A 12-hour clock has its hour, minute and second hands all pointing at '12' at midday, as it should. Sometime between 4 and 5 pm the minute hand precisely covers the hour hand. Where is the second hand at this time? (Most clocks make the second hand move in jumps but for this puzzle you should assume smooth angular movement of the three hands).


Let the time when the two hands are coincident be 20 minutes and x seconds passed 4.

The minute hand moves at the rate of 360 degrees in 1 hour (3600 seconds of time) or 0.1 degrees per second. So during those extra x seconds the minute hand moves through 0.1x degrees.

The hour hand moves at the rate of 360 degrees in 12 hours which works out at 1/120 th degree per second. In 20 minutes and x seconds (1200+x seconds) it will therefore move through (1200+x)/120 degrees.

Coincidence of the two hands are when both are 0.1x degrees after the '4' on the clock so we can equate the two angles:

                (1200+x)/120 = x/10
multiplying 120:    1200 + x = 12x
from which              1200 = 11x
so                         x = 1200/11
                             = 109.091 seconds
Subtracting 60 seconds leaves the second hand at 49.091 which is the answer: the time is 21 minutes and 49.091 seconds past 4 pm.

A more elegant method is to consider that the two hands of the clock are coincident 11 times in every 12 hours, and spaced at equal intervals, so the answer must be at one of these occasions. The obvious one to choose is the fourth so the time must be 4/11ths of 12 hours or 4.363636.. hours which converts to 4 hours, 21 minutes and and 49.01 seconds.

I received four answers, all correct, from John Stafford, Richard Birkhill, Michael Hodge and Clem Robertson. John Stafford won the draw and the £5 book token. Thank you for your entries.

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