The Clock Problem

I recall in my primary school days, we were taught to solve the clock problem in arithmetic, an example of which is as follows:

"Between the hour 8 and 9 o'clock, what time will the hour-hand and the minute-hand overlap with each other?"

We know that at 8 o'clock, the minute-hand is at the 0-minute position while the hour-hand is at the 40-minute position. As the minute-hand moves to catch up with the hour-hand, the hour-hand also moves albeit at a much slower pace. In fact, when the minute-hand moves 60 minutes (1 round), the hour-hand moves 5 minutes (1 hour). So they are moving at different speeds in the ratio of 12 : 1. When the minute-hand reaches the 40-minute position, the hour-hand would have moved a little bit further. We also know that the minute-hand will catch up with the hour-hand somewhere between the 40-minute and 45-minute position (They cannot go beyond the 45-minute position otherwise the hour-hand would have gone beyond 9 o'clock already).

In those school days, we were taught to obtain the answer arithmetically (frankly without a full understanding) as:

"40 x 12/11 or 43.64 minutes after 8 o'clock"

As we advanced into secondary school, we learnt Algebra. With the power of Algebra, the problem could be easily (and logically) understood and solved :

Let m be the position (in minutes) of the minute-hand and h be the position (in minutes) of the hour-hand.
As the minute-hand moves m minutes, the hour-hand moves m/12 minutes.
At 8 o'clock, the hour-hand is at 40 while the minute-hand is at 0 (i.e. the hour-hand has a head start of 40 before the minute-hand tries to catch up).
Hence, the position of the hour-hand (h) at any time is related to the position of the minute-hand (m) as follows:

h = 40 + m/12     ............... (1)

When the position of the hour-hand overlaps with the minute-hand,

h = m

Putting this back into the relation (1),

m = 40 + m/1 2

Solving this equation, we have:

m = 40 x 12/11 or 43.64 minutes

Hence, the time between 8 and 9 o'clock when the hour-hand and minute-hand overlap with each other is 43.64 minutes past 8.

Naturally, the problem can be re-phrased to ask for the same situation in any hour rather than between the hour 8 to 9 o'clock. In that case, we just replace the head start position 40 by the corresponding starting position of the hour-hand accordingly.

In addition, there are other variations of the clock problem: e.g.

A. "Between the hour 8 and 9 o'clock, what time will the hour-hand and the minute-hand form a straight line with each other?"

As from above, when the hour-hand forms a straight line with the minute-hand,

h = m + 30

Putting this back into the relation (1),

m + 30 = 40 + m/12        

Solving this equation, we have:

m = 10 x 12/11 or 10.91 minutes

Hence, the time between 8 and 9 o'clock when the hour-hand and minute-hand form a straight line with each other is 10.91 minutes past 8.

B. "Between the hour 8 and 9 o'clock, what time will the hour-hand and the minute-hand form a right angle  with each other?"

As the hour-hand can be right-angled in front of or behind the minute-hand, so when the hour-hand forms a right angle with the minute-hand,

h = m + 15 or h = m - 15

Putting this back into the relation (1),

m + 15 = 40 + m/12 or m - 15 = 40 + m/12

Solving this equation, we have:

m = 25 x 12/11 or 27.27 minutes

or

m = 55 x 12/11 or 60 minutes

Hence, the time between 8 and 9 o'clock when the hour-hand and minute-hand form a right angle with each other is either 27.27 minutes past 8 or 9 o'clock.

If at any point of the calculation, the number becomes negative, add 60 to the number before going on, reason being the minute-hand runs only between 0 and 60 in the clock face.