I gave a mental ability test today in which this question was asked.

The hour and minute hands of a clock are indistinguishable. How many minutes are there in a day (24 hours) when it is not possible to tell what time it is from the clock?

What I tried: I pondered a lot over this question... since both hands are same, the only difference lies in their speed. The hour hand turns 30 degrees in an hour and the minute hand turns a whole 360 degrees (one complete circle). So, in order to count the number of minutes, i need to calculate sum(probably) both of their rotations. The $24*60 = 1440$ of actual minute hand is to be counted, and hours hand's extra rotation has to be added.

But, here I got struck, how to proceed was a big problem.
Can someone please explain on how to solve this question in detail.

  • $\begingroup$ Is the clock a 12-hour or 24-hour clock? $\endgroup$ – 2012rcampion May 28 '16 at 19:08
  • $\begingroup$ I presume that it's a typical analog clock/watch, so 12 hour clock. $\endgroup$ – ABcDexter May 28 '16 at 19:10
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    $\begingroup$ That question is really ambiguous. I mean, what are they asking, exactly? My first reply would be that it's a trick question. The number of minutes in a day doesn't change just because your clock's broken. $\endgroup$ – Mike M. May 28 '16 at 19:20
  • $\begingroup$ @MikeM. Exactly! You are right, and using the same logic I wrote 1440 as my answer. $\endgroup$ – ABcDexter May 28 '16 at 19:22

I can clarify what the question is asking. The top row of the below picture has two clocks; the left is a bit after 1:15, the other at around 3:06. Below each clock is the what happens when you trim its minute hand to the length of its hour hand. Notice the results are identical, so at either of these times, you wouldn't be able to tell what time it was; you would only know it was one of those two possibilities.

The question is asking how many times a day this occurs. A better way of phrasing the question is:

"How many times are there in a day when it is not possible to tell what time it is by looking at the clock at that instant?".

enter image description here

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    $\begingroup$ Well, that makes a lot more sense. Visual aids help so much. That question is just horribly worded. $\endgroup$ – Mike M. May 28 '16 at 21:17
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    $\begingroup$ Ironically enough you were the one who asked the question a year ago. $\endgroup$ – Joe Z. May 28 '16 at 22:12
  • $\begingroup$ That's really helpful, wish they used simple words to ask questions :) $\endgroup$ – ABcDexter May 29 '16 at 2:24

Multiple interpretations

If we have no idea of the time of day there are

$24*60=1440$ unidentifiable minutes
since even if we can identify a time in $12$hr format we still do not know if it is a.m. or p.m.

If the hands move smoothly, we only need to read a $12$hr format, and we may observe the clock for some portion of the minute in question there are

$0$ unidentifiable minutes
since we may observe which hand is moving faster and identify it as the minute hand

If the hands move smoothly, we only need to read a $12$hr format, and we may only observe the clock for an instant there are

$264$ possibly unidentifiable minutes, depending on the instant we observe during that minute and the accuracy with which we can measure angles
since, the hands cross $22$ times ($11$ times for each $12$ hours) and there are $12$ ambiguous instants between each crossing, $12\times22=264$

There are ways to implement a clock such that the hands do not move smoothly too (e.g. see the answer by Dan Russell)

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  • $\begingroup$ A good answer, so what should be the answer as there were no options but a blank to fill and submit ? $\endgroup$ – ABcDexter May 29 '16 at 2:21
  • $\begingroup$ I'm not a mind reader but I imagine they are after my third version $\endgroup$ – Jonathan Allan May 29 '16 at 6:14

Somewhat surprisingly, if you take the clock to operate at single-second intervals, there are no times during the day when you can't tell what time it is.

To do this, I defined 43200 discrete positions on the clock, because 12 hours = 43200 seconds. So each time the second hand ticks, the hour hand moves one position and the minute hand jumps 12 positions (since the minute hand moves 12x as fast as the hour hand).

Then, you can track the positions of the hour hand (0, 1, 2, 3, 4, ...) and the minute hand (0, 12, 24, 36, 48, ...) (minutes are mod 43200) and make a big list of pairs of coordinates.

Ambiguous times would occur when:

  1. The hour and minute positions are not equal AND
  2. The reversed positions also appear in the list.

For example, the (hour, minute) position (1,12) is found in the list, but (12,1) is not. And if you scan the list (which I did in Python), there's just one position that appears reversed in the list: (0,0), which doesn't satisfy condition 1 and thus is not ambiguous.

Now I see JoeZ. has posted a much more elegant and mathematical answer on the duplicate question, but I'll still pop this up for those of us less math-capable.

Quick python check:

# Positions for the hours hand, 0 though 43199
hours = range(0,43200)
# The minute hand moves 12 positions per second
minutes = [i*12%43200 for i in hours]
# Create a list of (hour position, minute position)
positions = zip(hours, minutes)
for position in positions:
    # Flip this instance around
    flipped = (position[1], position[0])
    # Check if the flipped version is also present
    if flipped in positions:
        print "Values: %s, %s" % (str(position[0]), str(position[1]))
        print "Present in both orientations.\n"
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  • $\begingroup$ Great, can you please post the link to Python code. $\endgroup$ – ABcDexter May 29 '16 at 2:19
  • $\begingroup$ Are you sure that this code works? As its output is empty, and using a variable to count no. of flipped positions, the value of the variable comes out to be 0. $\endgroup$ – ABcDexter May 29 '16 at 6:35
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    $\begingroup$ @ABcDexter Yeah it works using Python2.7. You should immediately see "Values: 0, 0" as it runs. If you didn't copy and paste, make sure all of the position and positions variables have the correct pluralization. $\endgroup$ – Dan Russell May 29 '16 at 12:41

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