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Topic: Analog clock V (Read 1136 times) |
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inexorable
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Analog clock V
« on: Oct 22nd, 2005, 11:34pm » |
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The minute hand of an analog clock overtakes the hour hand at intervals of 64 minutes of correct time. how much a day does the clock gain or lose?
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Neelesh
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Re: Analog clock V
« Reply #1 on: Oct 23rd, 2005, 1:28am » |
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thirty-two minutes Gained by the inaccurate clock?
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| « Last Edit: Oct 23rd, 2005, 9:57am by Neelesh » |
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inexorable
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Re: Analog clock V
« Reply #2 on: Oct 23rd, 2005, 2:25am » |
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can you please explain neelash?
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Neelesh
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Re: Analog clock V
« Reply #3 on: Oct 23rd, 2005, 4:13am » |
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Actually I should have explained it there itself.
Anyways here is how I went about it :
at 12.00 , hour and minute hands are exactly on digit 12. After this point, they will meet soon after 1.05 and before 1.10.
So lets assume that they meet at 1.x , i.e. x minutes after 1o clock.
Now consider that its 1o clock. The hour and minutes hands will thus meet after exactly x minutes.
since in 60 minutes the minute hand travels one full circle (i.e. 360 degrees) hence in x minutes it will travel 6x degrees (starting from the position 12.)
Hence minute hand will travel 6x degrees (starting from number 12) before it meets the hour hand.
The hour hand is at number 1, thus it has already travelled 30 degrees (starting from 12). In next x minutes, it will travel x/2 degrees.
(This is beacuse in 60 minutes hour hand travels 30 degrees. So in x minutes it will travel x/2 degrees.)
So the total angle covered by hour hand (from 12 to meeting point) 30+x/2
But since minute and hour hand are at same place, hence the angles covered (from 12) must be same.
Hence
6x = x/2 + 30
Which gives x = 60/11
What this means is that after 1oclock, the hour and minute hands meet after exactly 60/11 minutes
Since previously the two hands met at 12.00, hence the total time between two overlaps is
60 + 60/11 = 720/11 = a little more than 65.4 minutes.
But the incorrect clock covers up this time in only 64 minutes.
Hence the incorrect clock runs faster by
(720/11 - 64) minutes in the span of 720/11 minutes.
That is, the incorrect clock runs faster by 16/11 minutes in the span of 720/11 minutes.
Hence to find how much it will gain in one day (24*60 minutes) -
16/11 minutes gain in 720/11 minutes
?? in 24*60 minutes?
Solving this we get the answer as
(11*24*60*16) / (11 *720)
= 32 minutes.
Thus the analog clock V will gain 32 minutes per day.
[edit]
Thanks to Padzok, I have edited this post to make it accurate.
[/edit]
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| « Last Edit: Oct 23rd, 2005, 9:59am by Neelesh » |
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Padzok
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Re: Analog clock V
« Reply #4 on: Oct 23rd, 2005, 4:44am » |
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I think it can also be done in slightly fewer steps (although there isn't that much difference).
Say both accurate clock and inaccurate clock start with both hands at 12.
The next time that both hands are pointing back up at 12 again, it is the eleventh time they have been on top of each other.
For the accurate clock, that is 12 hours or 720 minutes.
For the inaccuarate clock that is 11 periods of 64 minutes each, which is a total of 704 minutes.
So in each 12 hour period, 16 minutes are lost - which obviously is 32 minutes per "accurate" day.
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Neelesh
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Re: Analog clock V
« Reply #5 on: Oct 23rd, 2005, 4:59am » |
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Great!! This is much cleaner and simpler way to get the answer. Elegant solution. 
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Padzok
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Re: Analog clock V
« Reply #6 on: Oct 23rd, 2005, 6:53am » |
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Actually, I think it is misleadiing for me to say that the minutes are lost .
According to my solution, the inaccurate clock would be getting back to 12 oclock faster and so would run | hidden: | | 32 minutes fast per 24 hours |
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Neelesh
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Re: Analog clock V
« Reply #7 on: Oct 23rd, 2005, 9:56am » |
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on Oct 23rd, 2005, 6:53am, Padzok wrote:
getting back to 12 oclock faster
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Yes I agree with you, it was a stupid mistake on my part too to say it lost.
I have edited my post(s) to avoid confusion.
Thanks for pointing that out.
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