rookie1ja 11 Report post Posted March 30, 2007 Clock - Back to the Cool Math Games At noon the hour, minute, and second hands coincide. In about one hour and five minutes the minute and hour hands will coincide again. What is the exact time (to the millisecond) when this occurs, and what angle will they form with the second hand? (Assume that the clock hands move continuously.) This old topic is locked since it was answered many times. You can check solution in the Spoiler below. Pls visit New Puzzles section to see always fresh brain teasers. Clock - solution There are a few ways of solving this one. I like the following simple way of thinking. The given situation (when the hour and minute hands overlay) occurs in 12 hours exactly 11 times after the same time. So itâ€™s easy to figure out that 1/11 of the clock circle is at the time 1:05:27,273 and so the seconds hand is right on 27,273 seconds. There is no problem proving that the angle between the hours hand and the seconds hand is 131 degrees. On every clock we can see that at noon the hour, minute and second hand correctly overlay. In about one hour and five minutes the minute and hour hand will overlay again. Can you calculate the exact time (to a millisecond), when it will occur and what angle they will form with second hand? (Assume that the clock hands move continuously.) Share this post Link to post Share on other sites

Guest Report post Posted June 11, 2007 I disagree. By the rules of the puzzle, the hands must overlap. At 1:05 and 27 seconds the seconds hand is no where near the hour and minute hand. The secound half of the puzzle is easy. The angle of the seconds hands to the hour and minutes hands must by definition be 0. Share this post Link to post Share on other sites

rookie1ja 11 Report post Posted June 11, 2007 I disagree. By the rules of the puzzle, the hands must overlap. At 1:05 and 27 seconds the seconds hand is no where near the hour and minute hand. The secound half of the puzzle is easy. The angle of the seconds hands to the hour and minutes hands must by definition be 0. Have a look at the puzzle again. "In about one hour and five minutes the minute and hour hand will overlay again." Nothing about seconds hand. Share this post Link to post Share on other sites

Guest Report post Posted June 11, 2007 Thank you - josie. Share this post Link to post Share on other sites

Guest Report post Posted June 11, 2007 Hey guys, I am getting a different answer. Solving it like a distance problem: The hour hand moves at 1 unit/hr (1/12 slice of the clock) The minute hand moves at 12 units/hr 1x = 12x - 12 x = 1.0909091 Which means that at 1.0909091 hours the hour hand will be at 1.0909091 hr * 60 min/hr = 65.45454minutes and the minutes hand will be at [(1.0909091 hr * 12) -12] * 60 min/hr = 65.45454minutes. I think that the difference in answers is due to the fact that you assumed that the other remaining 11 overlaps of the hands will happen at regular intervals but they don't the intervals increase from one hour to the next. (Edited: my mistake, the inervals are constant = 1:05:27. I forgot to convert 65.45454 min to non-decimal minutes = 65 mins 27 secs) Share this post Link to post Share on other sites

Guest Report post Posted July 15, 2007 Clock - Back to the Logic Puzzles On every clock we can see that at noon the hour, minute and second hand correctly overlay. In about one hour and five minutes the minute and hour hand will overlay again. Can you calculate the exact time (to a millisecond), when it will occur and what angle they will form with second hand? (Assume that the clock hands move continuously.) I don't understand the question. The way I answered it is that it will occur at noon! and the angle will be 0. Share this post Link to post Share on other sites

Guest Report post Posted July 25, 2007 Because the hour hand moves forward as the minute hand moves, they only cross paths 11 times in a 12 hour period. 12 hours x 60 min x 60 sec = 43,200 seconds. 43,200/11 = 3927 & 3/11 seconds = 1 hour, 5 min, 27 & 3/11 seconds. So the second had will be at 27 & 3/11 seconds, or approx 27.2727. (By the way, this is exactly 5/11 of the way around the clock face, which will help us with the next problem). As to the angle formed between the hour/minute hands compared to the second hand, we need only realize that the hour and minute hand have traveled exactly 1/11 of the circumference of the clock face, while the second hand is exactly 5/11 of the same circumference. Since the difference is 4/11, and a complete circle is 360 degrees, the distance is 360 x 4 / 11 = 130 & 10/11 or approx 130.90909 degrees. Share this post Link to post Share on other sites

Guest Report post Posted July 31, 2007 The only way to figure this out is to first determine where each hand is individually. In an hour, the hour hand should be exactly on the 1:00. In one hour and five minutes, it will be 2.5 degrees past the one, since the hands are moving continuously. On a 360 degree clock, the hour hand moves 2.5 degrees every five minutes. Now, the minute hand, which is easy. It will be exactly on the one. There is a flaw with the question, since it will not overlap the hour hand exactly. The second hand will be on the 12. Also, easy, since it will make one full rotation for every minute, and therefore will end up back on the 12 after every full minute has passed. No matter how many minutes you have, in this case 65, the second hand will still be back on the 12. The second hand will be at a 30 degree angle to the minute hand, but at a 32.5 degree angle to the hour hand. Share this post Link to post Share on other sites

