Calendar Questions

NOTE: When a century year leaves a remiander 0, when divided by 400 then it is a leap year (366 days).

So, 1200 has 366 days.

1.12.91 is the first Sunday of December 91.
So, 3.12.91 is the first Tuesday of the month.
Clearly, 10.12.91, 17.12.91, 24.12.91 and 31.12.91 are also Tuesdays.
So, 24.12.91 is the fourth Tuesday.

Given that the month begins on a Friday and has 31 days

Sundays = 3rd, 10th, 17th, 24th, 31st
=> Total Sundays = 5

Every second & fourth Saturday is holiday.
2nd & 4th Saturday in every month = 2

Total days in the month = 31

Total working days = 31 - (5 + 2) = 24 days.

The year 1996 is a leap year. So, it has 2 odd days. So, the day on 17th March, 1997 will be 2 days beyond the day on 17th March,1996.

Here 17th March, 1997 is Monday. So, 17th March, 1996 is Saturday.

If yesterday were tomorrow, today is Friday => tomorrow of friday is Saturday.

Hence, it implies that today is Sunday if tomorrow is Saturday.

The day of the week repeats every 7 days.

Given today is Friday. Again Friday is repeated on the 7th day, 14th,... on 7 multiple days.

Hence, Friday is on the 63rd day, as 63 is multiple of 7.

Now, the required day of the week on the 65th day is Sunday.