 Introduction
 Concept: 7 Day cycle
 Concept: Day Gain Day loss
 Concept: Date Without reference Day
 Mock questions
Introduction
Article prepared with help of Mr. Deepak Singh.
There are two main types of questions from Calendar
when you’re given a reference day  when you’re given no reference “Day” 
What day of week was it on 5th November, 1989 if it was Monday on 4th April, 1988 ?

What was the day on 15^{th} august 1947?

Here question tells us that 4^{th} April 88 was Monday, so we have a ‘reference ‘day.  Here question doesn’t give us any reference “day”. 
can be solved with just two concepts

need to just mugup 2 table, and few steps. 
Although calendar question is not asked every year in every exam. But If and when calendar question is asked, just follow the given procedure and you’ll get the accurate answer= one mark is guaranteed=in that sense, cost: benefit is great.
Concept: 7 Day cycle
Name of the day will repeat after seven days.
1^{st} August 2013= Thursday. Therefore 1^{st} August + 7 =8^{th} August also has to be Thursday.
Then what day will be 10^{th} August 2013?
1^{st} August =Thursday => 1+7=8 August is also Thursday
And 8 August + 2 =10 August will be Thursday + 2 days =>Saturday.
let’s try a simple question from SSC
SSCCGL 2000 Question on 7Day cycle
If 9^{th} of the month falls on the day preceding Sunday, then on what day will 1^{st} of the month fall?
 Friday
 Saturday
 Sunday
 Monday
As per the question 9^{th} of the Month=Saturday. (day preceding Sunday)
Day name repeats after 7 days.
Therefore 9 minus 7=2^{nd} of the given month is also Saturday.
Then 1^{st} of the given month= Saturday Minus one day = Friday, Ans A
Concept: Day Gain Day loss
+1 year  When we proceed forward by one year, then 1 day is gained and viceversa.Example,9^{th} August 2013 is Friday, therefore 9^{th} August 2014 has to be Friday+1=Saturday.Reverse is also true. When we move backward by one year, then 1 day is lost.
9^{th} August 2014 is Saturday, therefore 9^{th} August 2013 has to be Saturday minus 1=Friday. 
+2 Leap Year  When we proceed forward by one leap year, then 2 days are gained and viceversa. Example If it is Wednesday on 10th august 2011 … then it would be Friday (Wednesday +2) on 10th august 2012 [because 2012 is a leap year]. The reverse is also correct: If 22nd April, 1988 = Friday , then 22nd April,1987 = Wednesday. (2 days) 
Exception to Leap Year day gain day loss– the date must have crossed 28th February if the coming year is a leap year for adding 2 days otherwise add 1 day.
For e.g.
 If 26th January 2011 is Wednesday … then 26th January 2012 would be Thursday (even if 2012 is leap year we have added +1 day, because 28th of February is not crossed).
 If 23rd March 2011 is Wednesday … then 23rd march 2012 would be Friday (+2 days. As 28th February of leap year is crossed).
What is Leap year?
A normal year has 365 days. A leap year has 366 days (the extra day is the 29th of February)
A year which is divisible by 4 is a leap year except, if it is divisible by 100 then we have to check it by dividing by 400. For e.g.
 1988, 2008, 2012, 2016 etc. all are leap years as divided by 4.
 2000, 2400, 1600 etc. all are leap as divided by 400 (100 and 4 too).
 1700, 1800, 2100, 2200 etc. are not leap years as they are not divisible by 400 (even if they are divisible by 4).
Let’s try a sum
SSC Investigator 2010 (Anniversary)
On 5^{th} December 1993, Nirmala and Raju celebrated their anniversary on Sunday. What will be the day on their anniversary in 1997?
