Aptitude - Calendar - Discussion

Centuries Table

1700-1799 4
1800-1899 2
1900-1999 0
2000-2099 6
2100-2199 4
2200-2299 2
2300-2399 0
2400-2499 6
2500-2599 4
2600-2699 2

Months Table:

January 0 (in leap year 6)
February 3 (in leap year 2)
March 3
April 6
May 1
June 4
July 6
August 2
September 5
October 0
November 3
December 5
Days Table:

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


For example

April 24,1982.

1. Look up the 1900s in the centuries table: 0
2. Note the last two digits of the year: 82
3. Divide the 82 by 4: 82/4 = 20.5 and drop the fractionalpart: 20
4. Look up April in the months table: 6
5. Add all numbers from steps 1-4 to the day of the month (in this case, 24): 0+82+20+6+24=132.
6. Divide the sum from step 5 by 7 and find the remainder: 132/7=18 remainder 6
7. Find the remainder in the days table: 6=Saturday.

Pavithra You told us a very good way to solve the calendar question but only few things you left i.e related to leap year, this formula doesn't work for leap year, it works only for non-leap year.

@Dg1234.

You can solve leap year by this formula as told by Pavithra but only few things I have changed. Check out.

Formula For Leap Year:

0 - Apr, jul 0 - Sat
1 - Jan 1 - Sun
2 - May 2 - Mon
3 - Aug 3 - Tue
4 - Feb, mar, nov 4 - Wed
5 - Jun 5 - Thu
6 - Sep, dec 6 - Fri
7 - Sat

Note:- To find the odd days for leap year we have to add 6 only with this formula.

Formula: 6 + Day + Month + (Year-1900) + (year-1900)/4.

Eg. 4-feb-2004.

= 6+4+4+(2004-1900)+(2004-1900)/4.
= 6+8+104+(104/4).
= 118+26.
= 144.

= 144/7.

= Remainder = 4.
= Wednesday.

Codes for the calendar.

DAYCODES.

Sun = 0.
Mon = 1.
Tue = 2.
Wed = 3.
Thurs = 4.
Fri = 5.
Sat = 6.

CENTURY CODE.
1600-1699 = 6.
1700-1799 = 4.
1800-1899 = 2.
1900-1999 = 0.
2000-2099 = 6.
2100-2199 = 4.
-
-
Repeats.

MONTH CODE:.
Jan = 0.
Feb = 3.
Mar = 3.
April = 6.
May = 1.
June = 4.
July = 6.
Aug = 2.
Sept = 5.
Oct = 0.
Nov = 3.
Dec = 5.

A leap year January code=-1 and Feb = 2.

For example:- Find the day of the week on 26-jan-1956.

Solution:-.

Date + last two digits of the year + (last two digits/4) + monthcode + centuarycode.

= 26 + 56 + (56/4) + (-1) + 0 (since, 56/4 remainder is' 0' it is a leap year) (when leap year comes above).
=26 + 56 + 14 - 1 + 0 codes we said that jan=-1).
= 95 (we should divide by 7 because we have to find the day of the week).
= 95/7 (remainder '4').
= 4 (Thursday).

You can also check the calendar on this date.

Hi Friends,



Odd days of a year X = Remainder of [No. of days in the year X / 7].



[Where 7 --> For every 7 days, the day will be repeated, thats why we are dividing the total number of days of an year by 7. If it has any remainder then it will be considered as an odd day for that year.]





The required remainders (odd days) for the different years are given below:



For 2006: 365/7 = 1 (Total no. of days in 2006 is 365)

For 2007: 365/7 = 1

For 2008: 366/7 = 2 (Leap year, hence Feb month has 29 days)

For 2009: 365/7 = 1



Therefore, total number of odd days = 5.



Given day is 'Sunday' and total no of odd days is '5'.



The required date = Sunday + 5



= (The 5^th day next to Sunday)



= Sunday + 1-Monday + 2-Tuesday + 3-Wednesday + 4-Thursday + 5-Friday.



