Calendar Questions

We already know that the calendar after a leap year repeats again after 28 years.

Here 1988 is a Leap year, then the same calendar will be in the year = 1988 + 28 = 2016.

As month begins on Saturday, so 2nd, 9th, 16th, 23rd, 30th days will be Sundays. While 8th and 22nd days are second Saturdays. Thus, there are 7 holidays in all.

Hence, no. of working days = 30 – 7 =23

We know that, After every 400 years, the same day occurs.

Thus, if 9th August 2016 is Saturday, before 400 years i.e., on 9th August 1616 has to be Saturday.

First,we count the number of odd days for the left over days in the given period.Here,given period is 09-07-2013 to 07-01-2014

 Therefore, No. of Odd Days =  1 + 3 + 2 + 3 + 2 + 3 + 0 = 14 = 0 odd days

So, given day Sunday + 0 = Saturday is the required result.

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.

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.

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.