How many times do the hands of a clock overlap in a day?

Answer 1

A Better Approach (with reasoning)
There are 2 cases depending on the working of the watch.

Case 1: the movement of the second, minute and hour hands are continuous (not step-wise or click-based)
Answer is: the hour and minute hands overlap every hour.

Case 2 (very unusual): the hour hand jumps from 1 to 2, 2 to 3, ... and so on, as soon as the minute hand crosses (or reaches 12) and the minute hand jumps from 1 to 2, 2 to 3, ... and so on, as soon as the second hand crosses (or reaches 12).
Answer: every 65 minutes. Interviewers often expect this answer cos they do not think accurately. The exact times are:

0000 (12:00 AM)
0105 (01:05 AM)
0211 (02:11 AM)
0316 (03:16 AM)
0422 (04:22 AM)
0527 (05:27 AM)
0633 (06:33 AM)
0738 (07:38 AM)
0844 (08:44 AM)
0949 (09:49 AM)
1055 (10:55 AM)
1200 (12:00 PM)
1305 (01:05 PM)
1411 (02:11 PM)
1516 (03:16 PM)
1622 (04:22 PM)
1727 (05:27 PM)
1833 (06:33 PM)
1938 (07:38 PM)
2044 (08:44 PM)
2149 (09:49 PM)
2255 (10:55 PM)

Reasoning for Case 1:
-----------------------------
When do they overlap? At every (n + (n/11)) hours where n = 0, 1, 2, 3, ..., 24.
How did I find this out? The following is the reasoning i used:

At 0000, the hour and minute hands overlap. So number of overlaps now is 1. The minute hand races away and never again overlaps during the next one hour.

Now, the minute hand moves at 360o/hour and the hour hand moves at 30o/hour.

At 0100, the hour hand would be the 1 mark (or 30o from the 12 mark) and the minute hand would be at the 12 mark. Starting at this position (at 0100), they would overlap when the number of degrees moved by both the minute and the hour hand are the same. Let them overlap at some time, say T, then I can write them in equation form as:

30o + (30o)x(T) = (360o)x(T)

How did I get to this equation? Note that, at 0100, when the minute hand starts moving from the 12 mark, the hour hand is already ahead of the minute hand by 30o. If the minute hand moves at a speed of 360o/hour, then in some time (T), it would cover (360o)(T) degrees. If the hour hand hand moves at a speed of 30o/hour, then in the same time (T), it would cover (30o)(T) degrees. Since the hour hand is already 30o ahead from the 12 mark, the total degrees covered by the hour hand from the 12 mark would then be (30o + the number of degrees covered in time T) which is (30o + (30o)x(T)).

Now the condition when the two hands will overlap is that they should have covered the same number of degrees at a moment (or)

No. of degrees covered by minute hand = No. of degrees covered by hour hand
(or)

30o + (30o)x(T) = (360o)x(T)

If you solve this equation to find the value of T, you would get

30o = (360o)x(T) - (30o)x(T)
30o = (360o - 30o) x T
30o = 330o x T
(or)
T = 30o/330o
T = 1/11

At 0200, the hour hand would be the 2 mark (or 60o from the 12 mark) and the minute hand would be at the 12 mark. Starting at this position (at 0200), they would overlap when the number of degrees moved by both the minute and the hour hand are the same. Let them overlap at some time, say T, then I can write them in equation form as:

60o + (30o)x(T) = (360o)x(T)

Solving this equation, you will the value of T = 2/11

At 0300, using the same reasoning (the hour hand at 90o past the 12 mark) and modifying the equation accordingly (90o + (30o)x(T) = (360o)x(T)), you would get the answer for T = 3/11.

In general, the value for T for every hour is T = n/11 where n = 0, 1, 2, 3, ..., 24. So the exact time when the two hands overlap can be written as:
the hour (n) + the time taken during that hour (T)

(or)
n + n/11
where n = 0, 1, 2, 3, ..., 24.

AM
12:00
1:05
2:11
3:16
4:22
5:27
6:33
7:38
8:44
9:49
10:55 PM
12:00
1:05
2:11
3:16
4:22
5:27
6:33
7:38
8:44
9:49
10:55
22 is correct.

The hands overlap about every 65 minutes, not every 60 minutes. In a day, the hands would only overlap 22 times, as illustrated in the table above.


I would propose that the hands always overlap, as they're both attached at the center of the dial.

If you didn't want to be facetious (or, at least, less facetious), you would still have to ask how many hands were on the clock. It may have a second hand, for example, or be digital (no hands at all).