MAH MBA CET Calendars Questions

In every year, 29th date appears 11 times i.e. in every month except February.
This will amount to 11*400 = 4400 times.
However, in every leap year, 29th date appears 1 more time in February.
So, the number of leap years:
Years divisible by 4: 400 / 4 = 100
Subtract non-leap century years (divisible by 100 but not 400): -3 (e.g., 100, 200, 300)
Total leap years = 100 - 3 = 97 occurrences.
So, total occurrences = 4400 + 97 = 4497.

We will calculate the odd number of days i.e. remainder when 363 is divided by 7. 
The odd number of days will be 6. 
So, the day on 363rd day will be Thursday + 6 = Wednesday.

Last day of a non leap year = first day + 364 .
Now, we calculate the odd days which is the remainder when 364 is divided by 7 = 0.
So, the day on last day of the year = Friday + 0 = Friday.

We will count the number of odd days.
Odd days per year = 1 and odd days per leap year = 2.
As both 1997 and 1998 are not leap years, the number of odd days = 1 + 1 = 2.
Thus, the day on 18th Feb 1999 will be Tuesday + 2 = Thursday i.e. Option A.

To celebrate her wedding on the exact same day, the year of next celebration should have either 0 or 7 or 14...odd days.
It means that if there are N days to the next day of celebration, then remainder when N is divided by 7 should be 0.
We know that every non leap year has +1 odd days and every leap year has +2 odd days.
30th Sept. 1998 --> +1
30th Sept. 1999 --> +1
30th Sept. 2000 --> +2
30th Sept. 2001 --> +1
30th Sept. 2002 --> +1
30th Sept. 2003 --> +1
Sum of all these odd days = 7 which means that day on 30th Sept. 2003 will be Tuesday + 7 = Tuesday.