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.
Hitesh Handke said:
6 days ago
365 + 366 + 365 + 365 + 1 = 1461.
1461%7 = 5 and 5 it means Friday.
1461%7 = 5 and 5 it means Friday.
(1)
Dancan said:
4 months ago
Here Explanations are great. Thanks.
(7)
Sai said:
7 months ago
Solution:
Day = date+ month code + year code +no.of normal years + no.of leap years-1/7
01 + 01 + 06 + 10 + 10/4 - 4 = 7 - 1 = 6.
6 is Friday,
Day = date+ month code + year code +no.of normal years + no.of leap years-1/7
01 + 01 + 06 + 10 + 10/4 - 4 = 7 - 1 = 6.
6 is Friday,
(14)
Dhevahi said:
1 year ago
1 year = 365 days.
Leap year = 366 days.
2006=365
2007=365
2008 is a leap year.
So = 366
2009 = 365 days.
We add 4 years date,
Hence, (365+365+366+365)= 1461.
1461/7 = remainder 5
Add 5 days from Sunday = 0.
Monday = 1.
Tuesday =2.
Wednesday =3.
Thursday = 4.
Friday = 5
Saturday = 6.
So, the answer is Friday.
Leap year = 366 days.
2006=365
2007=365
2008 is a leap year.
So = 366
2009 = 365 days.
We add 4 years date,
Hence, (365+365+366+365)= 1461.
1461/7 = remainder 5
Add 5 days from Sunday = 0.
Monday = 1.
Tuesday =2.
Wednesday =3.
Thursday = 4.
Friday = 5
Saturday = 6.
So, the answer is Friday.
(213)
Deepti lodhi said:
1 year ago
How to find odd days? Please explain.
(49)
Azharuddin Alam said:
1 year ago
Firstly we will count the total number of odd days till 2009.
2009 = 2000+9.
Since, 2000 is a leap year so there is no odd days and 9 has 2 leap years and 7 normal years,
Hence.
No of odd days = 0 + 7 + 4 (one leap year contains the 2 odd days).
= 11 odd days.
Now, 11 + 1 (1 jan 2006).
Total odd days = 12 days.
12 is divided by 7 for getting the weekdays.
12/7= 5 remainders.
Now 5th day of the week is Friday.
C option is correct.
2009 = 2000+9.
Since, 2000 is a leap year so there is no odd days and 9 has 2 leap years and 7 normal years,
Hence.
No of odd days = 0 + 7 + 4 (one leap year contains the 2 odd days).
= 11 odd days.
Now, 11 + 1 (1 jan 2006).
Total odd days = 12 days.
12 is divided by 7 for getting the weekdays.
12/7= 5 remainders.
Now 5th day of the week is Friday.
C option is correct.
(33)
Anshul rawat said:
2 years ago
Solution:
We know that in NL year :
we have the same day on 1 Jan and 31 Dec
And on L year :
we have 1 day ahead of the day in 1 Jan
Therefore:
2006 -> dec 31 -> Sun , Hence :
2007 -> Jan 1 -> Mon , (NL year)
2007 -> Dec 31 -> Mon , Hence:
2008 -> Jan 1 -> Tues , (L Year)
2008 -> Dec 31 -> Wed , Hence:
2009 -> Jan 1 -> Thurs (NL Year)
2010 -> Jan 1 -> Fri.
Therefore, 2010 Jan 1 will be -> Friday.
We know that in NL year :
we have the same day on 1 Jan and 31 Dec
And on L year :
we have 1 day ahead of the day in 1 Jan
Therefore:
2006 -> dec 31 -> Sun , Hence :
2007 -> Jan 1 -> Mon , (NL year)
2007 -> Dec 31 -> Mon , Hence:
2008 -> Jan 1 -> Tues , (L Year)
2008 -> Dec 31 -> Wed , Hence:
2009 -> Jan 1 -> Thurs (NL Year)
2010 -> Jan 1 -> Fri.
Therefore, 2010 Jan 1 will be -> Friday.
(60)
Vikram said:
2 years ago
How to find the odd days? Please explain to me.
(28)
Arpit Kumar said:
2 years ago
Thanks, everyone for solving this clearly.
(36)
SUBASHINI said:
2 years ago
No.of days in 1 week = 7 days.
No.of days in x week = 7 * x = 7x.
No.of days in x week x days = 7x + x = 8x.
No.of days in x week = 7 * x = 7x.
No.of days in x week x days = 7x + x = 8x.
(27)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers