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.

Correct Option: A

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.

When we divide 79 by 7 we will get remainder 2 so we have 2 odd days, so required day must be 2 days back from today (i.e Sunday) and that day should be Friday.

Correct Option: B

When we divide 79 by 7 we will get remainder 2 so we have 2 odd days, so required day must be 2 days back from today (i.e Sunday) and that day should be Friday.

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.

Correct Option: B

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.

Same as previous question Option A is correct.

Correct Option: A

Same as previous question Option A is correct.

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.

Correct Option: A

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.