 Wednesday
 Thursday
 Friday
 Tuesday
Normal year jump +1 gain; Leap year jump +2 gain.
years  day gain 
1994  1 
1995  1 
1996 (leap year)  2 
1997  1 
total days  5 
Sunday plus five days
 Monday
 Tue
 Wed
 Thursday
 Friday
Answer (C) Friday
let’s try one more with a bit difference:
SSC CGL 2011 (Anniversary)
Mrs. Susheela celebrated her wedding anniversary on Tuesday 30^{th} September 1997. What will she celebrate her next wedding anniversary on same day (Tuesday)?
 30 Sept 2003
 30 Sept 2004
 30 Sept 2002
 30 October 2003
After 7 days=> day name will repeat
30^{th} September  day gain 
1998  1 
99  1 
2000 (leap)  2 
2001  1 
2002  1 
2003  1 
total days  7 
Meaning whatever was the day on 30^{th} sep. 1997 it’ll repeat on 2003
Therefore, after 1997, next time Mrs.Susheela’s anniversary will be on same Tuesday on 30^{th} September 2003 Ans A.
Now, How about a CAT level question:
CAT2001 Question on calendar
Dec 9, 2001 is Sunday then what was the day on Dec 9, 1971?
 Thursday
 Wednesday
 Saturday
 Sunday
Same day gain, Day loss
1971 to 2001=how many jumps?
20011971=Total 30 years jump
Out of those 30 years, how many leap years?
72, 76,….,’00 (multiples of of 4 and 2000 is also leap year because It is multiple of 400)
but no need to manually count leap years.
if you observe
18 x 4 =(19)72
25 x 4 =(20)00
so from 18 to 25 = total 8 leap years. (plz note: 2518=7 years, but we’ve to include both years as well…therefore.. in such counting, it’ll be 2518+1=8 leap years)
back to the start: 30 years jump and out of them 8 were leap years.
Meaning 22 normal years + 8 leap years = total 30 years
22 normal years (+1 day gain)  22 x 1 =22 days 
8 leap years (+2 days gain)  8 x 2= 16 days 
total gain  22+16=38 days 
38/7=(7×5)+3 remainder
Meaning whatever was the day on Dec 9, 1971, it’ll move to +3 days on dec 9 2001
Reverse is also true: Whatever was the day on Dec 2001, it’ll be three days less on Dec 9 1971:
Since Dec 2001 was Sunday therefore,
Sunday Minus 3 days= (just count in your head): Saturday…Friday…Thursday.
Final answer is A: Thursday
Question 1988 to 1989
What day of week was it on 5th November,1989 if it was Monday on 4th April, 1988 ?
 Monday
 Tuesday
 Saturday
 Sunday
Recall our Day gain Day loss principle
For NonLeap year, When we proceed forward by one year, then 1 day is gained and viceversa.
If 4th April, 1988 = Monday, then 4th April, 1989 = Tuesday (Because 1989 is a nonleap year)
Remaining days until 5^{th} Nov.89
Month  Days 
April (4 to 30)  26 
May  31 
June  30 
July  31 
Aug  31 
Sep  30 
Oct  31 
Nov  5 
total  215 
=Tuesday + 215
Divide 215 with 7 to find the remainder
215=(30*7)+5 Hence five is the remainder
Back to the problem
= Tuesday + [5 days]
What is the 5^{th} day after Tuesday? Count on your fingertip: Wed, Thurs., Fri, Sat., Sunday
=Sunday (Final answer D)
Speed Technique tip#1
in above problem, if you don’t want to waste time in adding the days like 26+31+30…. then use following approach
30 days=(7*4)+2=> 2 remainder
31 days=(7*4)+3=> 3 remainder.
Month  Days  Remainder with 7 
April (4 to 30)  26  5 (because 26=7*3+5) 
May  31  3 
June  30  2 
July  31  3 
Aug  31  3 
Sep  30  2 
Oct  31  3 
Nov  5  5 
total  don’t need  26 
Divide this by 7 and find remainder: 26=(7*3)+5 so remainder is 5
=Tuesday + 5
= Tuesday + 5 days
=Sunday
Speed Technique tip#2
in speed tech #1, if you don’t want to waste time in adding the remainders of seven: 5+3+2…. then do following
pickup remainders that add upto 7 and cancel them.
for example 5+2=7. So I’ll cancel each such pair in the table. Observe
Month  Days  Remainder with 7after cancelling numbers that add upto “7” (e.g. 5+2) 
April (4 to 30)  26  5 
May  31  3 
June  30  2 
July  31  3 
Aug  31  3 
Sep  30  2 
Oct  31  3 
Nov  5  5 
total  don’t need  12 
After cancelling the two pairs of (5+2), I’m left with four 3s= 12 (because 4 x 3=12)
=Tuesday + 12
now divide 12 with 7 find remainder: 12=7*1 +5
= Tuesday + 5 days
=Sunday
Concept: Date Without reference Day
Example: What was the day on 15^{th} August 1947?
To solve this type of questions, you’ve to first mug up these two tables:
Table#1: The odd days
100  5 
200  3 
300  1 
400, 800, 1200, 1600 etcmultiples of 400  0 
^ok but what is the use of above table? It tells us the number of “odd” days in that given year. I don’t want to bore you with the theory so just mug up those values.
Tablet #2: NumberDay
0, 7  Sunday 
1  Monday 
2  Tuesday 
3  Wednesday 
4  Thursday 
5  Friday 
6  Saturday 
Just remember that 1 to 6 is Monday to Saturday and 0 or 7=Sunday.
^ok but what is the use of above table? just hang on for a few more paragraphs and you’ll know! Now let’s try a sum
What day of week was on 15th ugust,1947?
Step 1:
Subtract the given year by 1.
1947 1 = 1946
Step 2:
Break into relevant years as in table #1.
100  5 
200  3 
300  1 
400, 800, 1200, 1600 etc (multiples of 400)  0 
Therefore,
1946 = 1600 + 300 + ( 46 )
Step 3:
Now write corresponding values from the table: 1600=0 and 300=1
1946=>0+1+(46)
Step 4:
Now for the number in bracket (46), divide it by “4” and add quotient in same line.
in this case 46=(11x4)+2 therefore, 11 is quotient and 2 is remainder. We’re concerned with number and quotient here
1946=0+1+(46+11)
Step5:
Now add all these numbers and divide by 7
0+1+(46+11)
=58
And when you divide 58 by 7, you get 58= (7*8)+2. Therefore remainder is 2.
we got the number “2” = we’ll call this our “31Dec Number”
Observe second table
0, 7  Sunday 
1  Monday 
2  Tuesday 
3  Wednesday 
4  Thursday 
5  Friday 
6  Saturday 
From this table we can see that “2”=>Tuesday
It means 31^{st} December 1946 was Tuesday. Now we apply our “7day cycle” concept to find out the day on 15^{th} August 1947 using the following formula
Given day= (our 31Dec Number “2”)+ *how many days till we reach 15^{th} August?*
Step 6
from 1^{st} Jan 1947 till we reach 15^{th} August 1947  days in given month 
Jan  31 
feb (not leap year)  28 
march  31 
april  30 
may  31 
june  30 
july  31 
august (upto 15^{th} Aug)  15 
total  227 days 
back to the “bold part”
(Our 31Dec Number)+ *how many days till we reach 15^{th} August?*
=2 + 227
=229
Step 7
Divide this number (229) by 7 and whatever remainder you get= that is our day from table #2
229/7=(32*7)+5
so 5 is the remainder and as per table#2
0, 7  Sunday 
1  Monday 
2  Tuesday 
3  Wednesday 
4  Thursday 
5  Friday 
6  Saturday 
5=>Friday.
That means 15^{th} August 1947 was Friday.
Speed increase tip#1:
in above 15^{th} August example, Back to the step6
31Dec Number+ *how many days till we reach 15^{th} August?*
after that, we did following:
from 1^{st} Jan 1947 till we reach 15^{th} August 1947  days in given month 
Jan  31 
feb (not leap year)  28 
march  31 
april  30 
may  31 
june  30 
july  31 
august (upto 15^{th} Aug)  15 
total  227 days 
^as you see we had to sum of 31+28+31….=lot of time taken in doing the addition (+).
So, it is better to just add remainders for each month with “7”
30=(7*4)+2=> 2 remainder
31=(7*4)+3=> 3 remainder.
Now observe again
from 1^{st} Jan 1947till we reach 15^{th} August 1947  days in given month  remainder with 7 
Jan  31  3 
feb (not leap year)  28  0 
march  31  3 
april  30  2 
may  31  3 
june  30  2 
july  31  3 
august (upto 15^{th} Aug)  15  1 (because 14+1) 
total  227 days  17 
now instead of 227 we can simply write 17
Our 31Dec Number+ *how many days till we reach 15^{th} August?*
=2+17
=19
Therefore 19/7 = its remainder will tell us the final day
19=(7*2) + 5=therefore remainder is 5 and as per table#2, “5“ means Friday.
Find the day on 10th May, 1857?
Subtract “1” from the given year
18571=1856
Now breakup “1856” as per our table #1
100  5 
200  3 
300  1 
400, 800, 1200, 1600 etc (multiples of 400)  0 
1856
=1600+200+(56)
write corresponding values from table#1
=0+3+(56)
Divide the bracket number by “4” and write quotient along with number
56=14*4 + 0 therefore quotient is 14
1856
=0+3+(56+14)
add these numbers and divide by 7 find remainder
=73/7
=remainder 3
Speed Technique #3
after doing the division with 4 to find quotient step, you got following
=0+3+(56+14)
Whenever you see the multiple of 7, just scratch that number (because it’ll give remainder zero anyways)
for example
=0+3+(56+14)
Here both 56 and 14 are multiples of 7 so even if you divide them by 7, you’re going to get zero as remainder.
=0+3+0+0=3 is 31Dec Number
as per table#2: “3” means Wednesday. Meaning 31^{st} December 1856 was Wednesday
now “our 31Dec Number”+ ”how many days till 10th May, 1857?”
from 1^{st} Jan 1857till we reach 31^{th} May 1857  days in given month  remainder with 7 
Jan  31  3 
feb (not leap year in 1857)  28  0 
march  31  3 
april  30  2 
may  10  3, Because 10=(7 x 1)+3 
total  *not needed*  11 
back to our problem
“our 31Dec Number”+ ”how many days till 10th May, 1857?”
=3 + 11
=14
when 14 is divided by 7 we get zero as remainder.
as per table#2
0, 7  Sunday 
1  Monday 
2  Tuesday 
3  Wednesday 
4  Thursday 
5  Friday 
6  Saturday 
Zero means Sunday. so final answer 10^{th} May 1857 was Sunday.
Extra facts:
 The 1st day of a century must be Tuesday, Thursday, or Saturday.
 The last day of a century cannot be Tuesday, Thursday, or Saturday.
Mock questions
 In 2013, Gandhi’s birthanniversary is on Wednesday. In which nearest future year, will his birthanniversary be on Monday?
 If 29^{th} April 2013 is Monday then what is the day on 30^{th} November 2013?
 If June 11, 2013 is Monday, what was the day on July 11, 2000
 What was the day on 9/11 attacks in 2001
 What was the day on 26/11 attacks in 2008
Answers
 2017
 30/nov/13=Saturday
 11/jul/00=Monday
 11/09/01=Tuesday
 26/11/08=Wednesday
For more articles on aptitude, visit Mrunal.org/aptitude
165 Comments on “[Reasoning] Calendar Questions: Finding day or date, concepts, shortcuts explained”
This is brilliant. All tricks in one post. Thank you!!