NOTE :

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2:  Add +6 to the Given Year.

Solution : 

Given Year is 1993, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 1993 + 11) = 2004, Which is a leap year. 

Step 2 : Add +6 to the given year (i.e 1993 + 6) = 1999

Therfore, The calendar for the year 1993 will be same for the year 1999

How to find the years which have the same Calendars : 

 Leap year calendar repeats every 28 years. 

Here 28 is distributed as 6 + 11 + 11.

Rules:

a) If given year is at 1st position after Leap year then next repeated calendar year is Given Year + 6.

 b) If given year is at 2nd position after Leap year then next repeated calendar year is Given Year + 11.

 c) If given year is at 3rd position after Leap year then next repeated calendar year is Given Year + 11. 

Now, the given year is 2018

We know that 2016 is a Leap year.

2016  2017  2018  2019  2020

 LY    1st     2nd    3rd    LY

Here 2018 is at 2 nd position after the Leap year.

According to rule ( b) the calendar of 2018 is repeated for the year is 2018 + 11 = 2029.

We will show that the number of odd days between last day of February and last day of October is zero. .

March April May June July Aug. Sept. Oct.

31 + 30 + 31 + 30 + 31 + 31 + 30 + 31

= 241 days = 35 weeks = 0 odd day. ,Number of odd days during this period = 0.

Thus, 1st March of an year will be the same day as 1st November of that year. Hence, the result follows

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.

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

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

15th August, 1947 = (1946 years + Period from 1st Jan., 1947 to 15th )

Counting of odd days:

 1600 years have 0 odd day. 300 years have 1 odd day.

 47 years = (11 leap years + 36 ordinary years)= [(11 x 2) + (36 x 1) ]odd days = 58 odd days = 2 odd days.

Jan  Feb  Mar  Apr  May  Jun  Jul  Aug.

31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 days = (32 weeks + 3 days) = 3,

Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.

Hence, the required day was '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.

NOTE :

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2:  Add +6 to the Given Year.

Solution : 

Given Year is 2005, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 2005 + 11) = 2016, Which is a leap year. 

Step 2 : Add +6 to the given year (i.e 2005 + 6) = 2011

Therfore, The calendar for the year 2005 will be same for the year 2011

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.