Clearly it can be understood from the question that 9 days ago was a Thursday Number of odd days in 9 days = 2 (As 9 – 7 = 2, reduced perfect multiple of 7 from total days)
Hence today = (Thursday + 2 odd days) = Saturday

Given that 2.12.91 is the first Tuesday
Hence we can assume that 4.12.91 is the first Thursday

If we add 7 days to 4.12.91, we will get second Thursday
If we add 14 days to 4.12.91, we will get third Thursday
If we add 21 days to 4.12.91, we will get fourth Thursday

=> fourth Thursday = (4.12.91 + 21 days) = 25.12.91.

As given, Satuday falls on 26th January and we have to find the day on 14th February.
Clearly, 2nd, 9th and 16th February each is a Saturday.
Thus, 14th February was a Thursday.

This can be illustrated with an example.
Ex: 1896 is a leap year.The next leap year comes in 1904 (1900 is not a leap year).

Explanation : The length of the solar year, however, is slightly less than 365 1/4 days-by about 11 minutes. To compensate for this discrepancy, the leap year is omitted three times every four hundred years.

In other words, a century year cannot be a leap year unless it is divisible by 400. Thus 1700, 1800, and 1900 were not leap years, but 1600, 2000, and 2400 are leap years.

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.

On 31 Dec, 2005 it was Saturday.
Number of Odd days for 1 normal year 365/7=1
Number of Odd days for leap year 2
so 2006=1odd day
2007=1odd day
2008leap year 2 odd days
2009=1 odd day
Number of Odd days from the year 2006 to the year 2009 = 1 +1+2+1 =5 On 31st Dec 2009 it was Thursday So, 1st Jan 2010 is Friday

5th is Monday, then 12, 19, 26th also Monday
Then 3 days after  is 29th, which is Thursday

For a year to have the same calendar with 1990, total odd days from 1990 should be 0. Take the year 1992 from the given choices.
Total odd days in the period 1990-1991 = 2 normal years
⇒ 2 x 1 = 2 odd days Take the year 1995 from the given choices.
Number of odd days in the period 1990-1994 = 4 normal years + 1 leap year
⇒ 4 x 1 + 1 x 2 = 6 odd days Take the year 1996 from the given choices.
Number of odd days in the period 1990-1995 = 5 normal years + 1 leap year
⇒ 5 x 1 + 1 x 2 = 7 odd days = 0 odd days
(As we can reduce multiples of 7 from odd days which will not change anything) Though number of odd days in the period 1990-1995 is 0, there is a catch here.
1990 is not a leap year whereas 1996 is a leap year.
Hence calendar for 1990 and 1996 will never be the same.Take the year 2001 from the given choices. Number of odd days in the period 1990-2000 = 8 normal years + 3 leap years
⇒ 8 x 1 + 3 x 2 = 14 odd days = 0 odd days
Also, both 1990 and 2001 are normal years.
Hence 1990 will have the same calendar as that of 2001

First, we count the number of odd days for the left over days in the given period.
Here, given period is from 15.8.2012 to 11.6.2013
Aug Sept Oct Nov Dec Jan Feb Mar Apr May Jun
16   30   31  30   31   31   28  31   30   31   11(left days)
Therefore, 2 + 2 + 3 + 2 + 3 + 3 + 0 + 3 + 2 + 3 + 4 (odd days) = 6 odd days
So, the given day Thursday + 6 = Wednesday.

Since the month begins on Saturday, so 2nd, 9th, 16th, 23rd, 30th days are Sundays.
While 8th and 22nd days are second Saturdays.
Thus, there are 7 holidays in all.
Therefore number of working days = 30 – 7 = 23.

Total number of odd days from 1st Jan, 1991 to 1st Jan, 1998
Year 1991 1992 1993 1994 1995 1996 1997
Odd days I + 2 + 1 + 1 + 1 + 2 + 1 = 9 odd days
2 odd days
The day is 2 days beyond the day on 1st Jan, 1991.
i.e. The required day must be Thursday.