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 2 of 26.
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.
Ram said:
1 decade ago
There is a another Method of to calculate this.
We memorize this,
Jan-0
Feb-3
mar-3
Apr-6
May-1
Jun-2
Jul-6
Aug-2
Sept-5
Oct-0
Nov-3
Dec-5
Given 1 Jan 2010.
Add date and year of the last 2 digits.
1+10=11 this is the equation (1).
Divide by 4 last 2 digits of the year.
10/4=2 equation(2).
Here we take only coefficient. Don't take reminder.
In 1st the codes to take the code and add equation 1 & 2.and divide by 7.
11+2+0/7=5.
Here we take reminder.
Answer is Friday.
We memorize this,
Jan-0
Feb-3
mar-3
Apr-6
May-1
Jun-2
Jul-6
Aug-2
Sept-5
Oct-0
Nov-3
Dec-5
Given 1 Jan 2010.
Add date and year of the last 2 digits.
1+10=11 this is the equation (1).
Divide by 4 last 2 digits of the year.
10/4=2 equation(2).
Here we take only coefficient. Don't take reminder.
In 1st the codes to take the code and add equation 1 & 2.and divide by 7.
11+2+0/7=5.
Here we take reminder.
Answer is 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.
Sumanth geras said:
1 decade ago
Hello guys according to the formula.
Jan=1.
Feb=4.
Mar=4.
Apr=0.
May=2.
Jun=5.
Jul=0.
Aug=3.
Sep=6.
Oct=1.
Nov=4.
Dec=6.
Days:
S=1.
M=2.
T=3.
W=4.
Th=5.
F=6.
Sat=7 or 0.
Formula = Day+month no. + (x-1900) + (x-1900) /4.
X = The year we have to calculate.
Answer/7 = Day we require. We can calculate it for leap year also.
If the given date is before 1st mar' of leap year. The obtained answer is given date if it is. On and after 1st march leap year you should subtract 1 from answer.
Jan=1.
Feb=4.
Mar=4.
Apr=0.
May=2.
Jun=5.
Jul=0.
Aug=3.
Sep=6.
Oct=1.
Nov=4.
Dec=6.
Days:
S=1.
M=2.
T=3.
W=4.
Th=5.
F=6.
Sat=7 or 0.
Formula = Day+month no. + (x-1900) + (x-1900) /4.
X = The year we have to calculate.
Answer/7 = Day we require. We can calculate it for leap year also.
If the given date is before 1st mar' of leap year. The obtained answer is given date if it is. On and after 1st march leap year you should subtract 1 from answer.
Mangesh said:
8 years ago
Odd days -consider no of weeks in 15 days=15/7=2 week plus1additional day
This additional day is an odd day.
To solve given Q : ordinary yr =52
weeks plus 1 odd day=365 days
Leap yr=52 plus 2 odd days=366 days
2006 ordinary-1 odd
2007 ordinary-1 odd
2008 (leap bcoz divided by 4)-2 odd
2009 ordinary-1 odd
2010 ordinary-1 odd
Total odd days=1+1+2+1+1=6.
Go six day ahead of Sunday there is Friday.
Leap yr concerned with Feb month ordinary month 28 days.
Leap Feb 29 days.
This additional day is an odd day.
To solve given Q : ordinary yr =52
weeks plus 1 odd day=365 days
Leap yr=52 plus 2 odd days=366 days
2006 ordinary-1 odd
2007 ordinary-1 odd
2008 (leap bcoz divided by 4)-2 odd
2009 ordinary-1 odd
2010 ordinary-1 odd
Total odd days=1+1+2+1+1=6.
Go six day ahead of Sunday there is Friday.
Leap yr concerned with Feb month ordinary month 28 days.
Leap Feb 29 days.
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
Faiza said:
1 decade ago
1 ordinary year = 365days = 52weeks + 1day
so, each ordinary year has 1 odd day.
1 leap year = 366 days = 52 weeks + 2 days (since 52 weeks=364days so, in leap year , to get 366 days we add 2 in 364)
so, each leap year has 2 odd days
100 years = 76 ordinary years + 24 leap years
= 76 multiply with its odd day i.e 1 + 24 multiply with its odd days i.e 2
= 76 + 48
=124 odd days
This means that every 100 years contain 124 odd days.
so, each ordinary year has 1 odd day.
1 leap year = 366 days = 52 weeks + 2 days (since 52 weeks=364days so, in leap year , to get 366 days we add 2 in 364)
so, each leap year has 2 odd days
100 years = 76 ordinary years + 24 leap years
= 76 multiply with its odd day i.e 1 + 24 multiply with its odd days i.e 2
= 76 + 48
=124 odd days
This means that every 100 years contain 124 odd days.
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.
(62)
Meethi said:
1 decade ago
//
How many times does the 29th days of the month occur in 400 consecutive years
1) 97 times 2) 4400 times 3) 4497 times 4) none
Sol: In 400 consecutive years there are 97 leap years. Hence in 400 consecutive years, February has the 29th day 97 times, and the remaining 11 months have the 29th day 400 x 11 or 4400 times.
Therefore, 29th day of the month occurs (4400 + 97) or 4497 times.
//
How to calculate leap years in 400 consecutive years ?
How many times does the 29th days of the month occur in 400 consecutive years
1) 97 times 2) 4400 times 3) 4497 times 4) none
Sol: In 400 consecutive years there are 97 leap years. Hence in 400 consecutive years, February has the 29th day 97 times, and the remaining 11 months have the 29th day 400 x 11 or 4400 times.
Therefore, 29th day of the month occurs (4400 + 97) or 4497 times.
//
How to calculate leap years in 400 consecutive years ?
Pavithra said:
1 decade ago
Hi there is a formula for this type of problem.
Plz memorize this
jan=1
feb=4
mar=4
apr=0
may=2
jun=5
jul=0
aug=3
sep=6
oct=1
nov=4
dec=6
days:
s=1
m=2
t=3
w=4
th=5
f=6
sat=7 or 0
formula=day+mon no.+(x-1900)+(x-1900)/4
x= the year we have to calculate
Eg:
16-may-2011
for=16+2+(2011-1900)+(2011-1900)/4
=16+2+111+111/4
=16+2+111+27
(here we should not take the no behind decimal i.e 111/4=27.75)
=156
156/7=2(reminder)
mon =2
Plz memorize this
jan=1
feb=4
mar=4
apr=0
may=2
jun=5
jul=0
aug=3
sep=6
oct=1
nov=4
dec=6
days:
s=1
m=2
t=3
w=4
th=5
f=6
sat=7 or 0
formula=day+mon no.+(x-1900)+(x-1900)/4
x= the year we have to calculate
Eg:
16-may-2011
for=16+2+(2011-1900)+(2011-1900)/4
=16+2+111+111/4
=16+2+111+27
(here we should not take the no behind decimal i.e 111/4=27.75)
=156
156/7=2(reminder)
mon =2
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