# Solution

I left home shortly after 6pm and returned back shortly before 7pm. So, let's say I left home at 6 hours and x minutes, and returned back at 6 hours y minutes.

First and foremost, let me convert everything into hours for ease of calculations. Now, 60 minutes make an hour, so x minutes = x/60 hours. Similarly y minutes = y/60 hours.

Therefore, 6 hours and x minutes = 6 + x/60 = (360 + x)/60 hours. Similarly, 6 hours and y minutes = (360 + y)/60 hours.

Let's take 12 noon as the starting point. At 12 noon, the hands of the clock make zero angle. Then, the hour hand moves 6 hours and x minutes (or, y minutes) and the minute hand also moves 6 hours and x minutes (or, y minutes) and the hands form an angle, which in this case is 110 degrees.

Now, angle is actually a measure of angular distance. Suppose my friend and I, starting together at the same time, same direction, and same starting point, run for an hour. My friend, however runs a distance of 7 kilometres and I run a distance of 5 kilometres. Then, the distance between the two of us is 7 - 5 = 2 kilometres. Similarly, if the hour hand of the clock moves, say "h" degrees, in (360 + x)/60 hours, and the minute hand moves, say "m" degrees, during the same period, then the distance, or angle, between them is (h - m) degrees. It is like the minute and hour hand of a clock running a race along a circular track. We are interested in the positive degree, so we actually consider the modulus.

Okay, now getting back to our clock. I left home at (360 + x)/60 hours. Now, the hour hand moves 360 degrees in twelve hours. Therefore, in (360 + x)/60 hours it moves (360 + x)/60 * 360/12 = (360 + x)/2 degrees. The minute hand, on the other hand, moves 360 degrees in one hour; so, in 6 hours and x minutes it will have completed 6 revolutions (appearing as though it had not moved at all!) and will effectively appear to have moved only by x minutes. So, for the minute hand we need to consider only the minutes (poor fellow, all the trouble it has taken to complete the six complete rounds has to be ignored!). The minute hand moves 360 degrees in 60 minutes (one hour), so in x minutes it moves 6x degrees.

Therefore, the angle (the distance) made by the clock hands at the time I left home was [(360 + x)/2 - 6x]. That, of course, is 110. However, the distance has to be positive; you can't run -10 kilometres! So, we have to consider the modulus. Thus, |(360 + x)/2 - 6x| = 110. Similarly, the angle made by the hands when I return home is |(360 + y)/2 - 6y| = 110. But this is the same equation; only the variable is different. So, let's consider only one of the equations; that should provide us with two solutions.

So, let's consider |(360 + x)/2 - 6x| = 110.

Now, |(360 + x)/2 - 6x| = (360 + x)/2 - 6x, or |(360 + x)/2 - 6x| = - [(360 + x)/2 - 6x].

So, (360 + x)/2 - 6x = 110, or -[(360 + x)/2 - 6x] = 110.

Therefore, (360 - 11x) = 220, or -360 + 11x = 220.

So, 11x = 140, or 11x = 580. Thus, x = 140/11 or x = 580/11. Therefore, x = 12.727 or x = 52.727.

So, I get two values for x; this means I left home at 6 hours and 12.727 minutes (ever heard of such a time?) and returned at 6 hours and 52.727 minutes. So, I have been away for 52.727 - 12.727 = 40 minutes.

I am certain I must have made too many mistakes along the way. Did I do it right?

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