Kiss those Math Headaches GOODBYE!


The problem, once again, reads as follows. Before you go out to lunch, you glance at the clock above your desk. When you come back from lunch, you glance at the clock again, and you notice something strange: the minute and the hour hand have exchanged places from the positions they had just before you went to lunch.

The question is:  how long were you away?

I received several answers to this problem, but the first person who got it right was Adrian W. Langman, of Port Angeles, TX.

Here is Adrian’s answer, in his own words:

It’s probably true that the hour hand is near the 12 at the beginning and near the 1 at the end. So it’s about 5 minutes after noon at the beginning, and just a bit after 1 at the end.

(Or, it could be a tad before 11 in the morning at the beginning, and about 5 minutes before noon at the end, if the worker is an early riser.  But this problem is just the geometric mirror image of the one hypothesized above, so the duration of the lunch break will be exactly the same.)

Obviously the lunch break is about 55 minutes. But to find the exact length, let M be the number of minutes past noon at the beginning. I’ll use the obvious coordinate system – the origin at the center of the clock, the clock hands radial lines, the 12 at 0 degrees, and the 3 at 90 degrees.

At M minutes past noon, the minute hand is at M/60 x 360 degrees, i.e. 6M degrees, and the hour hand is at M/60 x 30 degrees (since it’s M/60 of the way from the 12 to the 1, which is at 30 degrees), i.e. M/2 degrees.

So at the end of lunch, since they’ve switched places, the hour hand is at 6M degrees and the minute hand is at M/2 degrees.

Since the minute hand is at M/2 degrees, it is 1/6(M/2) minutes past 1 o’clock, i.e. M/12 minutes past 1 o’clock.

Since the hour hand is at 6M degrees, it’s at 6M-30 degrees past the numeral 1, so it’s 2(6M-30) minutes after 1o’clcock.

Setting M/12 = 2(6M-30) and solving for M yields 720/143 (which is approximately 5.035).

So you left at 720/143 minutes after 12, and returned at 60/143 minutes after 1 o’clock.

So you were gone for 60 + 60/143 – 720/143 minutes,

i.e. 7920/143 minutes, i.e. 55 and 55/143 minutes, which reduces to 55 and 5/13 minutes.


Comments on: "Answer to Fun Math Problem #2" (2)

  1. Cool word problem. I use these a lot in my classroom and my students enjoy them!

    Jim the Teacher
    individualized learning math technology


  2. Note that you don’t have to assume anything about time of day: hand-switched positions are always a multiple of 1/13 of a clock cycle (12/13 of an hour) apart.


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