Answer: Option A. -> Monday
Answer: (a)
15th August 1949 means,
1948 complete year + First 7 months of the year 1949 + 15 days of August
Number of odd days in 1600 yrs = 0
Number of odd days in 300 yrs = 1
Number of odd days in 48 yr (36 non - leap years + 12 leap years)
= 36 × 1 + 12 × 2
= 60 = 7 × 8 + 4 = 4
odd days From 1st January 1949 to 15th August 1949
Number of odd days in 1949,
January
3
February
0
March
3
April
2
May
3
June
2
July
3
August
(15 ÷ 7) = 1
Total number of odd days in 1949 = 3 + 0 + 3 + 2 + 3 + 2 + 3 + 1 = 17
= 7 × 2 + 3 = 3 odd day
Total odd days = 1 + 4 + 3 = 8
= 1 odd days
Since, 1 is the code for Monday.
Therefore, the required day was Monday.

Answer: Option D. -> Wednesday
Answer: (d)
19th August 1992 means,
1991 complete years + First 7 months of 1992 + 19 days of August
Number of odd days in 1600 years = 0
Number of odd days in 300 yrs = 1
Number of odd days in 91 yrs (22 leap year + 69 non-leap years)
= 22 × 2 + 69 × 1
= 44 + 69 113
= 7 × 16 + 1 = 1 odd day
From 1st January, 1992 to 19th August, 1992
Number of odd days in 1992,
January
3
February
1
March
3
April
2
May
3
June
2
July
3
August
5
= 3 + 1 + 3 + 2 + 3 + 2 + 3 + 5
= 22 = 7 × 3 + 1
= 1 odd day
∴ Number of odd days till 19th August, 1992 = 0 + 1 + 1 + 1 = 3
So, the required day was Wednesday.

Answer: Option C. -> Monday
Answer: (c)
30th June 1980 means 1979 complete years + 6 months of 1980
Number of odd days in 1600 yrs = 0
Number of odd days in 300 yrs = 1
Number of odd days in 79 yrs = (19 leap yrs + 60 ordinary years)
= 19 × 2 + 60 × 1 = 38 + 60 = 98
⇒ o odd days
January
3
February
1
March
3
April
2
May
3
June
2
Number of odd days in 1980 = 3 + 1 + 3 + 2 + 3 + 2 = 14
⇒ 0 odd days
Total number of odd days till 30th June, 1980 = 0 + 1 + 0 + 0 = 1
So, the required day was Monday.

Answer: Option A. -> Wednesday
Answer: (a)
18th September 1991 means,
1990 complete years + 8 months of 1991 + 18 days of September
Number of odd days in 1600 yrs = 0
Number of odd days in 300 yrs = 1
Number of odd days in 90 yrs (22 leap year + 68 ordinary years )
= 22 × 2 + 68 × 1
= 44 + 68 = 112
⇒ 0 odd days
Number of odd days in 1991,
January
3
February
0
March
3
April
2
May
3
June
2
July
3
August
3
September
4
= 3 + 0 + 3 + 2 +3 + 2 + 3 + 3 + 4
= 23 = 7 × 3 + 2
= 2 odd days
Total number of odd days till 18th September, 1991
= 0 + 1 + 0 + 2 = 3
So, the required day was Wednesday.

Answer: Option B. -> 1, 8, 15, 22, 29
Answer: (b)
First of all, we have to find the day on 1st April, 2012 1st April 2012 means
(2011 years 3 months and 1 day)
Now, 2000 years have 0 odd days 11 years have
(2 leap years and 9 ordinary years)
= ( 2 × 2 + 9 × 1 ) odd days
= (4 + 9) odd days = 13
= 6 odd days
3 months and 1 day
January
31
February
29
March
31
April
1
= 92 days = 1 odd day
Total number of odd days = ( 6 + 1 ) = 7
⇒ 0 odd day
Hence, it was Sunday on 1st April 2012. (1st Sunday).
Subsequently, Sundays of the month were on 1st, 8th, 15nd, 22nd, and 29th.

Answer: Option D. -> 4th, 11th, 18th, 25th
Answer: (d)

Answer: Option A. -> Monday
Answer: (a)
1st January 2001 means 2000 complete years + 1 day of January 2001
Number of odd days in 2000 yrs = 0
Number of odd days in January 2001 = 1
Total number of odd days = 0+1 = 1
So, the required day was Monday

Answer: Option C. -> Wednesday
Answer: (c)

Answer: Option D. -> Sunday
Answer: (d)

Answer: Option B. -> Friday
Answer: (b)
Odd days in 1600 yrs = 0
Odd days in 300 yrs = 1
46 yrs = (11 leap year + 35 ordinary year)
= (11x2 + 35 x 1) = 1 odd day
∴ Odd days in 1946 yrs = (0+ 1+ 1) = 2
Month
Odd days
January
3
February
0
(ordinary year) 
March
3
April
2
May
3
June
2
July
3
August
1 i.e, (15  ÷ 7)
Total  17  17 ÷ 7= remainder 3 odd days
Total odd days =2 + 3 = 5
∴ Required day = Friday