The correct answer is  3th, 10th, 17th, 24th

Sol.
If day before yesterday was Saturday, then today is Monday.
Thus tomorrow will be Tuesday and day after tomorrow will be Wednesday.

Clearly, Nine days ago, it was Thursday.
.’. Today is Saturday.

15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)

Odd days in 1600 years = 0

Odd days in 400 years = 0

9 years = (2 leap years + 7 ordinary years) = (2 × 2 + 7 × 1) = 11 = 4 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 odd days.

Total number of odd days = (0 + 0 + 4 + 3) = 7 = 0 odd days.

The given day is Sunday.

Hence, the correct answer is Sunday.

Explanation

16 July, 1776

= (1775 years + period from 1.1.1776 to 16.7.1776.

Checking of odd days.

No. of odd days in 1600 years =0

No. of odd days in 100 years = 5

75 years= 18 jump years+57 conventional years

= (18*2+57*1) odd days = 93 odd days

= (13 weeks +2 days) = 2 odd days.

∴ 1775 years have= (0+5+2) odd days= 7 days= 0 days

Jan + Feb. + March + April + May + June + July

31 +29 + 31 + 30 + 31 + 30 +16 = 198 days

198 days = (28 weeks+2 days) = 2 odd days

=Total no. of odd days = (0+2) = 2

Thus, the required day is Tuesday.

The year 2004 is a leap year. So, it has 2 odd days.
But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.
∴ The day on 6^th March, 2005 will be 1 day beyond the day on 6^th March, 2004.
Given that, 6^th March, 2005 is Monday.
∴ 6^th March, 2004 is Sunday (1 day before to 6^th March, 2005).

Number of days from 26-Jan-1996 to 15-May-1996 (both days included)
= 6(Jan) + 29(Feb) + 31 (Mar) + 30(Apr)+ 15(May)
= 111

Here we have to count the number days from 3^rd February, 2012 to 18^th April 2012 ( both inclusive)
The given year is leap year, So February month has 29 days
From 3^rd to 29^th February = 27 days
In March = 31 days
From 1^st to 18^th April = 18 days
Total number of days = 27 + 31 + 18 = 76 days

In 400 consecutive years, there are 97 leap years. Hence, in 400 consecutive years, February has the 29th day 97 times and the remaining eleven months have the 29th day 400 × 11 or 4400 times.
Thus the 29th day of the month occurs
= 4400 + 97
= 4497 times.

Given that 25th August = Thursday
Hence 29th August = Monday
So 22nd,15th and 8th and 1st of August also will be Mondays
Number of Mondays in August = 5

After every 400 years, the same day occurs. (Because a period of 400 years has 0 odd days)
Thus, 18th April 1603 is Thursday, After 400 years i.e., on 18th April 2003 has to be Thursday.

Each day of the week is repeated after 7 days.
So, after 63 days, it will be Monday.
∴ After 61 days, it will be Saturday.

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.

29th Feb, 2016 = Monday => 28th Feb, 2012 = Sunday
28th Feb, 2017 = Tuesday (because 2016 is a leap year, there will be 2 odd days)
Therefore» Feb 28th 2018 (Wednesday), Feb 28th 2019 (Thursday), Feb 28th 2020 (Friday), Feb 29th 2020 (Saturday)
Or, Feb 29th to Feb 29th after 4 years, we have 5 odd days.
So, every subsequent birthday, would come after 5 odd days.
2020 birthday – 5 odd days
2024 birthday – 10 odd days = 3 odd days
2028 birthday – 8 odd days = 1 odd day
2032 birthday – 6 odd days
2036 birthday – 11 odd days = 4 odd days
2040 birthday – 9 odd days = 2 odd days
2044 birthday – 7 odd days = 0 odd days. So, after 28 years, his birthday would fall on Monday.
The next birthday on Monday would be in year 2072 (further 28 years later), the one after that would be in year 2100. But we are told that she lives upto year 2099.
So, there are 2 occurrences of his birthday falling on Monday – 2044 & 2072.

The year 2007 is an ordinary year. So, it has 1 odd day.

1^st day of the year 2007 was Monday.

1^st day of the year 2008 will be 1 day beyond Monday.

Hence, it will be Tuesday.

Number of days = (6 + 29 + 31 + 30 + 15) = 111.
Note : 1988 is a leap year. So, number of days in February = 29.

100 years contain 5 odd days.(it includes leap years also)
2000 years contain 20*5=100 odd days
400,800,1200,1600,200 are leap years because these are divisible by 400 so another 5 odd days
whereas 100,200,300,500..etc are not leap years
total of 105 odd days
105/7=0
2000 Dec 31 is Sunday next day is Monday

The year 2006 is an ordinary year. So, it has 1 odd day.

So, the day on 8^th Dec, 2007 will be 1 day beyond the day on 8^th Dec, 2006.

But, 8^th Dec, 2007 is Saturday.

 8^th Dec, 2006 is Friday.