Interesting problem?

Question

construct a simple formula to find out the times of the day when the hour hand and the minute hand of a clock would be exactly opposite to each other.

Answer 1

For time less than or equal to 6 'O Clock, it should be
M = H + 30 minutes
For time after 6 until 12 'O Clock, it should be
M = H - 30 minutes

Answer 2

Let T represent the time in hours. The phase of each hand progresses as some fraction of a full revolution. For the hour hand it would be T / 12 for time from 0 to 12 hours. For the minute hand it would be T mod 1 which is to say the remainder of T when divided by 1. Example at 4.6 hours the minute phase would be 0.6 or 60% of a revolution. Another way to express the mod function is like this: T - n where n is an integer. We want the hands to differ by 1/2 revolutiopn. So the equation is:

T / 12 = T - n + 1/2
T = 12T - 12 n + 6
12 n + 6 = 11 T

12 n + 6
T = ----------------
11

When n = 0 then T = 6/11 hours or 12:32:44
When n = 10 then T = 11 5/11 hours or 11:27:16

n can vary from 0 to 10 during a 12 period.

Answer 3

This is a [time] = [distance] / [rate] problem. Both hands travel for the same amount of time.

Let
d = distance travelled by hour hand
d + 12 = distance traveled by minute hand
x = rate of hour hand
12x = rate of minute hand

We have

t = d/x = (d + 12)/12x
12dx = dx + 12x
12d = d + 12
11d = 12
d = 12/11 hours

Every 12/11 hours the hands will once again be opposite each other.

12/11 hour = 1 hour, 5 minutes, 27 3/11 seconds.

So the hands are opposite each other at times

6:00:00 + (1 hour, 5 minutes, 27 3/11 seconds)*n
where n is an integer.

Answer 4

X = Y + 30.

X is the hour hand.Y is the minute hand

what is your age?, why do you find it is interesting?

Answer 5

X = Y + 30.

X is the hour hand.Y is the minute hand

Answer 6

hour + 30 min

Answer 7

Why do you think it's interesting?