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 3 of 26.
Azharuddin Alam said:
2 years 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.
(34)
Calnd said:
4 years ago
@All.
Here is the clarification.
Most of us have no prob in calculating no. Off odd days.
But confusion is there when considering the day code.
Simple;
1jan 2000 -sundays.
To
1jan 2006.
Odd days = 6.
2006 = 1.
2007 = 1.
2008 = 2.
2009 = 1.
1 jan 2006 so one day =one odd day.
Total=6
Odd day=7-6=1.
Starting day is sunday so again start from sunday then keep increasing by one
For odd 2=
1 = Monday,
2 = Tuesdays .
So on.
Here is the clarification.
Most of us have no prob in calculating no. Off odd days.
But confusion is there when considering the day code.
Simple;
1jan 2000 -sundays.
To
1jan 2006.
Odd days = 6.
2006 = 1.
2007 = 1.
2008 = 2.
2009 = 1.
1 jan 2006 so one day =one odd day.
Total=6
Odd day=7-6=1.
Starting day is sunday so again start from sunday then keep increasing by one
For odd 2=
1 = Monday,
2 = Tuesdays .
So on.
(15)
Desu said:
6 years ago
On a fix particular date, as the year increases by one unit, the day also increases by one unit, Except for leap year where the day is increased by two units.
So coming to the problem:
On particular date 1 Jan.
Day- sun and year-2006.
So on increasing year by one unit.
Year - Day
2007 -.Monday.
2008. - Tuesday.
2009 - Thursday (when leap year is increased by one unit days Will be increased by two units)
2010 - Friday.
So coming to the problem:
On particular date 1 Jan.
Day- sun and year-2006.
So on increasing year by one unit.
Year - Day
2007 -.Monday.
2008. - Tuesday.
2009 - Thursday (when leap year is increased by one unit days Will be increased by two units)
2010 - Friday.
Ramesh said:
7 years ago
Formula:day + month code + no of years(last 2 digits) + no of leap years + century code.
Month code :1 4 4 0 2 5 0 3 6 1 4 6.
Centuty code: 1900-1999=0.
2000-2099=6.
2100-2199=4.
2200-2299=2.
Day code: Sat-0,Sun-1,Mon-2,Tue-3,wed-4, Thu-5, Fri-6.
Month code :1 4 4 0 2 5 0 3 6 1 4 6.
Centuty code: 1900-1999=0.
2000-2099=6.
2100-2199=4.
2200-2299=2.
Day code: Sat-0,Sun-1,Mon-2,Tue-3,wed-4, Thu-5, Fri-6.
Jyoti said:
8 years ago
Please, memorise 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.
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.
Dhara said:
1 decade ago
Easy style to calculate this question.
jan 1, 2006 sunday.
jan 1, 2007 monday (+1 in day because 2007 is not leap year) .
jan 1, 2008 wendsay (+2 in day because 2008 is divisible by 4).
jan 1, 2009 Thursday (+1 in day because it is not leap year).
jan 1, 2010 Friday (+1 because it is not leap year).
If leap year then add 2 days otherwise 1 day will be add if and only if year is only changed and date will be same.
jan 1, 2006 sunday.
jan 1, 2007 monday (+1 in day because 2007 is not leap year) .
jan 1, 2008 wendsay (+2 in day because 2008 is divisible by 4).
jan 1, 2009 Thursday (+1 in day because it is not leap year).
jan 1, 2010 Friday (+1 because it is not leap year).
If leap year then add 2 days otherwise 1 day will be add if and only if year is only changed and date will be same.
Vijaya said:
1 decade ago
Given :.
Jan 1, 2006 is sunday.
Jan 1, 2010 =?.
Consider completed year before 2010 i.e 2009 (since 2010 was starting).
2009=2000+9.
2009=0+9 (since 400, 800, 1200, 1600. Contains '0' odd days).
9/4=2 (2 leap years are there. Here it is divided by 4 because for every 4 years we get leap year).
Leap year=2*2=4.
Non leap =7*1=7.
Total = 11.
Jan 1st=1 day.
11+1=12.
12/7 = 5.
Friday.
Jan 1, 2006 is sunday.
Jan 1, 2010 =?.
Consider completed year before 2010 i.e 2009 (since 2010 was starting).
2009=2000+9.
2009=0+9 (since 400, 800, 1200, 1600. Contains '0' odd days).
9/4=2 (2 leap years are there. Here it is divided by 4 because for every 4 years we get leap year).
Leap year=2*2=4.
Non leap =7*1=7.
Total = 11.
Jan 1st=1 day.
11+1=12.
12/7 = 5.
Friday.
(1)
Salini said:
7 years ago
Hi guys.
For this 1/1/2006 Sunday then 1/1/2010 ___day.
For this, from 2006 to 2009 the odd days will be 1+1+2+1.
So, 5 days.
Sunday - 0, Monday - 1, Tuesday - 2, Wednesday - 3,Thursday - 4, Friday - 5.
So, till 31st December. We counted.
And my doubt is 1/1/2010 is another day which we have not counted. And it will be Saturday. Please, anyone, clear my doubt.
For this 1/1/2006 Sunday then 1/1/2010 ___day.
For this, from 2006 to 2009 the odd days will be 1+1+2+1.
So, 5 days.
Sunday - 0, Monday - 1, Tuesday - 2, Wednesday - 3,Thursday - 4, Friday - 5.
So, till 31st December. We counted.
And my doubt is 1/1/2010 is another day which we have not counted. And it will be Saturday. Please, anyone, clear my doubt.
Kundan K said:
1 decade ago
Hi friends,
There are 52 weeks in a year.And there are 7 days a week.So, the closest factor of 7 is 49.And 52-49=3.So after 49 days again there will be Sunday.So,add 3 days to Sunday that is Wednesday.Now similarly continuously add 3 days for 4 times as there is difference of 4 years between 2010 & 2006.
Therefore the ans is "FRIDAY".
There are 52 weeks in a year.And there are 7 days a week.So, the closest factor of 7 is 49.And 52-49=3.So after 49 days again there will be Sunday.So,add 3 days to Sunday that is Wednesday.Now similarly continuously add 3 days for 4 times as there is difference of 4 years between 2010 & 2006.
Therefore the ans is "FRIDAY".
Raaj said:
1 decade ago
@kamatchi the answer for ur question is( to find number of odd days )
be clear for every 7 days a cyle(week) wii be compleated so just divide the number of days with 7 the remainder will be the number of odd days for example number of odd days in leap year 366 divided by 7 so the remainder will be 2 days which is the number of odd days in a leap year
be clear for every 7 days a cyle(week) wii be compleated so just divide the number of days with 7 the remainder will be the number of odd days for example number of odd days in leap year 366 divided by 7 so the remainder will be 2 days which is the number of odd days in a leap year
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