= Friday (Required answer).

Hi Friends,

Odd days of a year X = Remainder of [No. of days in the year X / 7].

[Where 7 --> For every 7 days, the day will be repeated, thats why we are dividing the total number of days of an year by 7. If it has any remainder then it will be considered as an odd day for that year.]

The required remainders (odd days) for the different years are given below:

For 2006: 365/7 = 1 (Total no. of days in 2006 is 365)
For 2007: 365/7 = 1
For 2008: 366/7 = 2 (Leap year, hence Feb month has 29 days)
For 2009: 365/7 = 1

Therefore, total number of odd days = 5.

Given day is 'Sunday' and total no of odd days is '5'.

The required date = Sunday + 5.

= (The 5th day next to Sunday).
= Sunday + 1-Monday + 2-Tuesday + 3-Wednesday + 4-Thursday + 5-Friday.
= Friday (Required answer).

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

Ans :-
Step 1 ; how many years we have between 2006 to 2010 = 2006 - 2010 =4.
Step 2 ; per year 365 so if we divide by weeks = 365/7 = 52.1 that means 52weeks 1ood days.
Step 3 ; So every 4th year we have an extra day that we call leap year means 366 days again if we divide by weeks = 366/7 = 52.2 in this 2 ood days we have.
Step 4; 2006 = 1.
For 2007 = 1.
For 2008 = 1.
For 2009 = 2 (1+1+1+2 = 5days).
Step 5;
1 jan 2006 = Sunday count 1 .
1 jan 2007 = Monday count 1.
1 jan 2008 = Tuesday count 1.
1 jan 2009 = Wednesday & Thursday count 2 because as per our count it is leap year(2days extra).
1 jan 2010 = answer is Friday.

Month code:

January -1.
February -4.
March -4.
April -0.
May -2.
June -5.
July -0.
August -3.
September-6.
October -1.
November -4.
December -6.

Days:

Saturday -0.
Sunday -1.
Monday -2.
Tuesday -3.
Wednesday -4.
Thursday -5.
Friday -6.

Century code:

1500-1599=0.
1600-1699=6.
1700-1799=4.
1800-1899=2.
1900-1999=0.
2000-2099=6.
2100-2199=4.

Formula:

Date+Month code+No. of years+No. of leap years+century code/7.

Problem:

That ''2004 august 15'' which day?

Solution:

15+3+4+1+6/7.

=>7) 29 (4.
28.
_.
1 = Sunday.

Answer is Sunday.

Note:

No. of leap year calculation is.

4) 4 (1=quotation value can be taken.
4.
-.
0.

Hi, you can do it easily follow my shortcut rule :-

jan = (1) = 3 jan = 31/7 = remain = 3
feb = (2)= 0 feb= 28/7 = remain = 0
march = (3)= 3
april = (4)= 2 april = 30/7 = reminder = 2
may = (5) = 3
june = (6) = 2
july = (7) =3 ====================================
aug = (8)= 3 (1 , 3, 5, 7, 8, 10, 12) = 3
sept = (9)= 2 ( 4, 6, 9, 11) = 2
oct = (10)= 3 2= 0 (Feb)
nov = (11) = 2
dec = (12) = 3 JUST REMEMBER ABOVE

It is very simple remeber these 2 basic terms:

1)whenever we go from general year to general year or leap year then add 1.
2)whenever we go from leap year to general year then add 2.

Now come to question:

Given Jan 1 , 2006 was Sunday then;
Jan 1, 2010 = ?.

So here we go;

2006(general year) to 2007(general year)=1
2007(general year) to 2008(leap year)=1
2008(leap year) to 2009(general year)=2
2009(general year) to 2010(general year)=1

Adding them=1+1+2+1=5.
now Jan 1, 2010 = sunday+5= Friday.

Hope you guys got it. It's very simple method.