Remember the "cloks" problems? "at what time around 11:00 will the TWO hands will be on top of each other?"?

Here's one method:

The minute hand goes a full 360° in 1 hour (60 minutes), so that's 6° per minute

The hour hand goes a full 360° in 12 hours (720 minutes), so that's ½° per minute

So each minute the minute hand gets 6° closer to the hour hand, but the hour hand also moves ½° further away. Thus the angle between them is closing at a net of 5½° per minute.

At 11:00 the two hands are 1/12 or 30° apart but the minute hand has already gone past the hour hand. We need the faster minute hand to go backwards to meet up with the slower hour hand.

The angle is 30°, the closure rate is 5.5°/minute, divide to get the time:

30° / 5.5°/minute

= 30/5.5 minutes

= 60/11 minutes

= 5 5/11 minutes

So the time will be 5 5/11 minutes before 11:00

--> 10 hours 54 6/11 minutes

If we want to expand that out to seconds --> 6/11 minutes x 60 second/min = 360/11 = 32.7272.. seconds

Answer:

10 hours 54 minutes 32.7272.. seconds

10 : 55

I've solved this by noting that when the minute hand is near 11, it's late in the hour. If the hour hand is near 11, that's specifically late in the 10:00-to-11:00 hour. Starting from 10:00 and measuring degrees clockwise from straight up:

The minute hand begins at 0 and travels 360 degrees in 60 minutes or 6 deg/min.

The hour hand begins at (10/12)*360 = 300 degrees and travels (1/12) the speed of the minute hand, or 1/2 deg/min.

For those positions to be equal you need:

6m = 300 + m/2 .... with m = minutes after 10:00

(11/2)m = 300

m = 600/11, or about 54.5454 minutes after 10:00

The hands are on top of each other 11 times in a 12 hour cycle.

i.e. minutes past the hour: 60/11, 120/11, 180/11,....600/11

Now, 60/11 = 5.45 minutes and 600/11 = 54.54 minutes

i.e. at 11 hours 5 mins and 27 seconds

and at 10 hours 54 mins and 33 seconds

:)>