Overlapping Clock Hands

It have formula to find the angle between Hours hand and Minutes hand as Angle = 30H - (11/2)M... Here, after 1'clk, then H=1. Angle = 0. Substitute it in this formula it gives, M = 5.4555... Hence, At 13:5.455 angle should be Zero :) Simple.

Over a twelve hour period the hands will cross over 11 times; at a little after 1.05, 2.10, etc. It is not twelve times because, in the hour starting 11.00, by the time the minute hand reaches the hour hand they are both on 12. Therefore the crossing points occur every 12/11 hours, which equals 1 hour 5 minutes and 27 3/11 seconds, or 1 hour 5.455 minutes.

1 hour on the minute hand is 360°. 1 hour on the hour hand is 30°. 1 minute on the minute hand is 6°. 1 minute on the hour hand is 0.5°. At least one hour must pass before the minute hand and the hour hand overlap. Write an equation, with x being the amount of minutes that pass after 1 pm. The left side will be the hour hand and the right will be the minute. When they are at equal angles from 12 they will overlap. 30 + 0.5x = 6x. ; 5.5x = 30 ; x = 60/11 (answer)

The hands overlap at 1:5.455.so time left=13:5.455-12:00

Note that after every minute, the hour and minute hand get closer together by $6^\circ-0.5^\circ=5.5^\circ$ . After one hour, the hour and minute hand are $30^\circ$ apart. $\frac{30}{5.5}=5.4545\cdots$ , so the answer is 1 hour and 5.455 minutes.

The hands are at the same point 11 times in a 12 hour period. 12/11 = 1.090909. 1.090909 X 60 = 65.454545 minutes

When the hour hands at 1, the minute hand is at 12. By the time the minute hand has reached 1, the hour hand has moved a twelth of the distance between 1 and 2. So the time is 1:05+(1/12+1/12^2+1/12^3....) which is 1:05.45454545

I was able to find the equation by making a ratio between minutes indicator speed and hours indicator speed as the ratio between the two-speed is 12 because when the first indicator moves 60 step , the second one moves 5 steps (step = min), so if we started from 12 am to approximately 1:05 we will find the equation as follows: (60 + x) / x = 12, where x = the number of minutes after 1 am, that is produced from the equation x = 5.4545

ITS EASY BECAUSE AS SECOND HAND MOVES HOUR HAND ALSO MOVES

Let x be the number of minutes needed to overlap the minute hand onto hour hand at the same angle 360deg / 60min = 6deg/min Thus, every digit(1 to 12) implies 30deg. We make our equation like this. (6x-30)/30 = x/60 /* 6deg/min times the x min we need - 30deg for we know that it is impossible to have an equal angle before 1 hour. Then divide the result to 30 to get the ratio and finally, equate it to the ratio of the x min to 60 to know if they are equal.*/ 2(6x-30) = x; 12x - 60 = x; 11x = 60; x = 60/11; x = 5.4545455 ;

the next overlapping of the hour and minute hand is in the time of 1:00 to 2:00

if the hands of the hours and minutes overlapped then the angle from 0 minute to the hour and minute hand is the same

the angle of the hour hand is $\angle H = 30^{o} * (H) + [30^{o} * (M/60)]$ where $M$ is the number of minutes passed from 0 minute and $H$ is number of hours passed from 0 minute

the angle of the minute hand is $\angle M = 360^{o} * (M/60)$

if we equate them and let $H=1$ because it only passes 1 hour we will have a formula of $30^{o} *(1)+ [30^{o} * (M/60)] = 360^{o} * (M/60)$ we will have $M = 5.4545....$

So we will have $1$ hour and $5.455$ minutes before the hands overlapped again.

just after 1 hr (i.e at 1:0 pm) hour hand will be at 1 and minute hand at 12. After then let us consider they meet after some time 't' , after time t angle made by hour hand and minute hand from vertical will be same. Since hour hand rotate by speed 0.5 degree/min and minute hand rotates by speed 6 degree/min. After time 't' angle made by min. hand from vertical=angle made by hour hand from vertical 6 t=30+0.5 t (where 30 degree is angle between hour and minute hands at 1:00 pm ) t=30/5.5=5.4545~5.55 minute total time=1hr 5.55 minute

Hour hand moves by 1/120 degrees per sec ( 360 / 12 x 60 x 60 ) while minute hand runs by 1/10 degrees per sec ( 360 / 60 x 60 ) 1 hour 5.455 minutes = 3926.7 seconds 5.455minutes = 327.3minutes Therefore 3926.7 x 1/120 = 32.7225 & 327.3 x 1/10 = 32.73

Consider the hr hand and min hand coincide after t sec. The hour hand takes 120sec for 1 degree. So it would cover t/120 degrees in t sec. Similarly the min hand takes 10sec for 1degree. So it would cover t/10 degrees in t sec. Time taken for min hand to cover t/10 degrees= time taken for hr hand to cover t/120 degrees=t But the min hand might have completed many rounds before it coincides with hour hand Equating the angular displacements, t/10 = t/120 + 360n where n is a positive integer which denotes the number of rounds completed. Simplifying we get, t = (360 x 120 x n)/11. Converting to he's, we get, t = (12/11)n. Since the first coincidence of the min and hr hand is asked, n=1 as 1is the least value. Hence the and is 1 hr 5.455 min

When the minute hand completes 1 cycle, the hour hand completes 1/12 of a cycle.
Difference in speed= 1- 1/12= 11/12
Distance to catch up= 60 scales out of 60(since the minute hand is directly over the hour hand at 12:00)
Time taken= (60/60)/ (11/12)= 1/ (11/12)= 12/11= 1 1/11 hour= 1hour 5.455 minutes

The hour hand speed is slower than minute hand 1/12 times So it's obviously that they can't overtlap if they start same position. Let the clock run 1 hour more(1:00 - hour hand at 1 and minute hand at 12). We can easily form this (x/12+5) = x (x=5.45), left side is the distance from hour hand to overlap-place, right hand is distance of minute hand. Hence the solution is 13:5.45