Calendar deals with odd days, leap year, ordinary year, counting of odd days, and day of the week related to odd days. To find the day of the week on a given date concept of odd days is used. The article **Calendar Quiz 2** lists important **Calendar** practice questions for competitive exams like **RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams** and etc.

Examples:

- Years like 1988, 2008 are leap year (divisible by 4).

Centuries like 2000, 2400 are leap year ( divisible by 400).

Years like 1999, 2003 are not leap year (not divisible by 4).

Centuries like 1700, 1800 are not leap year ( not divisible by 400).

In a century, there is 76 ordinary year and 24 leap year.

An ordinary year has one odd day.

10 years = [latex]8 \times 400 + 2 \times 450[/latex] days = 3200 + 900 days = 4100 days

4100 days = 585 weeks + 5 days

585 weeks will end on a Thursday

5 additional days will end on Tuesday.

29th February [latex]\Rightarrow[/latex] Monday

Therefore;

1st March [latex]\Rightarrow[/latex] Tuesday.

Hence Option C is correct

Therefore, today is Thursday + 4 = Monday

Tomorrow will be Friday

Day after tomorrow will be Saturday

Now, three days after tomorrow will be Tuesday.

30th September 1999 = Thursday

30th September 2000 = Saturday

Because, 2000 is a Leap Year (there is one extra day in the month of February.)

30th September 2001 = Sunday

30th September 2002 = Monday

30th September 2003 = Tuesday

An ordinary year has one Odd day.

Thus, she will celebrate her next wedding anniversary on 30th September 2003 as the same day.

Hence, option A is correct.

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