Guest Report post Posted August 1, 2007 Okay, laughing at myself now. I posted before looking at the solution. It says ABOUT one hour and five minutes, and I calculated for exactly one hour and five mminutes, so the hands can overlap, and the second hand would be at a different place at that exact moment. D-uh... Share this post Link to post Share on other sites

Guest Report post Posted August 1, 2007 Okay, laughing at myself now. I posted before looking at the solution. It says ABOUT one hour and five minutes, and I calculated for exactly one hour and five mminutes, so the hands can overlap, and the second hand would be at a different place at that exact moment. D-uh... And let that be a lesson for you Ms Peggy to read before you talk. Share this post Link to post Share on other sites

Guest Report post Posted November 6, 2007 it is very simple and as usual people are over complicating the issue. at 1 hr and 5 mins both the hour hand and the minute hand are on the 1 because that is 1 hr and 5 minutes after 12. now for it to be exact the second had has to be on the 12 at that exact second for it to be 1 hr and 5 minutes past 12 on a complete rotation. therefore the second hand is 30 degrees exactly from both the hour hand and the minute hand which are both exactly on each other. 1 Share this post Link to post Share on other sites

Guest Report post Posted November 8, 2007 its 1:05 and five seconds, 30 degrees then 2:10 and 10 seconds, 60 degrees next 3:15 and 15 seconds, 90 degrees. 1 Share this post Link to post Share on other sites

Guest Report post Posted November 9, 2007 oops, misread, original answers sound correct now. Isn't this more of a math, than logic puzzle Share this post Link to post Share on other sites

Guest Report post Posted November 21, 2007 I started to write this asking a question, but managed to answer it while typing. I was doing it mathematically, and it works like this: where 2 * pi is a 360 degree angle in radians t is the time in milli seconds so we want hours angle = minutes angle +/- 2 * pi * X (+/- 2 * pi * X is used to denote that we don't care how many more or less times the minutes turn full turns extra than the hours one, note that X can be an integer between 0 and 11 inclusive) so 2 * pi * t / (12 * 60 * 60 * 1000) = 2 * pi * t / (60 * 60 * 1000) +/- 2 * pi * X solving for t in terms of X yields: t = 12 * 60 * 60 * 1000 * X / 11 you can get the time for t at X = 0, 1, 2, 3, .. , 11 to figure out the time in milliseconds from 12 of course if you don't like pis and radians, you could have used degrees or nothing (parts of a circle) to simplify as it cancels out from the equation anyway. OK, so it is a bit more confusing but it's the way I went Share this post Link to post Share on other sites

Guest Report post Posted December 5, 2007 here it is since the hour and minute hand are moving calculate the time it takes for the minute hand to catch the hour hand second hand: 6 degree/sec minute hand: 1/10 degree/sec hour hand: 1/120 degree/sec for the first time t*1/10-(t*1/120+30deg)=0 solve for t t=3600/11 = 327.2727 seconds it will take 2*t for second time or 654.5454... etc angle of minute hand=hour hand= t*1/10 angle of second hand= t*6 subtract the difference to get the angle between the here is the table: N t Min Sec Ang SH Ang MH Angle 0 0.00 0.00 0.00 0.00 0.00 0.00 1 327.27 5.45 27.27 163.64 32.73 130.91 2 654.55 10.91 54.55 327.27 65.45 261.82 3 981.82 16.36 21.82 130.91 98.18 32.73 4 1309.09 21.82 49.09 294.55 130.91 163.64 5 1636.36 27.27 16.36 98.18 163.64 65.45 6 1963.64 32.73 43.64 261.82 196.36 65.45 7 2290.91 38.18 10.91 65.45 229.09 163.64 8 2618.18 43.64 38.18 229.09 261.82 32.73 9 2945.45 49.09 5.45 32.73 294.55 261.82 10 3272.73 54.55 32.73 196.36 327.27 130.91 11 3600.00 60.00 0.00 0.00 0.00 0.00 Share this post Link to post Share on other sites

Guest Report post Posted March 17, 2008 I calculate it like the below way: For hour hand, 60 minutes = 30 degrees (1/12 of 1 round), so the rate = 30 / 60 = 1/2 deg per minute For minute hand, 60 minutes = 360 deg ( 1 full round), so the rate = 360 / 60 = 6 deg per minute let t = the time we interest, so in any time the formula will be: for hour hand = 6 * t - n * 360 ( n = whole number, so that if the number of round > 1, just deduct accordingly so that the angle within 0 to 360) and for minue hand = 1/2 * t when hour hand and minute hand overlay: 6t - n(360) = t/2 so, (11/2)t = n(360) t = (2/11) * n(360), where n = 0, 1, 2, 3, 4, .... t = 0/11 min, 720/11 min, 1440/11 min, 2160/11 min ...... etc since we talking about the time when around 1 hour and 5 minuts, the answer for t should be very near to 65 minutes, which is the second one, t = 720/11 min or t = 65 min and 27.273 sec the angle by that time is 720/11 * 1/2 = 32.73 deg, and by that time, the second hand is making a degree of ( 6 * 27.273) = 163.64 deg, and the angle formed is 163.64 - 32.73 = 130.9 deg. Correct? I didn't see the answer from Spoilers! Share this post Link to post Share on other sites

