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 3 of 26.
Kushi said:
5 years ago
Thank you all.
(2)
Vijaya said:
1 decade ago
Given :.
Jan 1, 2006 is sunday.
Jan 1, 2010 =?.
Consider completed year before 2010 i.e 2009 (since 2010 was starting).
2009=2000+9.
2009=0+9 (since 400, 800, 1200, 1600. Contains '0' odd days).
9/4=2 (2 leap years are there. Here it is divided by 4 because for every 4 years we get leap year).
Leap year=2*2=4.
Non leap =7*1=7.
Total = 11.
Jan 1st=1 day.
11+1=12.
12/7 = 5.
Friday.
Jan 1, 2006 is sunday.
Jan 1, 2010 =?.
Consider completed year before 2010 i.e 2009 (since 2010 was starting).
2009=2000+9.
2009=0+9 (since 400, 800, 1200, 1600. Contains '0' odd days).
9/4=2 (2 leap years are there. Here it is divided by 4 because for every 4 years we get leap year).
Leap year=2*2=4.
Non leap =7*1=7.
Total = 11.
Jan 1st=1 day.
11+1=12.
12/7 = 5.
Friday.
(1)
Veer singh pal said:
1 decade ago
If any year divided by 4 with out reminder its year known as leep year.
E.g: 2008/4 = 502 (no reminder)
And in this year were 29 days in Feb.
E.g: 2008/4 = 502 (no reminder)
And in this year were 29 days in Feb.
(1)
Basavaraj said:
1 decade ago
I want clear method to calculate odd days.
(1)
Dhanagopal said:
1 decade ago
I want briefly to find odd days.
(1)
Bipin said:
1 decade ago
Pavithra You told us a very good way to solve the calendar question but only few things you left i.e related to leap year, this formula doesn't work for leap year, it works only for non-leap year.
@Dg1234.
You can solve leap year by this formula as told by Pavithra but only few things I have changed. Check out.
Formula For Leap Year:
0 - Apr, jul 0 - Sat
1 - Jan 1 - Sun
2 - May 2 - Mon
3 - Aug 3 - Tue
4 - Feb, mar, nov 4 - Wed
5 - Jun 5 - Thu
6 - Sep, dec 6 - Fri
7 - Sat
Note:- To find the odd days for leap year we have to add 6 only with this formula.
Formula: 6 + Day + Month + (Year-1900) + (year-1900)/4.
Eg. 4-feb-2004.
= 6+4+4+(2004-1900)+(2004-1900)/4.
= 6+8+104+(104/4).
= 118+26.
= 144.
= 144/7.
= Remainder = 4.
= Wednesday.
@Dg1234.
You can solve leap year by this formula as told by Pavithra but only few things I have changed. Check out.
Formula For Leap Year:
0 - Apr, jul 0 - Sat
1 - Jan 1 - Sun
2 - May 2 - Mon
3 - Aug 3 - Tue
4 - Feb, mar, nov 4 - Wed
5 - Jun 5 - Thu
6 - Sep, dec 6 - Fri
7 - Sat
Note:- To find the odd days for leap year we have to add 6 only with this formula.
Formula: 6 + Day + Month + (Year-1900) + (year-1900)/4.
Eg. 4-feb-2004.
= 6+4+4+(2004-1900)+(2004-1900)/4.
= 6+8+104+(104/4).
= 118+26.
= 144.
= 144/7.
= Remainder = 4.
= Wednesday.
(1)
Naveenamargam said:
9 years ago
Codes for the calendar.
DAYCODES.
Sun = 0.
Mon = 1.
Tue = 2.
Wed = 3.
Thurs = 4.
Fri = 5.
Sat = 6.
CENTURY CODE.
1600-1699 = 6.
1700-1799 = 4.
1800-1899 = 2.
1900-1999 = 0.
2000-2099 = 6.
2100-2199 = 4.
-
-
Repeats.
MONTH CODE:.
Jan = 0.
Feb = 3.
Mar = 3.
April = 6.
May = 1.
June = 4.
July = 6.
Aug = 2.
Sept = 5.
Oct = 0.
Nov = 3.
Dec = 5.
A leap year January code=-1 and Feb = 2.
For example:- Find the day of the week on 26-jan-1956.
Solution:-.
Date + last two digits of the year + (last two digits/4) + monthcode + centuarycode.
= 26 + 56 + (56/4) + (-1) + 0 (since, 56/4 remainder is' 0' it is a leap year) (when leap year comes above).
=26 + 56 + 14 - 1 + 0 codes we said that jan=-1).
= 95 (we should divide by 7 because we have to find the day of the week).
= 95/7 (remainder '4').
= 4 (Thursday).
You can also check the calendar on this date.
DAYCODES.
Sun = 0.
Mon = 1.
Tue = 2.
Wed = 3.
Thurs = 4.
Fri = 5.
Sat = 6.
CENTURY CODE.
1600-1699 = 6.
1700-1799 = 4.
1800-1899 = 2.
1900-1999 = 0.
2000-2099 = 6.
2100-2199 = 4.
-
-
Repeats.
MONTH CODE:.
Jan = 0.
Feb = 3.
Mar = 3.
April = 6.
May = 1.
June = 4.
July = 6.
Aug = 2.
Sept = 5.
Oct = 0.
Nov = 3.
Dec = 5.
A leap year January code=-1 and Feb = 2.
For example:- Find the day of the week on 26-jan-1956.
Solution:-.
Date + last two digits of the year + (last two digits/4) + monthcode + centuarycode.
= 26 + 56 + (56/4) + (-1) + 0 (since, 56/4 remainder is' 0' it is a leap year) (when leap year comes above).
=26 + 56 + 14 - 1 + 0 codes we said that jan=-1).
= 95 (we should divide by 7 because we have to find the day of the week).
= 95/7 (remainder '4').
= 4 (Thursday).
You can also check the calendar on this date.
(1)
Rohan Roy said:
8 years ago
Month code- 033614625035 or 144025036146?
Please explain it.
Please explain it.
(1)
Kummari Hemanth said:
8 years ago
1600+400=6odd days.
10 =2+8
=4+8=12
=4odd days
Total=4+6/7=10/7=3odd days.
Jan=2.
All odd days =3+2=5=Friday.
10 =2+8
=4+8=12
=4odd days
Total=4+6/7=10/7=3odd days.
Jan=2.
All odd days =3+2=5=Friday.
(1)
Aryan said:
7 years ago
If Friday is on 15 September 2000 then what would be the day on 15 September 2001?
Please explain me.
Please explain me.
(1)
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