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:
214 comments Page 7 of 22.
Rabin said:
1 decade ago
Why there is 1900 in above mentioned explanation?
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?
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?
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
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.
Abuzar Siddiqui said:
1 decade ago
The fast and easy solution according to me is like this:
Days code:- S=1,M=2,T=3,W=4,TH=5,F=6,S=7/0
Month code
JFM AMJ JAS OND
144 025 036 146
CENTURY CODE
100/ 200/ 300/ 400
500 /600/700/ 800
--------------------
4 / 2 / 0 / 6
Ex. 15/8/1947 = day ?
Sol: date+month code+ century code + leap year days + normal year days= odd days/7
15+3 + 0 +11*2 + 36 = 76/7, odd days = 6 by matching odd days, it is FRIDAY
Days code:- S=1,M=2,T=3,W=4,TH=5,F=6,S=7/0
Month code
JFM AMJ JAS OND
144 025 036 146
CENTURY CODE
100/ 200/ 300/ 400
500 /600/700/ 800
--------------------
4 / 2 / 0 / 6
Ex. 15/8/1947 = day ?
Sol: date+month code+ century code + leap year days + normal year days= odd days/7
15+3 + 0 +11*2 + 36 = 76/7, odd days = 6 by matching odd days, it is FRIDAY
Shashank said:
1 decade ago
Guyz we always calculate from 31st DEC 200.. to 31 st DEC 200..
So here, 31 st DEC 2000 to 31st DEC 2005 we gonna calculate
Now year 2001, odd days => 1
2002 => 1
2003 => 1
2004 => 2
2005 => 1
6 odd days and rest of the calculation you can see above.
So here, 31 st DEC 2000 to 31st DEC 2005 we gonna calculate
Now year 2001, odd days => 1
2002 => 1
2003 => 1
2004 => 2
2005 => 1
6 odd days and rest of the calculation you can see above.
Anjali said:
1 decade ago
Too many different month codes used to do the calculation in the above answer figures !
Adishrey said:
1 decade ago
Odd days= (that particular year) /7, the remainder that you get is the no.of odd days in it.
Mrugesh said:
1 decade ago
Important Points:
An ordinary year has 365 days = 52 weeks and 1 odd day.
A leap year has 366 days = 52 weeks and 2 odd days.
Century = 76 Ordinary years + 24 Leap years.
Century contain 5 odd days.
200 years contain 3 odd days.
300 years contain 1 odd day.
400 years contain 0 odd days.
Last day of a century cannot be Tuesday, Thursday or Saturday.
First day of a century must be Monday, Tuesday, Thursday or Saturday.
Explanation:
100 years = 76 ordinary years + 24 leap years
= 76 odd days + 24 x 2 odd days
= 124 odd days = 17 weeks + 5 days
100 years contain 5 odd days.
No. of odd days in first century = 5
Last day of first century is Friday.
No. of odd days in two centuries = 3
Wednesday is the last day.
No. of odd days in three centuries = 1
Monday is the last day.
No. of odd days in four centuries = 0
Sunday is the last day.
Since the order is continually kept in successive cycles, the last day of a century cannot be Tuesday, Thursday or Saturday.
So, the last day of a century should be Sunday, Monday, Wednesday or Friday.
Therefore, the first day of a century must be Monday, Tuesday, Thursday or Saturday.
Working Rules:
Working rule to find the day of the week on a particular date when reference day is given:
Step 1: Find the net number of odd days for the period between the reference date and the given date (exclude the reference day but count the given date for counting the number of net odd days).
Step 2: The day of the week on the particular date is equal to the number of net odd days ahead of the reference day (if the reference day was before this date) but behind the reference day (if this date was behind the reference day).
Working rule to find the day of the week on a particular date when no reference day is given
Step 1: Count the net number of odd days on the given date
Step 2: Write:
For 0 odd days " Sunday
For 1 odd day " Monday
For 2 odd days " Tuesday
. . . .
. . . .
. . . .
