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 15 of 26.
Divya said:
9 years ago
Hi, I had calculated the day for 23 Jan 1988 and got the answer as Sunday which was wrong in actual calendar. It was Saturday. Could anyone help me to solve this?
Anand said:
9 years ago
1st Jan 2006 - Sunday.
Next, 1/1/07- Monday.
2008 is a leap year so here we have 2 days extra, so 01/01/08 - Wednesday.
Next 01/01/2009 - Thursday.
so then,
01/01/2010 - Friday.
Next, 1/1/07- Monday.
2008 is a leap year so here we have 2 days extra, so 01/01/08 - Wednesday.
Next 01/01/2009 - Thursday.
so then,
01/01/2010 - Friday.
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)
Aishu said:
9 years ago
How everyone is calculating month codes? Can anyone explain?
Jose kappani said:
9 years ago
@Divya.
When you calculated leap year 88/4 you got 22 but the date is in January, so any day before February 28 there will be this problem.
When you calculated leap year 88/4 you got 22 but the date is in January, so any day before February 28 there will be this problem.
Sai krishna chari said:
9 years ago
Your solution is very nice. Thank you @Pavithra.
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?
Bhavya said:
9 years ago
How can you guess the leap year? Please teach me.
Sheela said:
9 years ago
Please help me to solve in a simple way.
Moon said:
9 years ago
How two same calendar are found? please help me.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers