Find the times?

Question

In a 12 hour period there are 11 times when the hour and minute hand are ontop of each other (form a 0-degree angle) find the first two times.

Answer 1

Let m = minutes elapsed during hour (integer)
let h = hours elapsed after 12:00 (integer)
α = angle of the minute hand (respect to vertical)
β = angle of the hour hand (respect to vertical)

α = m*360/60 = 6*m
Note: The hour hand shifts slightly each minute
β = h*360/12 +m* 30/60 = 30*h + m/2

6m = 30*h + m/2
m = (60/11)*h

I suppose it's best to round here.

ACTUALLY: Technically speaking other than 12:00 the hands never do lie precisely on top of each other. In actuality they approach within a small angle for one minute and than the next minute (when they both shift) the minute hand jumps slightly ahead by a small angle.

First time
h = 0, m = 0 or 12:00
Second time
h = 1, m = 5 or 1:05
Third time
h = 2, m = 11 or 2:11
Fourth time
h = 3, m = 16 or 3:16
Fifth time
h = 4, m = 22 or 4:22
Sixth time
h = 5, m = 27 or 5:27
Seventh time
h = 6, m = 33 or 6:33
Eigth time
h = 7, m = 38 or 7:38
Ninth time
h = 8, m = 44 or 8:44
Tenth time
h = 9, m = 49 or 9:49
Eleventh time
h = 10, m = 55 or 10:55

Answer 2

It depends on the starting positions of the two hands. Here I will assume that the "zero" time is 12 o'clock.

speed of minute hand: 360°/60min=6°/min
speed of hour hand: 360°/(60*12)min=0.5°/min

every minutes the minute hand gets (6-0.5=)5.5 degrees further ahead of the hour hand.

at the beginning, one hand is on top of the other; if this situation is to happen again, the minute hand needs to go 360° more than the hour hand.

the time needed: 360°/5.5°=720/11 minutes

one hand is on top of the other every 720/11 minutes.

the first time is 720/11 minutes after 12:00, which is approx. 1:05;
the second time is 1440/11 minutesafter 12:00, which is spprox. 2:11.

Answer 3

1 x 11/12= 11/12 --> 55 minutes.

2 x 11/12= 11/6 ---> 110 minute = 1 hour and 50 minutes