Answer: Option B. -> 2001
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 daysTake 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 daysTake 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

Answer: Option B. -> 2
A period of 100 years has 5 odd days . In 26 years , 4 are leap, remaining are ordinary years
125 years = 100 years + 26 years
= 100 years + 6 leap years + 20 ordinary years
= 5 odd days + 12 odd days + 20 odd days
= 37 odd days = 5 x 7 +2 = 2 odd days

Answer: Option A. -> 1
Here we have to count the number days from 13th May, 2005 to 18rd August 2005 ( both inclusive)
From 13th to 31st May = 19 days
In June = 30 days
In July = 31 days
From 1st to 19th April = 19 days
Total number of days = 19 + 30 + 31 + 19 = 99 days
The number of odd days are = 14 x 7 + 1 = 99
So there is 1 odd day in the given period

Answer: Option B. -> Tomorrow
Day before yesterday was Thursday
⇒ Yesterday was a Friday
⇒ Today is a Saturday
⇒ Tomorrow is a Sunday

Answer: Option A. -> Same day
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

Answer: Option A. -> Saturday
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.

Answer: Option C. -> 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'.

Answer: Option C. -> 2016
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.

Answer: Option A. -> Saturday
350 days = $$\frac{{350}}{7}$$ = 50 weeks, i.e No odd days,
So it will be a Saturday.

Answer: Option D. -> 2018
Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.
Year
Odd Day
2007
1
2008
2
2009
1
2010
1
2011
1
2012
2
2013
1
2014
1
2015
1
2016
2
2017
1
Sum = 14 odd days ≡ 0 odd days.
∴ Calendar for the year 2018 will be the same as for the year 2007.