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 3 of 13.
Anand said:
1 decade ago
I have query regarding year 2002.
Years- 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013.
No of odd days for respective years.
1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2.
Till 2007 total odd days becomes 0 hence answer should be 2008 but it is 2013 why ?
Years- 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013.
No of odd days for respective years.
1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2.
Till 2007 total odd days becomes 0 hence answer should be 2008 but it is 2013 why ?
KATY said:
1 decade ago
X/4 (x=given year).
For any year divided by 4, the possibility of remainders are 0, 1, 2, 3.
If remainder is 0---->x+28.
If remainder is 1---->x+6.
If remainder is 2 (or) 3------>x+11.
So 2007 divide by 4 gives 3 as remainder.
So 2007+11 = 2018.
For any year divided by 4, the possibility of remainders are 0, 1, 2, 3.
If remainder is 0---->x+28.
If remainder is 1---->x+6.
If remainder is 2 (or) 3------>x+11.
So 2007 divide by 4 gives 3 as remainder.
So 2007+11 = 2018.
Shivam Padmani said:
5 years ago
Short trick for solving.
One has to divide the given year by Four.
If :
Remainder = 1 Then Add or Subtract 6 years in given years.
Remainder = 2 or 3 Then Add or Subtract 11 years in given years.
Remainder = 0 Then Add or Subtract 28 years in given years.
One has to divide the given year by Four.
If :
Remainder = 1 Then Add or Subtract 6 years in given years.
Remainder = 2 or 3 Then Add or Subtract 11 years in given years.
Remainder = 0 Then Add or Subtract 28 years in given years.
(7)
Anu said:
8 years ago
It applies to all years:
Leap year == add 28.
1 yr after leap year== add 6.
2 yr after leap year== add 11.
3 yr after leap year == add 11.
Here 2007, nearest leap year before 2007, 2004. Therefore 2007 is 3 yrs after 2004. Therefore add 11 to 2007=2017.
Leap year == add 28.
1 yr after leap year== add 6.
2 yr after leap year== add 11.
3 yr after leap year == add 11.
Here 2007, nearest leap year before 2007, 2004. Therefore 2007 is 3 yrs after 2004. Therefore add 11 to 2007=2017.
Riya said:
1 decade ago
Hii prasad.
I have an confusion,
Let suppose I have an year x and want to know the next same year, if I get sum of odd days 7 in this case => odd days=0,
Now what is the answer?
1) is the next year will be the same as x as you told in your comment?
I have an confusion,
Let suppose I have an year x and want to know the next same year, if I get sum of odd days 7 in this case => odd days=0,
Now what is the answer?
1) is the next year will be the same as x as you told in your comment?
Jesu said:
9 years ago
To find that the year is a leap year or an ordinary year. The year should be divided by 4 if you get the remainder as 0 then it must be a leap year or else if you get a remainder as 1 or 2 or 3 then it must be a ordinary year. Hope you understand.
Dfeverx said:
7 years ago
For every ordinary year number of odd days = 1 odd day.
For leap year number of odd days = 2 odd days.
2016-2 (sum=2)
2017-1 (sum=3)
2018-1 (sum= 4)
2019-1 (sum=5)
2020-2 (sum=7)
It implies next year i.e 2021 will have the same calendar as 2016.
For leap year number of odd days = 2 odd days.
2016-2 (sum=2)
2017-1 (sum=3)
2018-1 (sum= 4)
2019-1 (sum=5)
2020-2 (sum=7)
It implies next year i.e 2021 will have the same calendar as 2016.
Pratik said:
8 years ago
How to find answer this question?
Seema remembers that her mother's birthday falls on after 21st January but before 25th January while his brother remembers that it falls after 23rd January but before 28th January. When is Seema mother birthday?
Seema remembers that her mother's birthday falls on after 21st January but before 25th January while his brother remembers that it falls after 23rd January but before 28th January. When is Seema mother birthday?
SamruddhiA said:
7 years ago
@All.
A leap year calendar repeats itself in 28 years and an Ordinary year Calender repeats itself in 6 or 11 years.
Here, we can simply add 2007+6=2013 (which is not there in the option) and 2007+11= 2018 which is the answer.
A leap year calendar repeats itself in 28 years and an Ordinary year Calender repeats itself in 6 or 11 years.
Here, we can simply add 2007+6=2013 (which is not there in the option) and 2007+11= 2018 which is the answer.
Akshay chouhan said:
3 years ago
Only divide the year by 4 if the remaining is 0 then add 28 in years if the remaining is 1 then add 6.
And if it is 2/3 then add 11 so 2007/4 we get 3 as a remainder.
So we add 11 in the year.
The answer is 2007+11 = 2018.
And if it is 2/3 then add 11 so 2007/4 we get 3 as a remainder.
So we add 11 in the year.
The answer is 2007+11 = 2018.
(50)
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