Time from 9 pm in a day to 2 pm on the following day = 29 h.
24 h 10 min of this clock = 24 h of the correct clock
or 145/6 h of this clock = 24 h of the correct clock
29 h of this clock = 24 x (6/145) x 29 h of the correct clock
= 28 h 48 min of correct clock
? The correct time is 28 h 48 min after 9 am. This will be 48 min past 1.

Given that, time to overtake N = 70 min

The required result = [(720/11) - N) ] x [(60 x 24)/N) min
= [(720/11) - 70) ] x [(60 x 24)/70) min
= [(720 - 770) / 11] x [(60 x 24) / 70] min
= - (50/11) x [(6 x 24)] / 7 min = -7200/77 min
-ve sign indicates that there is a loss.

If the minute hand of a clock overtakes the hour hand at interval N min of the correct time, then clock losses or gains by
The required result = [(720/11) - N) ] x [(60 x 24)/N) min
Positive for gain and negative for loss Here, N = 64

required time = [(720/11) - 64) ] x [(60 x 24)/64) min
= 32⁸/₁₁ min

The year 2007 is an ordinary year, so it has 1 odd day.
3rd day of the year 2007 was Wednesday.
? 3rd day of the year 2008 will be one day beyond the Wednesday. Hence, it will be Thursday.

The year 2008 is a leap year, so it has 2 odd days, 3rd day of the year 2008 was Friday.
? 3rd day of the year 2009 will be 2 days beyond the Friday.
Hence, the required answer is Sunday.

5th March, 1999 is Friday.
Then, 5 March 2000 = Friday + 2 = Sunday
(2000 is leap year and it crosses 29th feb 2000, so 2 is taken as odd day}

? 5th March 2000 = Sunday
Then, 9th March 2000 = Thursday.

6th March 2005 = Monday.
Then, 6th March 2004 = Monday - 1 days = Sunday.
(2004 is a leap year but it does not cross 29th February of 2004, so only 1 is taken as odd day).
6th March 2004 = Sunday
7th March 2004 = Monday.

On 31 st December, 2005 it was Saturday, Number of odd days from the year 2006 to the year 2009
= 1 odd day of 2006 + 1 odd day of 2007 + 2 odd days of 2008 (leap year) + 1 odd day of 2009
= 1 + 1 + 2 + 1 = 5
? On 31st December 2009, it was Thursday
Thus, on 2nd January 2010, it was Saturday .