Aptitude - Calendar - Discussion
Discussion Forum : Calendar - General Questions (Q.No. 4)
4.
What will be the day of the week 15th August, 2010?
Answer: Option
Explanation:
15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)
Odd days in 1600 years = 0
Odd days in 400 years = 0
9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days 
4 odd days.
Jan. Feb. March April May June July Aug. (31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days
227 days = (32 weeks + 3 days) 3 odd days.
Total number of odd days = (0 + 0 + 4 + 3) = 7 0 odd days.
Given day is Sunday.
Discussion:
95 comments Page 10 of 10.
Ayesha said:
1 decade ago
Please explain how to take 9 years = 2 leap years + 7 ordinary years.
Arjun said:
1 decade ago
Year Code Month Code Day Code
1600-1699=6 Jan-0 Sun-0
1700-1799=4 Feb-3 Mon-1
1800-1899=2 Mar-3 Tue-2
1900-1999=0 Apr-6 Wed-3
2000-2099=6 May-1 Thu-4
2100-2199=4 Jun-4 Fri-5
July-6 Sat-6
Aug-2
Sept-5
Oct-0
Nov-3
Dec-5
Note: Day+Year+Leap Year+Year Code+Month Code
First we need to calculate leap year; it is calculated on 2010 and we need to take last two digits i.e;(10).
2)10(5
10
----
0 leave 0 and take only quotient i.e;5
Now come to formula
D+Y+L.Y+Y.C+M.C ==> 15+10+5+0+2 = 32.
2)32(16
2
------
12
12
------
0
The 0 shows day's code.
.'.2010 Aug 15th is Sunday.
Note: While calculating leap year, take last two digits from the year.
And considering year in the problem we need to take last two digits (i.e;10 in the problem).
Thank You & Regards,
Arjun.
1600-1699=6 Jan-0 Sun-0
1700-1799=4 Feb-3 Mon-1
1800-1899=2 Mar-3 Tue-2
1900-1999=0 Apr-6 Wed-3
2000-2099=6 May-1 Thu-4
2100-2199=4 Jun-4 Fri-5
July-6 Sat-6
Aug-2
Sept-5
Oct-0
Nov-3
Dec-5
Note: Day+Year+Leap Year+Year Code+Month Code
First we need to calculate leap year; it is calculated on 2010 and we need to take last two digits i.e;(10).
2)10(5
10
----
0 leave 0 and take only quotient i.e;5
Now come to formula
D+Y+L.Y+Y.C+M.C ==> 15+10+5+0+2 = 32.
2)32(16
2
------
12
12
------
0
The 0 shows day's code.
.'.2010 Aug 15th is Sunday.
Note: While calculating leap year, take last two digits from the year.
And considering year in the problem we need to take last two digits (i.e;10 in the problem).
Thank You & Regards,
Arjun.
Nishant said:
1 decade ago
@Sunita.
97 year == Divide 97 by 4:
= Quotient is 24.
= So 24 is leap year and in 1 leap year has 2 odd days.
= Than (97-24= 73): 73 is ordinary year and 1 ordinary year has 1 odd day.
= 24*2 + 73*1= 121 days.
=121/7 = than reminder equal to 2.
= So 2 is odd days.
97 year == Divide 97 by 4:
= Quotient is 24.
= So 24 is leap year and in 1 leap year has 2 odd days.
= Than (97-24= 73): 73 is ordinary year and 1 ordinary year has 1 odd day.
= 24*2 + 73*1= 121 days.
=121/7 = than reminder equal to 2.
= So 2 is odd days.
Sunita said:
1 decade ago
I am not able to understand that how you calculate ordinary and leap years in 100 years, 97 years.
Is there any formula for that?
Is there any formula for that?
Akhil said:
1 decade ago
@Gpvsai : 15+2+10+2+6 = 35.
So, dividing 35 by 7 we get 5 which is friday.
So, dividing 35 by 7 we get 5 which is friday.
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