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 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 ?

  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 year When 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 Year When 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.

years day gain
1994 1
1995 1
1996 (leap year) 2
1997 1
total days 5

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 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 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 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 ?

  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

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 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.

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 15th 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: Number-Day

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 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 1947 days in given month
Jan 31
feb (not leap year) 28
march 31
april 30
may 31
june 30
july 31
august (upto 15th Aug) 15
total 227 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, 7 Sunday
1 Monday
2 Tuesday
3 Wednesday
4 Thursday
5 Friday
6 Saturday

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 1947 days in given month
Jan 31
feb (not leap year) 28
march 31
april 30
may 31
june 30
july 31
august (upto 15th 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 1st Jan 1947till we reach 15th 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 15th 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 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

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 31st December 1856 was Wednesday

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

from 1st Jan 1857till we reach 31th 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 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