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The question in the textbook is:

The minute hand of a clock overtakes the hour hand at intervals of 65 minutes of correct time. How much a day does the clock gain?

My method:

The correct clock's minute hand gains over its hour hand in actual 65 minutes = $\dfrac {55}{60} \times 65$ minutes.

The incorrect clock's minute hand gains over its hour hand in actual 65 minutes = 60 minutes.

So the net gain of the incorrect clock over the correct clock in actual 65 minutes = $60 - \dfrac{55}{60} \times 65 = \dfrac {5}{12}$minutes.

So the net gain in 24 hours is $\dfrac{5}{11} \times \dfrac{60 \times 24}{65} = 10.07 $minutes

But the book says the correct answer is $10.2325$ minutes. Also the book uses a different method which I do not understand.

Question : Why is my method incorrect? In my method the incorrect clock's reading should be 5/12 minutes ahead to that of the correct clock's reading after 65 minutes from when both clocks started. But in book's method it is 5/11.

user103816
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    You have to consider that also the hour hand is moving. – Emilio Novati Jun 02 '15 at 13:09
  • @EmilioNovati Yes I guess the hour hand is moving faster than usual because of the faster minute hand. If this is true then net gain = (M_2-H_2)-(M_1-H_1) != M_2-M_1, M_2=minute hand's reading of 2nd clock, H_1= Hour hand's reading of 2nd clock and so on. – user103816 Jun 02 '15 at 13:12
  • In a normal clock, the minute hand overtakes the hour hand $22$ times per day. – Arthur Jun 02 '15 at 13:18
  • @Arthur You are right in 6024 actual minutes the minute hand gains 5524 minutes over the hour hand making (55*24)/60 overtakes. But how is this useful in solving this question? P.S: I guess I now understand the book's method. – user103816 Jun 02 '15 at 13:27
  • It helps you because then you know that when your clock thinks a day has passed, in reality only $22\cdot 65$ minutes have passed. – Arthur Jun 02 '15 at 14:12
  • @Arthur But the incorrect clock may not make exactly 22 overtakes in a day. Even if it does then 2265 is the real time when the incorrect clock shows a day -- we need to find the reading of incorrect clock when the real time is 2460 minutes. – user103816 Jun 02 '15 at 14:16
  • Yes, it does, as long as the relationship between minute hand and hour hand is correct. Very few clocks get this wrong. Most clocks, however, have a timing problem related to an imperfectly calibrated pendulum / quartz crystal. I would assume that that's the source of your clock's incorrectness. – Arthur Jun 02 '15 at 14:20
  • @Arthur How can we prove that every clock, whatever its rate(constant) is, makes 22 overlaps a day? – user103816 Jun 02 '15 at 14:23
  • Assuming 22 overtakes a day gives slightly wrong answer. CAlculation: In 2265 minutes the incorrect clock is above 10 minutes; in 2460 minutes the incorrect clock is above (10/1430)*1440 = 10.07 minutes. The right answer is 10.23 minutes. – user103816 Jun 02 '15 at 14:31
  • The minute hand moves $24$ rounds in a day, while the hour hand moves $2$. $24-2=22$. It happens $11$ times in the am, and at the exact same times in the pm. Those times are roughly 12:00, 1:05, 2:11, 3:16, 4:22, 5:27, 6:33, 7:38, 8:44, 9:49 and 10:55. – Arthur Jun 02 '15 at 14:32
  • @Arthur I didn't get this line while the hour hand moves 2. 24−2=22. If I get it right then this must not be true for the incorrect clock because it runs faster than usual. The incorrect clock overtakes every 65 minutes, so in 6024 minutes it overtakes (6024)/65 times, that is 22.1538 times not 22. – user103816 Jun 02 '15 at 14:47

1 Answers1

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On a normal clock, the hand cross each other 11 times in 12 hours, which means that they cross once every $\dfrac{12}{11} = 1.090909…$ hours or every 1 hour, 5 minutes, and 27.272727… seconds. The incorrect clock gains about 27 seconds at every crossing, so each day it gains $22(27.272727… seconds)= 10 minutes every day.

Note: These answers are all similar enough that different methods of calculating the answer could be the reason for the varying results.

PiGuy314
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