Aptitude - Calendar - Discussion
Discussion Forum : Calendar - General Questions (Q.No. 2)
2.
What was the day of the week on 28th May, 2006?
Answer: Option
Explanation:
28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2)
6 odd days
Jan. Feb. March April May (31 + 28 + 31 + 30 + 28 ) = 148 days
148 days = (21 weeks + 1 day)
1 odd day.
Total number of odd days = (0 + 0 + 6 + 1) = 7
0 odd day.
Given day is Sunday.
Discussion:
215 comments Page 16 of 22.
Chanakya said:
1 decade ago
Deena your wrong by the way gave some idea to do calender sum like this thank you.
Naveen your procedure is absolutely prefect and thanks boss.
Naveen your procedure is absolutely prefect and thanks boss.
Asa said:
1 decade ago
Solution: JFM AMJ JAS OND
1. Choose appropriate month code: 602 503 514 624
2. Leap years since 1/1/2000 (including the year 2000):
3. Add the month code from step 1, the number of leap years, the day of the month, and the amt of years since 2000. Divide this by 7 and take the remainder.
4. Sun = 1, Mon= 2, T = 3...Fri=6, Sat=0,7
Example: January 1,2050
1. month code 6 2. 50yrs/4=12r2. 12+1 (since Feb 2000 had a leap day) = 13
3. 6 + 13 + 1 + 50 = 70. 70/7=10r0
4. remainder of 0 -> January 1, 2050 will be a Saturday
*If the year you are calculating is a leap year and the date is BEFORE February 29, do add the 1*
Example of this: February 18, 2008
1. month code 2
2. 2 (Feb 2000 & 2004 BUT NOT 2008 yet)
3. 2 + 2 + 18 + 8 = 30. 30/7=4r2
4. remainder 2 -> Monday
1. Choose appropriate month code: 602 503 514 624
2. Leap years since 1/1/2000 (including the year 2000):
3. Add the month code from step 1, the number of leap years, the day of the month, and the amt of years since 2000. Divide this by 7 and take the remainder.
4. Sun = 1, Mon= 2, T = 3...Fri=6, Sat=0,7
Example: January 1,2050
1. month code 6 2. 50yrs/4=12r2. 12+1 (since Feb 2000 had a leap day) = 13
3. 6 + 13 + 1 + 50 = 70. 70/7=10r0
4. remainder of 0 -> January 1, 2050 will be a Saturday
*If the year you are calculating is a leap year and the date is BEFORE February 29, do add the 1*
Example of this: February 18, 2008
1. month code 2
2. 2 (Feb 2000 & 2004 BUT NOT 2008 yet)
3. 2 + 2 + 18 + 8 = 30. 30/7=4r2
4. remainder 2 -> Monday
Lalitha said:
1 decade ago
How can we say the odd day is 0 for 1600 and 400 years...i am getting 1 odd day for total 2000 years....hear is my calculation if wrong pls corret it..
2000/4=500
so 500 leap years..no of odd days for 500 years is (500*2)1000
and 1500 ordinary years having 1500 odd days
total=1500+1000=2500
no of odd days in these 2500 days: 2500/7 = 1 odd day
is it correct procedure?
2000/4=500
so 500 leap years..no of odd days for 500 years is (500*2)1000
and 1500 ordinary years having 1500 odd days
total=1500+1000=2500
no of odd days in these 2500 days: 2500/7 = 1 odd day
is it correct procedure?
Madhu said:
1 decade ago
How can we say 0 is sunday 1 is monday, 2 tues day, . How can we know the jan 1st 2006 week day?
Rabin said:
1 decade ago
Why there is 1900 in above mentioned explanation?
Rakhi said:
1 decade ago
I can't understand this step plz explain it?? 5=(5/4)leap yr+[5-1]ordinary day
=1 leap year+4 ordinary day
=1 leap year+4 ordinary day
Rakhi said:
1 decade ago
It can be calculated in easier way:
2006 -> January - 31 = 31/7 = 3 odd days.
Feb- 28=28/7 = 0 odd days.
March - 31 =31/7 = 3 odd days.
April 30 = 30/7=2 odd days.
May up to 27 dayth : 27 / 7 =6 odd days.
Total: 3+0+3+2+6= 14 -> 14 /7 = 0 odd days.
0 i.e. Sunday.
Hence: answer = Sunday.
2006 -> January - 31 = 31/7 = 3 odd days.
Feb- 28=28/7 = 0 odd days.
March - 31 =31/7 = 3 odd days.
April 30 = 30/7=2 odd days.
May up to 27 dayth : 27 / 7 =6 odd days.
Total: 3+0+3+2+6= 14 -> 14 /7 = 0 odd days.
0 i.e. Sunday.
Hence: answer = Sunday.
Ravindra reedy said:
1 decade ago
Suppose we have 23days at this time find odd days?
Ans:
No of days is divisible by 7 then we get odd days.
23/7=2.
2 odd days.
Ans:
No of days is divisible by 7 then we get odd days.
23/7=2.
2 odd days.
Chirag said:
1 decade ago
Rule 1: Find the remind from complete years by dividing 28
i.e(2005/28 gives 17)
So consider 17 years as left.. it has 4 leap year
as general procedure code for that is 0
Rule 2: Same for month and date calculation :
(31+28+31+30+28)/7 it gives remainder 1
So code is 1
Total code is 1+0=1
Now,
MAGIC rule: add 6 in total cade
i.e (1+6=7)
So code is 7 means 0 means Sunday ! !
It can apply for any type of question...
i.e(2005/28 gives 17)
So consider 17 years as left.. it has 4 leap year
as general procedure code for that is 0
Rule 2: Same for month and date calculation :
(31+28+31+30+28)/7 it gives remainder 1
So code is 1
Total code is 1+0=1
Now,
MAGIC rule: add 6 in total cade
i.e (1+6=7)
So code is 7 means 0 means Sunday ! !
It can apply for any type of question...
Shawisk said:
1 decade ago
@Deena.
What we write instead for 1900 if the date is 13 may 1666
For = 28+2+(2006-1900)+(2006-1900)/4.
What we write instead for 1900 if the date is 13 may 1666
For = 28+2+(2006-1900)+(2006-1900)/4.
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