Guest Report post Posted March 20, 2008 Because the hour hand moves forward as the minute hand moves, they only cross paths 11 times in a 12 hour period. 12 hours x 60 min x 60 sec = 43,200 seconds. 43,200/11 = 3927 & 3/11 seconds = 1 hour, 5 min, 27 & 3/11 seconds. So the second had will be at 27 & 3/11 seconds, or approx 27.2727. (By the way, this is exactly 5/11 of the way around the clock face, which will help us with the next problem). As to the angle formed between the hour/minute hands compared to the second hand, we need only realize that the hour and minute hand have traveled exactly 1/11 of the circumference of the clock face, while the second hand is exactly 5/11 of the same circumference. Since the difference is 4/11, and a complete circle is 360 degrees, the distance is 360 x 4 / 11 = 130 & 10/11 or approx 130.90909 degrees. VERY NICE EXPLANATION. Share this post Link to post Share on other sites

Guest Report post Posted April 24, 2008 Because the hour hand moves forward as the minute hand moves, they only cross paths 11 times in a 12 hour period. 12 hours x 60 min x 60 sec = 43,200 seconds. 43,200/11 = 3927 & 3/11 seconds = 1 hour, 5 min, 27 & 3/11 seconds. So the second had will be at 27 & 3/11 seconds, or approx 27.2727. (By the way, this is exactly 5/11 of the way around the clock face, which will help us with the next problem). As to the angle formed between the hour/minute hands compared to the second hand, we need only realize that the hour and minute hand have traveled exactly 1/11 of the circumference of the clock face, while the second hand is exactly 5/11 of the same circumference. Since the difference is 4/11, and a complete circle is 360 degrees, the distance is 360 x 4 / 11 = 130 & 10/11 or approx 130.90909 degrees. I agree with you up till the last part, if it takes an hour five minutes and approx. 27.2727 seconds, then you have to apply that to the one o'clock hour also. so the next time the hour and minute hands would overlay would be 2 hours ten minutes and approx. 54.5454 seconds from noon/midnight (0:0:0). If you are talking about the angle as the clock moves, from the ten minute mark to the ~54.5 second mark is between 253 and 254 degrees. Share this post Link to post Share on other sites

Guest Report post Posted May 3, 2008 My solution: The hour hand moves at 1/2deg per sec and the minute hand at 6deg/sec. The minute hand will move 360 more degrees than the hour hand when they meet. calling minutes x we see: 1/2 x+360=6x so (11/2)x=360 and x=720/11 minutes. The degree covered according to the hour hand is .5x or (1/2)*(720/11) = 360/11 Since 720/11=65minutes and 5/11 seconds, the second hand is 5/11(360) or 1800/11 1800/11-360/11=1440/11 degrees between the hour and second hand. Share this post Link to post Share on other sites

Guest Report post Posted May 11, 2008 (edited) if its 1 hour 5 mins after 12noon exactly its 13:05, theres no mention of the second hand...you are overcomplicationg it && the angle is 30degrees i think because a full circle is 360 degrees so you quarter it to make 90 and devide by 3...making the answer to the riddle 13:05 and 30 degrees.....thereis no need for long sums or pi...=S Edited May 11, 2008 by -x-babiiee-hezzahh-x- Share this post Link to post Share on other sites

Guest Report post Posted May 16, 2008 if its 1 hour 5 mins after 12noon exactly its 13:05, theres no mention of the second hand...you are overcomplicationg it && the angle is 30degrees i think because a full circle is 360 degrees so you quarter it to make 90 and devide by 3...making the answer to the riddle 13:05 and 30 degrees.....thereis no need for long sums or pi...=S What some people aren't taking into account is that the hour hand (just like the other hands on this clock) is moving continuously. In other words, it won't just be stationary over the 12 from 12:00 to 12:59 and then suddenly jump to the 1 at 1:00. It continues to move at a rate of 1/12 of the arc of the circle (30 degrees) per hour. I liked this riddle, a fun one to solve and a few different ways to do it (I used the speed of each hand around the clock). Share this post Link to post Share on other sites

Guest Report post Posted May 21, 2008 What some people aren't taking into account is that the hour hand (just like the other hands on this clock) is moving continuously. In other words, it won't just be stationary over the 12 from 12:00 to 12:59 and then suddenly jump to the 1 at 1:00. It continues to move at a rate of 1/12 of the arc of the circle (30 degrees) per hour. I liked this riddle, a fun one to solve and a few different ways to do it (I used the speed of each hand around the clock). so did i get it right? or no?...im confused i jus did it the logic answer bcuz its a logic question innit Share this post Link to post Share on other sites