Calendar Questions

The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 24th Nov, 2007 will be 1 day beyond the day on 24th Nov, 2006.
But, 24th Nov, 2007 is Thursday.
24 - 10 = 14 days.
Therefore, 2 weeks ago it is same day.
Thus, 10th Nov, 2006 is one day before 10th Nov, 2007 i.e. it is Wednesday.

The year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.

The day on 5th March, 2005 will be 1 day beyond the day on 5th March, 2004.

Given that, 5th March, 2005 is Monday.

5th March, 2004 is Sunday (1 day before to 5th March, 2005).

We know that,

1 year = 365 days

1 day = 24 hours

1 hour = 60 minutes

1 minute = 60 seconds.

Then, 1 year = 365 x 24 x 60 x 60 seconds.

= 8760 x 3600

1 year = 31536000 seconds.

Hence, 10 years = 31536000 x 10 = 315360000 seconds.

Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.

On 1st Jan 2006, it was Sunday.

Thus, On 1st Jan, 2010 it is Friday.

The year 1985 is an ordinary year. So, it has 1 odd day.

So, the day on 19th June, 1985 will be 1 day after the day on 19th June, 1984.

But, 19th June,1984 is Monday

S0, 19th June, 1985 is Tuesday.