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:
258 comments Page 6 of 26.
Shiva said:
8 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:
6 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:
6 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.
RAJIV RANJAN KUMAR said:
9 years ago
Sir, I want to general method for the year 1800 - 1900 and 1900 - 2000 and 2000 and above.
How can find if 30 Sept 1853 was Monday then what day will be on 18 March 1803 and 23 April 1879?
How can find if 30 Sept 1853 was Monday then what day will be on 18 March 1803 and 23 April 1879?
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