3
$\begingroup$

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.

$\endgroup$
11
  • 1
    $\begingroup$ You have to consider that also the hour hand is moving. $\endgroup$ Jun 2, 2015 at 13:09
  • $\begingroup$ @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. $\endgroup$
    – user103816
    Jun 2, 2015 at 13:12
  • $\begingroup$ In a normal clock, the minute hand overtakes the hour hand $22$ times per day. $\endgroup$
    – Arthur
    Jun 2, 2015 at 13:18
  • $\begingroup$ @Arthur You are right in 60*24 actual minutes the minute hand gains 55*24 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. $\endgroup$
    – user103816
    Jun 2, 2015 at 13:27
  • $\begingroup$ 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. $\endgroup$
    – Arthur
    Jun 2, 2015 at 14:12

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .