[Reasoning] Calendar Questions: Finding day or date, concepts, shortcuts explained

SubscribeAptitude159 Comments

IAS Mock Interviews
  1. Introduction
  2. Concept: 7 Day cycle
    1. SSC-CGL 2000 Question on 7-Day cycle
  3. Concept: Day Gain Day loss
    1. What is Leap year?
    2. SSC Investigator 2010 (Anniversary)
    3. SSC CGL 2011 (Anniversary)
    4. CAT-2001 Question on calendar
    5. Question 1988 to 1989
    6. Speed Technique tip#1
    7. Speed Technique tip#2
  4. Concept: Date Without reference Day
    1. Table#1: The odd days
    2. Tablet #2: Number-Day
    3. What day of week was on 15th ugust,1947?
    4. Speed increase tip#1:
    5. Find the day on 10th May, 1857?
    6. Speed Technique #3
  5. 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 daywhen you’re given no reference “Day”
What day of week was it on 5th November, 1989 if it was Monday on 4th April, 1988 ?

  1. Monday
  2. Tuesday
  3. Saturday
  4. Sunday
What was the day on 15th august 1947?

  1. Sunday
  2. Monday
  3. Tuesday
  4. Friday
Here question tells us that 4th 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

  1. 7 day cycle
  2. “day gain-Day loss” concept
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

aptitude calendar
Name of the day will repeat after seven days.

1st August 2013= Thursday. Therefore 1st August + 7 =8th August also has to be Thursday.

Then what day will be 10th August 2013?

1st 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

SSC-CGL 2000 Question on 7-Day cycle

If 9th of the month falls on the day preceding Sunday, then on what day will 1st of the month fall?

  1. Friday
  2. Saturday
  3. Sunday
  4. Monday

As per the question 9th of the Month=Saturday. (day preceding Sunday)

Day name repeats after 7 days.

Therefore 9 minus 7=2nd of the given month is also Saturday.

Then 1st of the given month= Saturday Minus one day = Friday, Ans A

Concept: Day Gain Day loss

+1 yearWhen we proceed forward by one year, then 1 day is gained and vice-versa.Example,9th August 2013 is Friday, therefore 9th August 2014 has to be Friday+1=Saturday.Reverse is also true. When we move backward by one year, then 1 day is lost.

9th August 2014 is Saturday, therefore 9th August 2013 has to be Saturday minus 1=Friday.

+2 Leap YearWhen we proceed forward by one leap year, then 2 days are gained and vice-versa. 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.

  1. 1988, 2008, 2012, 2016 etc. all are leap years as divided by 4.
  2. 2000, 2400, 1600 etc. all are leap as divided by 400 (100 and 4 too).
  3. 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 5th December 1993, Nirmala and Raju celebrated their anniversary on Sunday. What will be the day on their anniversary in 1997?

  1. Wednesday
  2. Thursday
  3. Friday
  4. Tuesday

Normal year jump +1 gain; Leap year jump +2 gain.

yearsday gain
19941
19951
1996 (leap year)2
19971
total days5

Sunday plus five days

  1. Monday
  2. Tue
  3. Wed
  4. Thursday
  5. 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 30th September 1997. What will she celebrate her next wedding anniversary on same day (Tuesday)?

  1. 30 Sept 2003
  2. 30 Sept 2004
  3. 30 Sept 2002
  4. 30 October 2003

After 7 days=> day name will repeat

30th Septemberday gain
19981
991
2000 (leap)2
20011
20021
20031
total days7

Meaning whatever was the day on 30th sep. 1997 it’ll repeat on 2003

Therefore, after 1997, next time Mrs.Susheela’s anniversary will be on same Tuesday on 30th September 2003 Ans A.

Now, How about a CAT level question:

CAT-2001 Question on calendar

Dec 9, 2001 is Sunday then what was the day on Dec 9, 1971?

  1. Thursday
  2. Wednesday
  3. Saturday
  4. Sunday

Same day gain, Day loss

1971 to 2001=how many jumps?

2001-1971=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: 25-18=7 years, but we’ve to include both years as well…therefore.. in such counting, it’ll be 25-18+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 gain22+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 ?

  1. Monday 
  2. Tuesday
  3. Saturday
  4. Sunday

Recall our Day gain Day loss principle

For Non-Leap year, When we proceed forward by one year, then 1 day is gained and vice-versa.

If 4th April, 1988 = Monday, then 4th April, 1989 = Tuesday (Because 1989 is a non-leap year)

Remaining days until 5th Nov.89

MonthDays
April (4 to 30)26
May31
June30
July31
Aug31
Sep30
Oct31
Nov5
total215

=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 5th 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.

MonthDaysRemainder with 7
April (4 to 30)265 (because 26=7*3+5)
May313
June302
July313
Aug313
Sep302
Oct313
Nov55
totaldon’t need26

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

MonthDaysRemainder with 7after cancelling numbers that add upto “7” (e.g. 5+2)
April (4 to 30)265
May313
June302
July313
Aug313
Sep302
Oct313
Nov55
totaldon’t need12

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 15th August 1947?

To solve this type of questions, you’ve to first mug up these two tables:

Table#1: The odd days

1005
2003
3001
400, 800, 1200, 1600 etcmultiples of 4000

^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: Number-Day

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

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.

1005
2003
3001
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, 7Sunday
1Monday
2Tuesday
3Wednesday
4Thursday
5Friday
6Saturday

From this table we can see that “2”=>Tuesday

It means 31st December 1946 was Tuesday. Now we apply our “7day cycle” concept to find out the day on 15th August 1947 using the following formula

Given day= (our 31Dec Number “2”)+ *how many days till we reach 15th August?*

Step 6

from 1st Jan 1947 till we reach 15th August 1947days in given month
Jan31
feb (not leap year)28
march31
april30
may31
june30
july31
august (upto 15th Aug)15
total227 days

back to the “bold part”

(Our 31Dec Number)+ *how many days till we reach 15th 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, 7Sunday
1Monday
2Tuesday
3Wednesday
4Thursday
5Friday
6Saturday

5=>Friday.

That means 15th August 1947 was Friday.

Speed increase tip#1:

in above 15th August example, Back to the step6

31Dec Number+ *how many days till we reach 15th August?*

after that, we did following:

from 1st Jan 1947 till we reach 15th August 1947days in given month
Jan31
feb (not leap year)28
march31
april30
may31
june30
july31
august (upto 15th Aug)15
total227 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 1st Jan 1947till we reach 15th August 1947days in given monthremainder with 7
Jan313
feb (not leap year)280
march313
april302
may313
june302
july313
august (upto 15th Aug)151 (because 14+1)
total227 days17

now instead of 227 we can simply write 17

Our 31Dec Number+ *how many days till we reach 15th 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

1857-1=1856

Now breakup “1856” as per our table #1

1005
2003
3001
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 31st December 1856 was Wednesday

now “our 31Dec Number”+ ”how many days till 10th May, 1857?”

from 1st Jan 1857till we reach 31th May 1857days in given monthremainder with 7
Jan313
feb (not leap year in 1857)280
march313
april302
may103, 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, 7Sunday
1Monday
2Tuesday
3Wednesday
4Thursday
5Friday
6Saturday

Zero means Sunday. so final answer 10th May 1857 was Sunday.

Extra facts:

  1. The 1st day of a century must be Tuesday, Thursday, or Saturday.
  2. The last day of a century cannot be Tuesday, Thursday, or Saturday.

Mock questions

  1. In 2013, Gandhi’s birth-anniversary is on Wednesday. In which nearest future year, will his birth-anniversary be on Monday?
  2. If 29th April 2013 is Monday then what is the day on 30th November 2013?
  3. If June 11, 2013 is Monday, what was the day on July 11, 2000
  4. What was the day on 9/11 attacks in 2001
  5. What was the day on 26/11 attacks in 2008

Answers

  1. 2017
  2. 30/nov/13=Saturday
  3. 11/jul/00=Monday
  4. 11/09/01=Tuesday
  5. 26/11/08=Wednesday

For more articles on aptitude, visit Mrunal.org/aptitude

Mrunal recommends

  1. (free) NCERT, NIOS, TN-Books 4 History,Geo,Sci
  2. Indian Polity M.Laxmikanth (Hindi | English)
  3. Spectrum: Modern History (Hindi | English)
  4. Maths: Quantam CAT Sarvesh Kumar
  5. Objective General English SP Bakshi
  6. Word Power made Easy -Norman Lowe
  7. Topic wise Solved Paperset by Disha

159 Comments on “[Reasoning] Calendar Questions: Finding day or date, concepts, shortcuts explained”

  1. the 26th december of a year falls on a sunday.on what day of the week did the 22nd july of the year fall ???
    can anyone solve this

    1. thursday on 22 nd july…’…..

