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 1 of 26.
Goutham said:
2 decades ago
What is odd days?
Alaguvel.c said:
2 decades ago
How to find odd days?
Sundar said:
2 decades ago
Hi Friends,
Odd days of a year X = Remainder of [No. of days in the year X / 7].
[Where 7 --> For every 7 days, the day will be repeated, thats why we are dividing the total number of days of an year by 7. If it has any remainder then it will be considered as an odd day for that year.]
The required remainders (odd days) for the different years are given below:
For 2006: 365/7 = 1 (Total no. of days in 2006 is 365)
For 2007: 365/7 = 1
For 2008: 366/7 = 2 (Leap year, hence Feb month has 29 days)
For 2009: 365/7 = 1
Therefore, total number of odd days = 5.
Given day is 'Sunday' and total no of odd days is '5'.
The required date = Sunday + 5
= (The 5th day next to Sunday)
= Sunday + 1-Monday + 2-Tuesday + 3-Wednesday + 4-Thursday + 5-Friday.
= Friday (Required answer).
Odd days of a year X = Remainder of [No. of days in the year X / 7].
[Where 7 --> For every 7 days, the day will be repeated, thats why we are dividing the total number of days of an year by 7. If it has any remainder then it will be considered as an odd day for that year.]
The required remainders (odd days) for the different years are given below:
For 2006: 365/7 = 1 (Total no. of days in 2006 is 365)
For 2007: 365/7 = 1
For 2008: 366/7 = 2 (Leap year, hence Feb month has 29 days)
For 2009: 365/7 = 1
Therefore, total number of odd days = 5.
Given day is 'Sunday' and total no of odd days is '5'.
The required date = Sunday + 5
= (The 5th day next to Sunday)
= Sunday + 1-Monday + 2-Tuesday + 3-Wednesday + 4-Thursday + 5-Friday.
= Friday (Required answer).
(2)
Rajeswari said:
1 decade ago
Nice Sundar.
Bhupendra said:
1 decade ago
31 dec 2005 = Saturday
365/7 = 52+1/7
2006= 1 odd day
2007= 1 odd day
2008=> 366/7 = 52+ 2/7=> 2 odd day
then 31 Dec 2008= Thursday
and 1 Jan 2009 = Friday.
365/7 = 52+1/7
2006= 1 odd day
2007= 1 odd day
2008=> 366/7 = 52+ 2/7=> 2 odd day
then 31 Dec 2008= Thursday
and 1 Jan 2009 = Friday.
Nilesh said:
1 decade ago
How do we know leap year of that specific year?
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)
Manu said:
1 decade ago
Everytime the odd days are considered from sunday onwards i.e.,if there are are 0 odd days then we take the day as sunday and the count follows likewise
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