For 6 odd days " Saturday
Examples:
1. If 11th January 1997 was a Sunday then what day of the week was on 10th January 2000?
Sol: Total number of days between 11th January 1997 and 10th January 2000
= (365 " 11) in 1997 + 365 in 1998 + 365 in 1999 + 10 days in 2000
= (50 weeks + 4 odd days) + (52 weeks + 1 odd day) +
(52 weeks + 1 odd day) + (1 week + 3 odd days)
Total number of odd days = 4 + 1 + 1 + 3 = 9 days = 1 week + 2 days
Hence, 10th January, 2000 would be 2 days ahead of Sunday i.e. it was on Tuesday.
2. What day of the week was on 10th June 2008?
Sol: 10th June 2008 = 2007 years + First 5 months up to May 2008 + 10 days of June
2000 years have 0 odd days.
Remaining 7 years has 1 leap year and 6 ordinary years2 + 6 = 8 odd days
So, 2007 years have 8 odd days.
No. of odd days from 1st January 2008 to 31st May 2008 = 3+1+3+2+3 = 12
10 days of June has 3 odd days.
Total number of odd days = 8+12+3 = 23
23 odd days = 3 weeks + 2 odd days.
Hence, 10th June, 2008 was Tuesday.
An ordinary year has 365 days = 52 weeks and 1 odd day.
A leap year has 366 days = 52 weeks and 2 odd days.
Century = 76 Ordinary years + 24 Leap years.
Century contain 5 odd days.
200 years contain 3 odd days.
300 years contain 1 odd day.
400 years contain 0 odd days.
Last day of a century cannot be Tuesday, Thursday or Saturday.
First day of a century must be Monday, Tuesday, Thursday or Saturday.
Explanation:
100 years = 76 ordinary years + 24 leap years
= 76 odd days + 24 x 2 odd days
= 124 odd days = 17 weeks + 5 days
100 years contain 5 odd days.
No. of odd days in first century = 5
Last day of first century is Friday.
No. of odd days in two centuries = 3
Wednesday is the last day.
No. of odd days in three centuries = 1
Monday is the last day.
No. of odd days in four centuries = 0
Sunday is the last day.
Since the order is continually kept in successive cycles, the last day of a century cannot be Tuesday, Thursday or Saturday.
So, the last day of a century should be Sunday, Monday, Wednesday or Friday.
Therefore, the first day of a century must be Monday, Tuesday, Thursday or Saturday.
Working Rules:
Working rule to find the day of the week on a particular date when reference day is given:
Step 1: Find the net number of odd days for the period between the reference date and the given date (exclude the reference day but count the given date for counting the number of net odd days).
Step 2: The day of the week on the particular date is equal to the number of net odd days ahead of the reference day (if the reference day was before this date) but behind the reference day (if this date was behind the reference day).
Working rule to find the day of the week on a particular date when no reference day is given
Step 1: Count the net number of odd days on the given date
Step 2: Write:
For 0 odd days " Sunday
For 1 odd day " Monday
For 2 odd days " Tuesday
. . . .
. . . .
. . . .
For 6 odd days " Saturday
Examples:
1. If 11th January 1997 was a Sunday then what day of the week was on 10th January 2000?
Sol: Total number of days between 11th January 1997 and 10th January 2000
= (365 " 11) in 1997 + 365 in 1998 + 365 in 1999 + 10 days in 2000
= (50 weeks + 4 odd days) + (52 weeks + 1 odd day) +
(52 weeks + 1 odd day) + (1 week + 3 odd days)
Total number of odd days = 4 + 1 + 1 + 3 = 9 days = 1 week + 2 days
Hence, 10th January, 2000 would be 2 days ahead of Sunday i.e. it was on Tuesday.
2. What day of the week was on 10th June 2008?
Sol: 10th June 2008 = 2007 years + First 5 months up to May 2008 + 10 days of June
2000 years have 0 odd days.
Remaining 7 years has 1 leap year and 6 ordinary years2 + 6 = 8 odd days
So, 2007 years have 8 odd days.
No. of odd days from 1st January 2008 to 31st May 2008 = 3+1+3+2+3 = 12
10 days of June has 3 odd days.
Total number of odd days = 8+12+3 = 23
23 odd days = 3 weeks + 2 odd days.
Hence, 10th June, 2008 was Tuesday.
(1)
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