### Aptitude-Problems on Calender

Calenders

Important Facts and Formulae:

a) Odd Days :
The number of days more than the complete number of weeks in a given period is number of odd days during that period.

b) Leap year : Every year which is divisible by 4 is called a leap year. Thus each one of he year 1992,1996,2004,2008,..etc, is a leap year.
Every 4th century is a leap year but no other century is a leap year thus each one of 400,800,1200,1600,2000,etc is a leap year. None of the 1900,2010,2020,2100,etc is a leap
year. An year which is not a leap year is called Ordinary year.

c)An ordinary year has 365 days.

d) A leap year has 366 days.

e)Counting of odd days :

i)1 ordinary year = 365 days =52 weeks+1 day,Therefore An
ordinary year has 1 Odd day.
ii)One leap year = 366 days =52 weeks+2 days, Therefore a leap
year has 2 Odd days.
iii) 100 years = 76 ordinary years+ 24 leap years
= [(76*52) weeks+76 days]+[(24*52)weeks+48 days]
= 5200 weeks+124days=[5217 weeks+5 days]
therefore 100 years contain 5 odd days
iv)200 years contain 10(1week+3days), i.e 3 odd days
v)300 years contain 15(2 weeks+1 day), i.e 1 odd day
vi)400 years contain (20+1), i.e 3 weeks,so 0 Odd days
similarly each one of 800,1200,1600,etc contains 0 odd days.

Note:(7n+m) odd days , where m less than or equal to 7
is equivalent to m odd days ,thus ,8 odd days = 1 odd day etc.

f) Some codes o remember the months and weeks:

a) Week

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

b) Month

jan – 1 july – 0
feb – 4 Aug – 3
Mar – 4 Sep – 6
Apr – 0 Oct – 1
May – 2 Nov – 4
june – 5 Dec – 6

Simple problems:

Shortcuts : This shortcut must be applied only starting
with 19 series.

Example:
What day of the week on 17th june , 1998?

Solution : 5 -> the given month code(august)
17 -> the given date
98->(19 th century after years)
24-> ((47/4) = 11 i.e how many leap years
——–
total = 144 ((144/7) = 20 and the remainder is 4)
therefore in the above week table the no 4 code
represents wednesday
so the required day is wednesday.

Problem 1:
The first republic day of the India was celebrated on 26th January,1950. It was

Solution : 01
26
50
12
———-
total = 89 ((89/7) = 12 and the remainder is 5)
therefore in the above week table represents the number 5
as thursday, so the required day was Thursday.

Problem 2:
Find on which day 15th august1947 ?

Solution :
03
15
47
11
———-
total = 76
Then (76)/7 = 6 odd days
6 indicates friday in the above week table.
Therefore required day is friday.

Problem 3:
Find on which day jan 26th 1956 ?

Solution :
01
26
56
14

-1 (-1 indicates leapyear(i.e 1956),so 1 reduce from the total)
———
total = 96
Then (96)/7 = 5 odd days
5 indicates thursday in the above week table
Therefore our required day is Thursday.

Problem 4:
Today is friday after 62 days,it will be :

Solution : Each day of the week is repeated after 7 days.
so, after 63 days,it will be friday. Hence ,after 62 days,
it will be thursday.
Therefore the required day is thursday.

Problem 5:
Find the day of the week on 25th december,1995?

Solution :
06
25
95
23
———
total = 149
Then (149)/7=(23)=2 odd days
Therefore the required day is “Monday”.

Medium Problems

Problem 1:
jan 1, 1995 was a sunday.what day of the week lies on jan 1,1996?

Solution :
01
01
96
24
-1(since 1996 was leap year)
———
total = 121
Then (121)/7 = (17) = 2 odd days
Therefore our required day wasMonday.

Problem 2:
On 8th feb,1995 it wednesday. The day of the week on 8th feb,1994 was?

Solution :
04
08
94
23
———
total = 129
Then (129)/7 = (18) = 3 odd days
Therefore the required day is Tuesday.

Problem 3:
may 6,1993 was thursday.what day of the week was on may 6,1992 ?

