21 replies to this topic

[rookie1ja]

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.

Spoiler for Solution

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 it’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.

Spoiler for old wording

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.)

[josiedacat]

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.

[rookie1ja]

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.

[josiedacat]

Thank you - josie.

[Garrek99]

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)

[domino9]

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.

[xandarr]

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.

[peggy]

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.

[peggy]

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...

[Garrek99]

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.