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.
$\therefore$ The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.
Given that, 6th March, 2005 is Monday.
$\therefore$ 6th March, 2004 is Sunday., 1 day before to 6th March, 2005.
Each day of the week is repeated after 7 days.
So, after 63 days, it will be Monday.
$\therefore$ After 61 days, it will be Saturday.
The year 2004 is a leap year. It has 2 odd days.
$\therefore$ The day on 8th Feb, 2004 is 2 days before the day on 8th Feb, 2005.
Hence, this day is Sunday.
Day before yesterday was Thursday
=>Yesterday was a Friday
=> Today is a Saturday
=> Tomorrow is a Sunday
Number of days from 26-Jan-1996 to 15-May-1996 [both days included]
= 6 Jan + 29 Feb + 31 Mar + 30 Apr+ 15 May = 111
Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.
The year 2008 is a leap year. So, it has 2 odd days.
1st day of the year 2008 is Tuesday Given
So, 1st day of the year 2009 is 2 days beyond Tuesday.
Hence, it will be Thursday.
15th August, 2010 = 2009 years + Period 1.1.2010 to 15.8.2010
Odd days in 1600 years = 0
Odd days in 400 years = 0
9 years = 2 leap years + 7 ordinary years = $(2 \times 2 + 7 \times 1)$ = 11 odd days $\equiv$ 4 odd days.
On 31st December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = $\left(1 + 1 + 2 + 1\right)$ = 5 days.
$\therefore$ On 31st December 2009, it was Thursday.
Thus, on 1st Jan, 2010 it is Friday.
$ x $ weeks $x$ days = 7$ x $ + $ x $ days = 8$ x $ days.
- Logical Sequence Test 1
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