Sol : Option C
100 years contain 5 odd days.
Therefore, Last day of 1st century is Friday
200 years contain (5 x2) is 3 odd days (approx.)
Last day of 2nd century is Wednesday
300 years contain ( 5x3) = 15 i.e. 1 odd day( approx.)
Last day of 3rd century is Monday
400 years contain 0 odd day
Last day of 4th century is Sunday.
Thus, the cycle is repeated.
Last day of a century cannot be Tuesday or Thursday or Saturday.

Sol : Option D
16th July, 1776 = (1775 years + time period from Jan 1, 1776 to July 16, 1776)
Counting of odd days:

• 1600 years: 0 odd day.
• 100 years: 5 odd days.
• 75 years = (57 ordinary years + 18 leap years) = [((57 × 1) + (18 × 2)] = 93 days (13 weeks + 2 days) = 2 odd days.
  1775 years have (0 + 5 + 2) odd days or 7 odd days = 0 odd day.
  January February March April May June July
  31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days= (2 days + 28 weeks) 
  ∴Total number of odd days = 2
  So, the required day was 'Tuesday'.

Sol : Option D
It was Saturday on 31st December, 2005.
Number of odd days from 2006 to 2009 = (1 + 1 + 2 + 1) = 5 days.
∴On December 31, 2009, it was Thursday. Thus, on Jan 3, 2010 it was Sunday.

Sol : Option B
15th August, 2010 = (2009 years + time period 1.1.2010 to 15.8.2010)
Odd days in 1600 years = 0
Odd days in 400 years = 0
9 years = 7 normal years + 2 leap years = (7 × 1 + 2 × 2) = 11 = 4 odd days.
January February March April May  June July August
(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days that is 3 odd days
Total number of odd days = (0 + 0 + 4 + 3) = 7 = 0 odd days.
Given day is Sunday.

Sol : Option D
We know year 2004 is a leap year. Therefore, it has 2 odd days.
Feb 2004 was not included as we are calculating from March 2004 to March 2005.
So it has 1 odd day.
The day on March 13, 2005 will be 1 day ahead of the day on March 13, 2004.
Given that, March 13, 2005 is Wednesday.
So March 13, 2004 is Tuesday (1 day before to 13th March, 2005).

Sol : Option C
The year 2007 is an ordinary year. So, it has 1 odd day.
1st day of the year 2007 was Monday
1st day of the year 2008 will be 1 day beyond Monday
Hence, it will be Tuesday.

Sol : Option B
The year 2006 was an ordinary year. So, it has 1 odd day.
The day on 7th Dec, 2007 will be 1 day beyond the day on 7th Dec, 2006.
But, 7th Dec, 2007 is Friday
So, 7th Dec, 2006 is Thursday.

Sol : Option A
Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.
Sum = 14 odd days 0 odd days
Calendar for the year 2018 will be the same as for the year 2007

Sol : Option C
15th August, 1947 = (1946 years + time 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 × 2) + (36 × 1)] odd days = 58 odd days
= 2 odd days.
Jan. Feb. March April May June July August
31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 days » 3 odd days,
Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.
Hence, the required day was 'Saturday'.

Sol : Option B
Let us find the day on 1st July, 2003.
2000 years have 0 odd day. 2 ordinary years have 2 odd days.
Jan Feb March April May June July
31 + 28 + 31 + 30 + 31 + 30 + 1 = 182 days = 0 odd day.
Total number of odd days = (0 + 0 + 2) odd days = 2 odd days.'
:. 1st July 2003 was 'Tuesday',-,-
Thus, 1st Sunday in July 2003 is 6th July.
So, during July 2003, Monday fell on 6th, 13th, 20th and 27th.