Aptitude - Calendar - Discussion
Discussion Forum : Calendar - General Questions (Q.No. 1)
1.
It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
Answer: Option
Explanation:
On 31st December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
On 31st December 2009, it was Thursday.
Thus, on 1st Jan, 2010 it is Friday.
Discussion:
261 comments Page 6 of 27.
Yilkunna Benedict said:
9 years ago
A question was asked that how can you calculate a no of leap year in 100 years but answers provided does not give a clear solution to the question. Please, tell me how can I know that 100 is the beginning of the year? Thanks.
Shiva said:
9 years ago
Thank you @Pavitra for your concept. And the only thing I found this for leap year. Especially if we calculate the month of Feb in any leap year. We have to use (answer-1) as the real answer. If I am wrong please let me know.
Srinivas said:
7 years ago
2006 = 1(odd day).
2007 = 1(odd day).
2008 = 2(odd days-it is a leap year).
2009 = 1(odd day).
Total = 5(odd days).
Monday=1
Tuesday=2
Wednesday=3
Thursday=4
Friday=5
Saturday=6
Sunday=7/0
So now we consider 5 as Friday.
2007 = 1(odd day).
2008 = 2(odd days-it is a leap year).
2009 = 1(odd day).
Total = 5(odd days).
Monday=1
Tuesday=2
Wednesday=3
Thursday=4
Friday=5
Saturday=6
Sunday=7/0
So now we consider 5 as Friday.
(1)
Kaveri said:
1 decade ago
Hi friends,
2005 1st Jan is sunday,
31st Dec 2005 is Saturday,
2006, 2007, 2008 (Leap Year), 2009.
2008 is leap year- add 2 No.
Other is 3 No(2006, 2007, 2009).
So you add + 5 days.
That mean Its day is Friday
2005 1st Jan is sunday,
31st Dec 2005 is Saturday,
2006, 2007, 2008 (Leap Year), 2009.
2008 is leap year- add 2 No.
Other is 3 No(2006, 2007, 2009).
So you add + 5 days.
That mean Its day is Friday
Somu said:
1 decade ago
Odd days in 2000 years = 0.
Odd days in 9 years = (2*2+7x1) Since (2 leap+7 ordinary)
= 4+7 = 11.
In 2010 January,
1 = 1 odd days.
Total odd days = 11+1 = 12.
Odd days = 12/7.
= 5.
5 mentions Fridays.
Odd days in 9 years = (2*2+7x1) Since (2 leap+7 ordinary)
= 4+7 = 11.
In 2010 January,
1 = 1 odd days.
Total odd days = 11+1 = 12.
Odd days = 12/7.
= 5.
5 mentions Fridays.
PRABHUCHANDU said:
1 decade ago
Centuries can be remembered like this :
1600-1699->6.
1700-1799->4.
1800-1899->2.
1900-1999->0.
2000-2099->6.
2100-2199->4.
2200-2299->2.
2300-2399->0.
.
.
.
.
.
.
This will continue.
1600-1699->6.
1700-1799->4.
1800-1899->2.
1900-1999->0.
2000-2099->6.
2100-2199->4.
2200-2299->2.
2300-2399->0.
.
.
.
.
.
.
This will continue.
Himanshu said:
1 decade ago
Method to find the leap year:
If any year divided by 4 without reminder its year known as leap year.
Example-: 2008/4 = 502(no reminder).
And in this year were 29 days in Feb.
And a year consist 366 days.
If any year divided by 4 without reminder its year known as leap year.
Example-: 2008/4 = 502(no reminder).
And in this year were 29 days in Feb.
And a year consist 366 days.
Pradyumna Kumar Tiwari said:
7 years ago
@Soumya.
6th Aug is Thursday it means 2nd Aug is Sunday.
After every 7 days, the same day will fall.
2nd Aug, 9th Aug,16th Aug, 23th Aug, 30th Aug are Sunday. It means a total of 5 Sunday in Aug.
6th Aug is Thursday it means 2nd Aug is Sunday.
After every 7 days, the same day will fall.
2nd Aug, 9th Aug,16th Aug, 23th Aug, 30th Aug are Sunday. It means a total of 5 Sunday in Aug.
(1)
Pinto said:
1 decade ago
2010=1600+400+10
1600=0 add days
400 =0 add days
10 =10/4=2 leap year + 8 ordinary year
=2*2 + 8*1
=4+8=12
=12/7(no days in a week)
=5 (Friday)
1600=0 add days
400 =0 add days
10 =10/4=2 leap year + 8 ordinary year
=2*2 + 8*1
=4+8=12
=12/7(no days in a week)
=5 (Friday)
Priya Gaikwad said:
1 decade ago
1 jan 2006 = Sunday count 1 .
1 jan 2007 = Monday count 1.
1 jan 2008 = Tuesday count 1.
1 jan 2009 = Thursday count 2 because 2008 is leap year(2008/4=0).
1 jan 2010 = Friday. Req. answer.
1 jan 2007 = Monday count 1.
1 jan 2008 = Tuesday count 1.
1 jan 2009 = Thursday count 2 because 2008 is leap year(2008/4=0).
1 jan 2010 = Friday. Req. answer.
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