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 20 of 26.
Raj said:
1 decade ago
@Mahesh your method is nice but if there is large variation in years we want to go for concept based solution for ex 15 aug 2012 is wednesday then what is the day of 15th aug 1935
Kamatchi said:
1 decade ago
Will give a clear explain to find odd day ?
Mahesh said:
1 decade ago
Hi Guys,
I don't know why guys are giving so much pressure to your brain its too simple what I know in my method, please check below logic and let me know its simple or complicate feel free to feedback on my answer.
i.e.,
1 Jan 2006 is Sunday.
1 Jan 2007 is Monday.
1 Jan 2008 is Tuesday.
1 Jan 2009 is Thursday (extra day for the 2008 FEB 29).
1 Jan 2010 is Friday.
Solution:
Add next day for the next year.
Ex: 1 Jan 2006 is Sunday in next year it will come on Monday.
In 2008 Feb 29 will come extra day for the year.
I don't know why guys are giving so much pressure to your brain its too simple what I know in my method, please check below logic and let me know its simple or complicate feel free to feedback on my answer.
i.e.,
1 Jan 2006 is Sunday.
1 Jan 2007 is Monday.
1 Jan 2008 is Tuesday.
1 Jan 2009 is Thursday (extra day for the 2008 FEB 29).
1 Jan 2010 is Friday.
Solution:
Add next day for the next year.
Ex: 1 Jan 2006 is Sunday in next year it will come on Monday.
In 2008 Feb 29 will come extra day for the year.
Somu said:
1 decade ago
Odd days in 2000 years = 0.
Odd days in 9 years = (2*2+7x1) Since (2 leap+7 ordinary)
= 4+7 = 11.
In 2010 January,
1 = 1 odd days.
Total odd days = 11+1 = 12.
Odd days = 12/7.
= 5.
5 mentions Fridays.
Odd days in 9 years = (2*2+7x1) Since (2 leap+7 ordinary)
= 4+7 = 11.
In 2010 January,
1 = 1 odd days.
Total odd days = 11+1 = 12.
Odd days = 12/7.
= 5.
5 mentions Fridays.
Narendra said:
1 decade ago
In each next year day will come next day.
2006 Sunday.
2007 Monday.
2008 Tuesday+Wednesday (leap year).
2009 Thursday.
2010 Friday.
2006 Sunday.
2007 Monday.
2008 Tuesday+Wednesday (leap year).
2009 Thursday.
2010 Friday.
Rajesham said:
1 decade ago
Normalyear-365days (365/7) 56weeks&1odd day.
Leapyear-366days (366/7) 56weeks&2odddays.
Leapyear-366days (366/7) 56weeks&2odddays.
Pinto said:
1 decade ago
2010=1600+400+10
1600=0 add days
400 =0 add days
10 =10/4=2 leap year + 8 ordinary year
=2*2 + 8*1
=4+8=12
=12/7(no days in a week)
=5 (Friday)
1600=0 add days
400 =0 add days
10 =10/4=2 leap year + 8 ordinary year
=2*2 + 8*1
=4+8=12
=12/7(no days in a week)
=5 (Friday)
Prasant said:
1 decade ago
Let's take a simple way:
1 Jan 2006 - Sunday (given)
1 Jan 2006 to 1 Jan 2007: 1 odd day
1 Jan 2007 to 1 Jan 2008: 1 odd day
1 Jan 2008 to 1 Jan 2009: 2 odd day
1 Jan 2009 to 1 Jan 2010: 1 odd day
Total odd days = 5 (i.e. 5th no. odd day is 1 Jan 2010)
Since given day is sunday, consider it as '0', then 1 for monday, 2 for tuesday....5 for Friday.
1 Jan 2006 - Sunday (given)
1 Jan 2006 to 1 Jan 2007: 1 odd day
1 Jan 2007 to 1 Jan 2008: 1 odd day
1 Jan 2008 to 1 Jan 2009: 2 odd day
1 Jan 2009 to 1 Jan 2010: 1 odd day
Total odd days = 5 (i.e. 5th no. odd day is 1 Jan 2010)
Since given day is sunday, consider it as '0', then 1 for monday, 2 for tuesday....5 for Friday.
K.velmurugan said:
1 decade ago
odd days
1-jan
4-feb
4-mar
0-apr
2-may
5-jun
0-july
3-agu
6-sep
1-oct
4-nov
6-dec
0-sunday.
1-monday.
2-tuesday.
3-wednesday.
4-thursday.
5-friday.
6-saturday.
20--=6 odddays
19--=0 oddays
ex : 27/03/2012
day=27 total=day+month+20--+ year
month=4
20--=6 =27+4+6+17=54/7
(year=--12/7= 5remainter) remainder=5
year=12+5=17
ans: friday
1-jan
4-feb
4-mar
0-apr
2-may
5-jun
0-july
3-agu
6-sep
1-oct
4-nov
6-dec
0-sunday.
1-monday.
2-tuesday.
3-wednesday.
4-thursday.
5-friday.
6-saturday.
20--=6 odddays
19--=0 oddays
ex : 27/03/2012
day=27 total=day+month+20--+ year
month=4
20--=6 =27+4+6+17=54/7
(year=--12/7= 5remainter) remainder=5
year=12+5=17
ans: friday
Kayal said:
1 decade ago
First we take the previous century why because every 4th century is leap year.
Then take the finding year with substract 1.
So 2009-2000=9.
Now find the how many leap year and how many ordinary year in this 9. So we divide.
9/4=2. So 2 leap year and 7 ordinary year.
= (2*2+7*1) =4.
Now take the months table what they ask to find out here jan 1.
So no months is there as to take 1.
=4+1.
=5 (friday).
Weeks table.
0-sunday.
1-monday.
2-tuesday.
3-wednesday.
4-thursday.
5-friday.
6-saturday.
Thats all.
Then take the finding year with substract 1.
So 2009-2000=9.
Now find the how many leap year and how many ordinary year in this 9. So we divide.
9/4=2. So 2 leap year and 7 ordinary year.
= (2*2+7*1) =4.
Now take the months table what they ask to find out here jan 1.
So no months is there as to take 1.
=4+1.
=5 (friday).
Weeks table.
0-sunday.
1-monday.
2-tuesday.
3-wednesday.
4-thursday.
5-friday.
6-saturday.
Thats all.
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