Solution :
02
06
92
23
-1
———-
total = 122
Then (122)/7 = (17) = 3 odd days
Therefore the required day is Tuesday

Problem 4:
jan 1, 1992 was wednesday. What day of the week was on jan 1,1993 ?

Solution :
01
01
93
23
———-
total = 118
Then (118)/7 = (16) = 6 odd days
Therefore the required day is Friday.

Problem 5:
January 1,2004 was a thursday,what day of the week lies on jan ,2005?

solution :
The year 2004 being a leap year, it has 2 odd days. so,
first day of the 2005 will be 2 days beyond thursday and
so it will be saturday
therefore the required day is Thursday.

Problem 6:
On 8th march,2005,wednesday falls what day of the week was it on 8th march,2004?

Solution : the year 2004 being a leap year,it has 2 odd days.
so, the day on8th march,2005 will be two days beyond the day
on 8th march,2004.but 8th march,2005 is wednesday. so,
8th march,2004 is monday.
Therefore the required day is Monday.

Problem 7:
what was the day of the week on 19th september ,1986 ?

Solution :
06
19
86
21
———
total = 132
Then ((132/7 = 18 and the remainder is 6)
In the above week table represents the number 6 is friday.
Therefore the required day is Friday.

Typical problems

Problem 1:
On what dates of october,1994 did monday fall ?

Solution : 01
01
94
23
——-
total = 119
Then (119)/7 = (17) = 0 odd days
so the day is saturday
Therefore in october first the day is saturday.so,
the monday fell on 3rd october 1994.During october 1994,
monday fell on 3rd ,10th,17th and 24th.

Problem 2:
How many days are there from 2nd january 1995 to 15 th march,1995 ?

Solution : Jan + Feb + March
30 + 28 + 15 = 73 days

Problem 3:
The year next to 1996 having the same calendar as that of 1996 is ?

Solution : Starting with 1996 , we go on countig the
number of odd days till the sum is divisible by 7.
Year 1996 1997 1998 1999 2000
odd days 2 1 1 1 2
2 + 1 + 1 + 1 + 2 = 7 odd days i.e odd day.
Therefore calendar for 2001 will be the same as
that of 1995.

Problem 4:
The calendar for 1990 is same as for :

Solution:
count the number of days 1990 onwards to get
0 odd day.
Year 1990 1991 1992 1993 1994 1995
oddd days 1 1 2 1 1 1
1 + 1 + 2 + 1 + 1 + 1 = 7 or 0 odd days
Therefore calendar for 1990 is the same as for the
year 1996.

Problem 5:
The day on 5th march of year is the same day on what date of the same year?

Solution:
In the given monthly code table represents the march code and november code both are same.that means any date in march is the same day of week as the corresponding date in november of that year, so the same day falls on 5th november.

1. priya says:

hai

2. Puzzy says:

Very very thank you for ur work, sir..

3. anil says:

your work for fresher is too good thank you very much.

4. pavan says:

what is the day on 31st oct 1984?how

5. vyas says:

hey.. great material..

6. vyas says:

hey.. ur solution for problem no 4 is wrong.. the calendar for 1990 and 1996 are not the same..

7. priyanka says:

great it helps us in many exms thnku very much

8. harshal says:

excellent.!! thank u very much

9. sankari says:

hey….. ur sln for pblm no 4 is wrong.the calender for 1990 and 1996 r not the same

10. raj says:

hai

Ok this is very helpfull for copmtative exam.

Regards
9560693583

11. raj says:

kevalam

12. gnanapriya says:

yenaku nenga fan thana

13. ms says:

I didnt understand the 4th line

14. nitin says:

there is no same calender 1990 1996
1996 and 2001
1995 and 2001
all you are cheater
copy question from R.S Agrawal and paste here

15. rajaanikaant says:

lund mera kuch nahi hoga isase

16. karthick-T says:

odd days assigning number rong.so change the assigning value

17. Himani says:

Could anyone tell me in problem1 no.12 how will come.

18. dev says:

thanks

19. fatima says:

hello frdz, 100 years=76 ordinary yrs + 24 leap yrs.how it happened?plz someone make me clear.

20. fatima says:

hi frndz,in ur shortcut method,17 june 1998
24=47/7=11 i.e how many leap year.plz give explanation of 24.plz ans fast as soon as possible.