Aptitude - Calendar - Discussion

On a fix particular date, as the year increases by one unit, the day also increases by one unit, Except for leap year where the day is increased by two units.

So coming to the problem:

On particular date 1 Jan.
Day- sun and year-2006.

So on increasing year by one unit.
Year - Day
2007 -.Monday.
2008. - Tuesday.
2009 - Thursday (when leap year is increased by one unit days Will be increased by two units)
2010 - Friday.

Month code for ordinary year 144 025 036 146.

Formula to find day = date + month code + last 2 digit of year + last 2 digit/4 ......note- take only whole number not after decimal.

Ex.. 15 Aug 1989.

15 + 3 + 89+ (89/4).
15 +3+ 89+22..........89/4=22.xxxx
= 129.

Divide 129 by 7.
129/7 = The remainder will be 3.

Then sunday =1
Monday=2
Tuesday=3
Wednesday=4
Thursday=5
Friday=6
Saturday=00
Aur remainder is 3 then answer will be Tuesday.

Pls, remember.

For 1900 series add in addition = 00
For series 2000 add +6
For series 2100 add +4
For series 2200 add +2
For series 2300 add +00
For series 1800 add +02
For series 1700 add +04
For series 1600 add +06

2006 = 1(odd day).
2007 = 1(odd day).
2008 = 2(odd days-it is a leap year).
2009 = 1(odd day).

Total = 5(odd days).

Monday=1
Tuesday=2
Wednesday=3
Thursday=4
Friday=5
Saturday=6
Sunday=7/0

So now we consider 5 as Friday.

3 April 2005.
6 August 2010.
5 December 2013.
15 August ?

How can we find that 1996 is a leap year?

How many Sundays are there in year if 1st December in that year is on Sunday?

November 25, 1950 was Saturday what day was November 2016?

If the first day of 1987 is Thursday, then what will be the last day of the year 1987?

@Deepika.

If the given year is a non-leap year then the first day is the same as the last day of the year.

For leap year last day will be the next day of the starting day of the year.

We can find odd days by year divided by 4.