There Are Two Main Types of Questions From Calendar

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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 5thNovember, 1989 if it was Monday on 4th

April, 1988 ?

a. Mondayb. Tuesdayc. Saturday

d. Sunday

What was the day on15 th august 1947?

a. Sundayb. Mondayc. Tuesdayd. Friday

Here question tells us that 4 April 88was Monday, so we have a „reference„day.

Here question doesn t giveus any reference “day”.

can be solved with just two concepts

1. 7 day cycle2. “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 Ifand when calendar question is asked, just follow the given procedureand you ll get the accurate answer= one mark is guaranteed=in thatsense, 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 alsohas 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

SSC-CGL 2000 Question on 7-Day cycle

If 9 th o f the mon th fa ll s on th e day preceding Sund ay, then on w hatday wi l l 1 s t o f the mon th fa ll?

a. Friday b. Saturday c. Sunday

d. 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.

http://www.flickr.com/photos/97816112@N02/9471602231/



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

Concept: Day Gain Day loss

+1year

When we proceed forward by one year, then 1 day is gainedand vice-versa.Example,9 th August 2013 is Friday, therefore9 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 beSaturday minus 1=Friday.

+2LeapYear

When we proceed forward by one leap year, then 2 days aregained and vice-versa. Example If it is Wednesday on 10thaugust 2011 … then it would be Friday (Wednesday +2) on 10thaugust 2012 [because 2012 is a leap year]. The reverse is alsocorrect: If 22nd April, 1988 = Friday , then 22nd April,1987 =Wednesday. (-2 days)

Exception to Leap Year day gain day loss – the date must havecrossed 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 2012would be Thursday (even if 2012 is leap year we have added +1day, because 28th of February is not crossed).

If 23rd March 2011 is Wednesday … then 23rd march 2012 wouldbe 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 isthe 29th of February)

A year which is divisible by 4 is a leap year except, if it is divisible by100 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 5 th Decemb er 1993, Nirm ala and Raju celebrated theiranniversary o n Sun day. What wi l l be the day on the i r anniversary in1997?

a. Wednesday b. Thursday c. Friday d. 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. Monday2. Tue



3. Wed4. Thursday5. Friday

Answer (C) Friday

let s try one more with a bit difference:

SSC CGL 2011 (Anniversary)

Mrs. Susheela celebrated her wedding anniversary on Tuesday30 th September 1997. What will she celebrate her next weddinganniversary on same day (Tuesday)?

1. 30 Sept 20032. 30 Sept 20043. 30 Sept 20024. 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 onsame Tuesday on 30 th September 2003 Ans A.

Now, How about a CAT level question:

CAT-2001 Question on calendar

Dec 9, 2001 is Sund ay then w hat w as the day o n Dec 9, 1971?

A. Thursday B. Wednesday C. Saturday D. 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 veto 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 ondec 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 i t on 5 th Novemb er,1989 i f i t was Mon day on4th A pr il , 1988 ?

1. Monday2. 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 dayis gained and vice-versa.

If 4th April, 1988 = Monday , then 4th April, 1989 = Tuesday (Because1989 is a non-leap 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 + 215Divide 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 like26+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 theremainders 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 thatadd upto “7” (e.g. 5+2)

April (4 to30) 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 tneed 12

After cancelling the two pairs of (5+2), I m left with four 3s= 12 (because4 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 justmug 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 or7=Sunday.

^ok but what is the use of above table? just hang on for a few moreparagraphs 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 insame line.

in this case 46=( 11 x4)+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 remainderis 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 “7daycycle” concept to find out the day on 15 th August 1947 using thefollowing formula

Given day= (our 31Dec Number “2”)+ *how many days till we reach15 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



15 August 1947 month 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 (because14+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“ meansFriday.

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 (becauseit 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 1856was Wednesday

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

from 1 st Jan 1857till we reach31 th May 1857

days in givenmonth 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 x1)+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:

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 whichnearest future year, will his birth-anniversary be on Monday?2. If 29 th April 2013 is Monday then what is the day on 30 th November

2013?3. If June 11, 2013 is Monday, what was the day on July 11, 20004. What was the day on 9/11 attacks in 2001



What was the day on 26/11 attacks in2008

Answers

1. 20172. 30/nov/13=Saturday3. 11/jul/00=Monday4. 11/09/01=Tuesday5. 26/11/08=Wednesday

There Are Two Main Types of Questions From Calendar

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