 (90 degrees, pointing up)


    The velocities of the hands are


    v_h = – 2*PI / (12*60*60) (it goes 360 degrees in 12 hours)
    v_m = – 2*PI / (60*60) (it goes 360 degrees in 1 hour)


    At the moment that the hands cross again for the first time, there is a difference of 360 degrees (=2*PI) between the angles of the hands. So equating h = m + 2*PI gives:


    PI/2 – 2*PI*t / (12*60*60) = PI/2 – 2*PI*t / (60*60) + 2*PI


    Solving this equation to t gives:
    t = 3927.27272727… seconds


    Which means that after 1h (3600s) 5 minutes (300s) 27.2727s the two hands are at the same position.


  21. Roel | Profile

    BTW, Dr. Bilbao, I solved this one in Bilbao! :o)


  22. Mashplum | PUZZLE MASTER | Profile

    1:05:27 (and 3/11 seconds)


  23. ludiavolo | Profile

    Okay,


    the correct time will occur shortly after 1 o’clock so I’ll start my problem when the hour hand at 1 o’clock and the minute hand at 12 o’clock.


    i thought of this as a race problem where the two hands are chasing each other around a circle. the hour hand has a 30deg head start because it’s starting at 1 o’clock.


    the angular frequency of the hour hand is 360deg/12hr and the angular frequency of the minute hand is 360deg/1hr.


    i need to set the angular positions of the two hands equal:


    30deg + 360deg*t/12hr = 360deg*t/1hr


    solving for t


    t = 1/11 hr (this is how long AFTER 1 o’clock the hands will coincide)


    doing some conversions, the time when the minute hand and hour hand will coincide is approximately: 1:05:27


  24. RK | Founder | Profile

    Very good, quite a few survived the 1st round :) Some old names I recognize have returned for the challenge, and several excellent explanations are above.


    #2 will be going up in just a couple minutes


  25. RK | Founder | Profile

    Wherefore art thou Suineg!?!


  26. APEX.JP | Profile

    The hands will next be superimposed at exactly 1:05:27.27


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