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 10 of 26.
Sreehitha said:
1 decade ago
Pavitra, can you please explain for 1st jan 2008 as it is a leap year how to approach it. The example 16 may 2011 which you have given is for non leap year right.
Nallathambim said:
1 decade ago
Please explain it with more details.
Shashi said:
1 decade ago
Nice one Sundar & Pavitra.
Mamatha said:
1 decade ago
Thanks Pavithra.
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.
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
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.
Rajesham said:
1 decade ago
Normalyear-365days (365/7) 56weeks&1odd day.
Leapyear-366days (366/7) 56weeks&2odddays.
Leapyear-366days (366/7) 56weeks&2odddays.
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.
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