Today is Sunday what day of the week was 79 days back.

Time from 12 p.m. on Sunday to 2 p.m. on the following Sunday = 7 days 2 hours.
= 24 x 7 + 2 = 170 hours.
The watch gain = (2 + 4 x 4/5) min = 34/5 minute in 170 hrs.
Since, 34/5 min are gained in 170 hrs.
2 min are gained in (170 x 5/34 x 2) hrs = 50 hours i.e., 2 days 2 hrs. after 12 p.m. on Sunday i.e., it will be correct at 2 p.m. on Tuesday.

Time from 10 a.m on a day to 3 a.m on 4th day = 24 x 3 + 17 = 89 hours.
Now 23 hrs 44 min. of this clock = 24 hours of correct clock.
89 hrs of faulty clock = (24 x 15/356 x 89) hrs = 90 hrs.
So, the correct time is 11 p.m

From Sunday noon to the following Sunday at 2 p.m. there are 7 days 2 hours or 170 hours. The watch gains 2 + 4⁴/₅ min in 170 hrs. Therefor, the watch gains 2 min in 2 x 170 hrs i.e., 50 hours 6⁴/₅. Now 50 hours Sunday noon = 2 p.m on Tuesday.

44

In 12 hours both hands are in straight line (either coincides or in opposite direction) 22 times. Hence in 24 hours both hands are in straight line 44 times.

Consider from 2^nd June 2010 to 2^nd June 2013 we have total 2 non leap year and one leap year so number of odd days are 1 + 1 + 2 = 4 so 2^nd June 2010 must be 4 days back from Sunday and that day is Wednesday.
From Zeller's Formula:
f = k + [13 x m - 1/ 5 ] + D + [D/4] + [C/4] - 2 x C.
In this case k = 2 (since 2^nd June)
Month m = 4 (As march = 1, April = 2, May = 3, June = 4 )
D is the last two digit of year here D = 10 (As year is 2010)
C is 1 st two digit of century here C = 20 (As year is 2010)
f = 2 + [13 x 4 - 1 / 5] + 10 + [10/ 4] + [ 20/4] - 2 x 20.
f = 2 + [51/5] + 10 +[2.5] + [5] - 40.
f = 2 + 10 + 10 + 2 + 5 - 40 = -11
This - ve value of f can be made positive by adding multiple of 7
So f = - 11 + 14 = 3
When divided by 7 we will get remainder 3, hence number of odd days is 3,
So 2^nd June 2010 is 3 days more than Monday, i.e Wednesday.

From Zeller's Formula:
f = k + [13 x m - 1 / 5] + D + [D/4] + [C/4] - 2 x C.
In this case k = 15 (since 15^th August)
Month m = 6 (As march = 1, April = 2, May = 3, August = 6)
D is the last two digit of year here D = 47 (As year is 1947)
C is the 1^st two digit of century here C = 19 (As year is 1947)
f = 15 + [13 x 6 - 1 / 5] + 47 + [47/4] + [19/4] - 2 x 19.
f = 15 + [77/5] + 47 + [11.75] + [4.75] - 38.
f = 15 + 15 + 47 + 11 + 4 - 38 = 54.
When divided by 7 we will get remainder 5, hence number of odd days is 3,
A remainder of 0 corresponds to Sunday, 1 means Monday,
So 15^th August 1947 is 5 days more than Sunday, i.e Friday.

From Zeller's Formula
f = k + [13 x m - 1 / 5] + D + [D/4] + [C/4] - 2 x C.
In this case k = 18 (since 18^th October)
Month m = 8 (As march = 1, April = 2, May = 3, October = 8)
D is the last two digit of year here D = 50 (As year is 2050)
C is the 1^st two digit of century here C = 20 (As year is 1950)
f = 18 + [13 x 8 - 1 / 5] + 50 + [50/4] + [20/4] - 2 x 20.
f = 18 + [103/5] + 50 + [12.5] + [5] - 40.
f = 18 + 20 + 50 + 12 + 5 - 40 = 65.
When divided by 7 we will get remainder 2, hence number of odd days is 2,
So 18^th October 2050 is 2 days more than Sunday, i.e Tuesday.

Here we have to find the number of odd days between, 5^th march and 5^th November,
Number of days in March is 26 or 5 odd days
(Here we have not included 5^th march)
Number of days in April is 30 or 2 odd days
Number of days in May is 31 or 3 odd days
Number of days in June is 30 or 2 odd days
Number of days in July is 31 or 3 odd days
Number of days in August is 31 or 3 odd days
Number of days in September is 30 or 2 odd days
Number of days in October is 31 or 3 odd days
Number of days in November is 5 or 5 odd days
(Here 5^th November is included)
So total number of odd days = 5 + 2 + 3 + 2 + 3 + 3 +2 + 3 + 5 = 28 when divided by 7 gives remainder 0 hence 5^th November will be same as that of 5^th march.

From Zeller's Formula we can find that 1^st March 2009 is Sunday so well have 5 Saturdays and 5 Sundays in total 10 weekends.