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:
121 comments Page 5 of 13.
Sonu kumar said:
3 years ago
@All.
Divide the given no by 4 if remainder is 0 add 28 to the year if remainder is 1 add 6 or if remainder is 2or3 add 11, in this case, the remainder is 3.
So, 2007 + 11 = 2018.
Divide the given no by 4 if remainder is 0 add 28 to the year if remainder is 1 add 6 or if remainder is 2or3 add 11, in this case, the remainder is 3.
So, 2007 + 11 = 2018.
(32)
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
Upendra said:
1 decade ago
Guys calender will be same if their number of odd days will be same
2007..6*1+2=8,1 odd day.
2014-3 odd days.
2017-0 odd days.
2018-1 days. So answer will be 2018.
2007..6*1+2=8,1 odd day.
2014-3 odd days.
2017-0 odd days.
2018-1 days. So answer will be 2018.
Pradyumna Kumar Tiwari said:
6 years ago
The two necessary conditions for the same calendar is;
1) Both years must either be normal or both leap.
2) A number of odd days between two years must be 0.
1) Both years must either be normal or both leap.
2) A number of odd days between two years must be 0.
Ghanshyamraut said:
2 years ago
Take different between 2007 and 2018.
2007 - 2018 = 11+3(odd day)=14 which is exactly divided by 7.
The reminder is 0.
So, 2007 is the same will be as 2018.
2007 - 2018 = 11+3(odd day)=14 which is exactly divided by 7.
The reminder is 0.
So, 2007 is the same will be as 2018.
(9)
Krishna kittu said:
9 years ago
For finding repeated calendars we used formulae of adding 28, 6, 11. for finding the past repeated calendars we have to subtract same 28, 6, 11. Am I right?
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)
Madhu said:
1 decade ago
@Akash
Actually 7 odd days = 0 odd days right??
And, even for 2012 2 odd days are assigned not for 2011 as the former is divisible by 4...
Actually 7 odd days = 0 odd days right??
And, even for 2012 2 odd days are assigned not for 2011 as the former is divisible by 4...
Venkatesh said:
9 years ago
2007 is a normal year and front year 2006 is also a normal year then we have to add 11years to the given year.
i.e; 2007 + 11years = 2018.
i.e; 2007 + 11years = 2018.
Prabhuchandu said:
1 decade ago
Easy method: 2000+28 = 2028.
2001+6 = 2007.
2002+11 = 2013.
2003+11 = 2015.
2004+28 = 2032.
2005+6 = 2011.
2006+11 = 2017.
2007+11 = 2018.
2001+6 = 2007.
2002+11 = 2013.
2003+11 = 2015.
2004+28 = 2032.
2005+6 = 2011.
2006+11 = 2017.
2007+11 = 2018.
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