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 16 of 26.
Sathiyaraj said:
9 years ago
Here, 2010-2006 = 4 days.
2008 is a leap year, so we get one more day => 5 days.
1-Jan-2006 is Sunday.
So 1-Jan-2010 = Sunday + 5 days = Friday.
2008 is a leap year, so we get one more day => 5 days.
1-Jan-2006 is Sunday.
So 1-Jan-2010 = Sunday + 5 days = Friday.
Sonali said:
9 years ago
Hello. Please solve my problem. I am a new learner to reasoning. If october1 is Sunday then 1st November will be what date? Solve this and tell the tricks.
Rishika said:
9 years ago
If Thursday falls on 1st January 2015, what day of the week will be on 1st January 2016?
Give me the clear solution.
Give me the clear solution.
Srijita biswas said:
9 years ago
Hi please solve this question.
If 2nd July 216 is Wednesday then find out the day of 4th June 2019.
If 2nd July 216 is Wednesday then find out the day of 4th June 2019.
Vinod said:
9 years ago
15 June 1987 how I can found which day of the week? Please let me know the solution.
Harsh said:
9 years ago
Nice explanation @Sundar.
Kundan said:
9 years ago
Hi,
A simple technique is;
2010 - 2006 + no. of leap year between them = 4 + 1 = 5.
[There is only a leap year between 2006 to 2010 is 2008]
Now,
Sunday + 5 = Friday.
A simple technique is;
2010 - 2006 + no. of leap year between them = 4 + 1 = 5.
[There is only a leap year between 2006 to 2010 is 2008]
Now,
Sunday + 5 = Friday.
Abhishek said:
8 years ago
It is very helpful, Thanks @Mahesh.
Prashant d said:
8 years ago
Really helpful. Thanks to all.
SHUBHAM GUPTA said:
8 years 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).
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