A Math Problem - Just for Kicks.

[quantyst]

I was late to a meeting the other day that was scheduled to begin at 2 pm. When I got there I looked at an analog clock on the wall and saw that I was late for more than 15 minutes but less than 20 minutes. When the meeting ended I noticed that the hour-hand and the minute-hand had precisely switched positions relative to the time when I arrived at the meeting. At exactly what point in time - to the nearest second - did the meeting end?

 

[Andy Nguyen]

Are you sure the question is not "how long is the meeting ?"
In any case, my calculation shows that the meeting lasted 55 minutes 23 seconds. It ends somewhere between 3:11PM and 3:14PM.

I probably can do some more calculation to nail down the ending time but it's for another time.

 

[Yuriy]

3:11:20 PM (maybe plus something small that I'm lazy to compute).

Are you sure the question is not "how long is the meeting ?"
In any case, my calculation shows that the meeting lasted 55 minutes 23 seconds. It ends somewhere between 3:11PM and 3:14PM.

I probably can do some more calculation to nail down the ending time but it's for another time.

The meeting could not last for 55 minutes he said he was late, so the meeting had started prior to his arrival. If it started at 2pm, then it was at least 1 hour 11 minutes

 

[Andy Nguyen]

that sounds about right. I was calculating how long he was in the meeting

 

[quantyst]

Hello Everyone,

Thank you for your responses. The issue of when the meeting began - whether at 2 pm or when I arrived - is immaterial. All you need to know is that two points in time are related by having the positions of the minute-hand and hour-hand PRECISELY switched.

Here's the solution:

Let MA denote the position of the minute-hand when I arrived, and ME the position of the minute-hand when the meeting ended. Similarly, let HA denote the position of the hour-hand when I arrived, and HE the position of the hour-hand when the meeting ended.

So HA=2+x and MA=12x, for some positive x less than 1. The variable x denotes the fraction of an hour that I was late. When the hour-hand has a certain displacement of x from the position 2, the minute-hand has a displacement of 12x from position zero (zero being same as position 12).

Similarly, HE=3+y since the minute-hand at the time of my arrival was between 3 and 4, which is when the meeting ended. Also ME=12y.

Now, according to the problem, we have HA=ME and MA=HE.

So, we have the system of equations {2+x=12y, 12x=3+y} to solve.

Once the system is solved, then the answer needs to be converted to HH:MM:SS with SS being to the nearest second.

A BONUS QUESTION: How many points in time in a 12-hour period are there that have the property that a precise switch of positions of the hour-hand and minute hand gives rise to 'legitimate' positions of the hands?

 

[Yuriy]

Well... what is the answer? New Year's laziness does not allow me to fully work out the problem.

 

[quantyst]

Yuriy,

You've already got the correct answer!

As you posted earlier, it is 03:11:20.

Congrats for the correct answer!

Now, how many times does 'that' happen in a 12-hour period?

I am curious: How did you get the answer?

 

[Yuriy]

Yuriy,

You've already got the correct answer!

As you posted earlier, it is 03:11:20.

Congrats for the correct answer!

Now, how many times does 'that' happen in a 12-hour period?

I am curious: How did you get the answer?

I did it manually drawing pictures.
1) Look at arrival: Arrival between 15 and 20 minutes implies that the hour hand was between 11 and 12 [since it moves from 10 to 15 in 60 minutes].
2) Look at leaving: (1) imples that you left 11 to 12 minutes after 3, so the hour hand was somewhere around 16 at the time of leaving.
3) Look at arrival again: (2) implies that the arrival was around 2:16. 16 minutes is 1 division and 1/3 of a division for the hour hand, making the hour hand landing at 11 + 1/3 minutes, which is 11 min 20 sec.
4) Look at leaving again: the hour hand is at 16 => 3pm, the minute hand is at 11:20 so 3:11:20 pm.