Aptitude - Calendar - Discussion
Discussion Forum : Calendar - General Questions (Q.No. 11)
11.
The calendar for the year 2007 will be the same for the year:
Answer: Option
Explanation:
Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.
Year : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 Odd day : 1 2 1 1 1 2 1 1 1 2 1
Sum = 14 odd days
0 odd days.
Calendar for the year 2018 will be the same as for the year 2007.
Discussion:
122 comments Page 2 of 13.
Dev said:
4 years ago
@All
Given;
Year +6
2007+6=2013.
Find out how many leap year there in between 2007 and 2013.
there is 2008 and 2012 (2 Leap year).
If there is 2 Leap year then add 11 to the given question. It will be your answer.
2007+11 = 2018.
Instead of;
If there will be 1leap year while adding 6 in given question the answer will be given year +6= answer.
If in question there will be Leap year. Then add +5 it will be your answer.
Given;
Year +6
2007+6=2013.
Find out how many leap year there in between 2007 and 2013.
there is 2008 and 2012 (2 Leap year).
If there is 2 Leap year then add 11 to the given question. It will be your answer.
2007+11 = 2018.
Instead of;
If there will be 1leap year while adding 6 in given question the answer will be given year +6= answer.
If in question there will be Leap year. Then add +5 it will be your answer.
(5)
Prasad said:
1 decade ago
For every ordinary year number of odd days = 1 odd day
For leap year number of odd days = 2 odd days.
If the odd days sum=7 implies odd days=0 then again loop starts from 1,2,3,4,..7 .
2008,2012,2016 are clearly leap years,
therefore
2007=1 odd day,
2008=2 (SUM=3),
2009=1 (SUM=4),
2010=1 (SUM=5),
2011=1 (SUM=6),
2012=2 odd days (SUM=8)
[HERE OBSERVE THAT SUM OF ODD DAYS EXCEEDING THE VALUE 7 THEREFORE CONTINUE ADDING THE ODD DAYS TILL WE GET A NUMBER WHICH IS DIVISIBLE BY 7],
2013=1 (SUM=9),
2014=1 (SUM=10),
2015=1 (SUM=11),
2016 = 2 odd days (SUM=13),
2017=1 (SUM=14)
upto 2017 SUM OF ODD DAYS = 14 divisible by 7 implies next year i.e.,2018 will have same calender as 2007 .
For leap year number of odd days = 2 odd days.
If the odd days sum=7 implies odd days=0 then again loop starts from 1,2,3,4,..7 .
2008,2012,2016 are clearly leap years,
therefore
2007=1 odd day,
2008=2 (SUM=3),
2009=1 (SUM=4),
2010=1 (SUM=5),
2011=1 (SUM=6),
2012=2 odd days (SUM=8)
[HERE OBSERVE THAT SUM OF ODD DAYS EXCEEDING THE VALUE 7 THEREFORE CONTINUE ADDING THE ODD DAYS TILL WE GET A NUMBER WHICH IS DIVISIBLE BY 7],
2013=1 (SUM=9),
2014=1 (SUM=10),
2015=1 (SUM=11),
2016 = 2 odd days (SUM=13),
2017=1 (SUM=14)
upto 2017 SUM OF ODD DAYS = 14 divisible by 7 implies next year i.e.,2018 will have same calender as 2007 .
(4)
Krushna said:
3 years ago
To find the total number of odd days.
Divide the total given days by 7 and it's remainder shows odd days.
Then, Maximum number of odd days is 6.
Divide the total given days by 7 and it's remainder shows odd days.
Then, Maximum number of odd days is 6.
(4)
Ramu said:
1 decade ago
@anand ques is to find a year whose calender wil b same as 2007
so you should go for coming years where the sum of odd days will be multiples of 7(which makes exact week)
for a regular year we wil be having 365/7 which gives remainder 1 will be having 1 odd day
and for leap years we will be havin 2 odd days
YEAR ODD DAYS
2007 1
2008(LEAP) 2
2009 1
2010 1
2011 1
2012(LEAP) 2
2013 1
2014 1
2015 1
2016(LEAP) 2
2017 1
here we will get the sum of odd days as 14 which again makes exact 2 weeks...so next year will get the same calender of 2007
so you should go for coming years where the sum of odd days will be multiples of 7(which makes exact week)
for a regular year we wil be having 365/7 which gives remainder 1 will be having 1 odd day
and for leap years we will be havin 2 odd days
YEAR ODD DAYS
2007 1
2008(LEAP) 2
2009 1
2010 1
2011 1
2012(LEAP) 2
2013 1
2014 1
2015 1
2016(LEAP) 2
2017 1
here we will get the sum of odd days as 14 which again makes exact 2 weeks...so next year will get the same calender of 2007
(3)
Mithilesh Mahatha said:
5 years ago
@All.
So, doubt here is how 14 = 0,
See,
No of days in a week is 7.
In 14 we can make to weeks,
14/7 completely divides no remainder.
Therefore 0 odd days.
All multiple of 7 will gives 0 odd days..
So, doubt here is how 14 = 0,
See,
No of days in a week is 7.
In 14 we can make to weeks,
14/7 completely divides no remainder.
Therefore 0 odd days.
All multiple of 7 will gives 0 odd days..
(2)
Priya said:
1 decade ago
This calendar method will take time to find the day of the week.... I have seen a new easy method in an quantitative aptitude book BEACON....if possible read that book
(1)
Hariharan said:
1 decade ago
Hi All
Pls explain me why 2018 is not included in this step.
Pls explain me why 2018 is not included in this step.
(1)
Jati n kamani said:
1 decade ago
Please explain me how to 14 odd days to equal 0 odd day.
(1)
Sudarshan said:
10 years ago
As per my calculations (all pl refer calendar) I have separate calculations for leap years and non leap years.
Eg: Take a leap year 2012. 5 years before this leap year and 6 years after this leap year you get the same calendar (except January and February as 2012 was a leap year and 2007 and 2018 were non leap years).
Hence you can arrive 2007 and 2018 as the same. Pl apply this logic to all leap years. I know this because I can say any days from years 1 A.D till infinity within seconds.
Eg: Take a leap year 2012. 5 years before this leap year and 6 years after this leap year you get the same calendar (except January and February as 2012 was a leap year and 2007 and 2018 were non leap years).
Hence you can arrive 2007 and 2018 as the same. Pl apply this logic to all leap years. I know this because I can say any days from years 1 A.D till infinity within seconds.
(1)
Navnath said:
9 years ago
The year next to 1973 having the same calendar as that of 1973 is ____.
1) 1977
2) 1976
3) 1979
4) 1978
Please solve this.
1) 1977
2) 1976
3) 1979
4) 1978
Please solve this.
(1)
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