      1. can not determine will be answer because we dont know the month ferbuary is 29 days or 28

        1. In such cases, where year is not given, consider it as ordinary year (i.e. 365 days)

      2. answer is Thursday year is not important on this problem.because same year between July and December.February not comes between July and December.

  2. Sir please guide 4 Mains sociology (hindi) preparation

  3. could u explain the question 3) If June 11, 2013 is Monday, what was the day on July 11, 2000

    1. I also have doubt about the answer. I am getting Sunday through my calculations but the answer mentions Monday. Can somebody please share the working of it.

  4. Hi, thanks for all the explanation. It really helps. I have a doubt on mock question number 3 (July 11,2000). I tried both the way and the answer seems to be different. Can u please check and if possible send the details?

    1. its sunday,,,
      11 june 2013 is monday ..than
      11june 2012 is sunday because it is a normal year it has one odd day.
      similarly 11 june 2011 its saturday than come to year
      2000 11 june must be friday
      but we have to find 11 july so count the days between them its 30
      thats means 4 weeks and 2 days..
      2 days are odd days so(it is friday+2)=sunday
      sunday is the answer

      1. I am also getting the same answer. I am still not sure if it’s the correct answers as it doesn’t match the above answer.

  5. Hi Ajay, I believe since the year is not mentioned we can find 22nd July just in the following way.
    Given 26th Dec is Sunday, let’s count backward till 22nd July. Like
    December 25 days (26th is excluded) (4 odd days)
    November 30 days (2 odd days)
    October 3 odd days
    September 2 odd days
    August 3 odd days
    July 10 days ( ’cause we have to find the day of week 22nd of the month and while going reverse we can count like this 31st-1,30th-2 so on..)
    Total odd days we got=4+2+3+2+3+10=24
    While dividing by 7 we get a reminder of 3.
    Since days were counted backward to find a past day, we have to substract 3 days from the day of December 26th i.e. Sunday. (It is Saturday… Friday.. Thursday)
    Answer is Thursday

  6. if 13 th march in a year is a Friday then on which day of the week will 13 th January of the next year fall
    a) Wednesday b) Friday c) sunday d) cannot be determined

    please tell me the answer…. I’m waiting for yur answer

    1. cant determine becase we dont know the next year is leap year or ordinary.

  7. What is your Message? Search before asking questions & confine discussions to exams related matter only. I want if one week mai how many day

  8. THANKS..It was a gr8 help…cheers:-)

  9. sir mock test question 4 where it is asked that 9/11 took place on which day then answer will be FRIDAY.Answer is given of 11/9 2001 which was tuesday.
    2001-1=2000
    400×5
    0 thus 31st dec of 2000 was sunday.
    from Jan to Nov 9= 31+28+31+30+31+30+31+31+30+31+9=313
    313=(44×7+5) thus 9 Nov 2001 was Friday.

  10. what was day on 31st march 1985?

  11. Sir please provide a method to find a REPEAT calender

  12. It was sunday on 31st march 1985.

  13. 9/11 means twin tower attack so here answer is according to the sept/11 date format.

  14. @surabhi , why are not counting 2000 though it’s a leap year ? Totally there are 4 leap years right .. please help ! Once I consider that way I’m getting answer as Sunday so please clarify
    This is in reference to 3rd question of the mrunal Sir’s mock q’s..

  15. @wrathod .. could you explain why are you not cosidering 2000 though it is a leap year … I followed the same like #rajat123 and I got the same answer as him ..

Leave a Reply

Your email address will not be published. Required fields are marked *