Aptitude - Calendar - Discussion

Hello everybody ,
You must know some codes first.

Years, Days, Century code(for every 400 years ).
Jan-0, sun-0, 0:99-6.
Feb-3, mon-1, 1:199-4.
Mar-3, tue-2, 2:299-2.
Apr-6, wed-3, 3:399-0.
may-1, thr-4,
Jun-4, fri-5,
Jul-6, sat-6,
Aug-2.
Sep-5.
Oct-0.
Nov-3.
Dec-5.

Now formula is ,

Day = (Date+month code+no.of years+no.of leap years + century code)/(7).

In the given question 15th Aug 2010.

Day = (15+2+10+2+6)/7 = 7/7 = 0. Which is 'Sunday'. :).

What the day on 16 may 1992?

1992 = (1991)+(days between 1/1/1992 to 16/5/1992).
1600 = 0.
300 = 1.

91 = (22 leap)+(69 ordinary day).
= (22*2)+(69*1).
= 113 days.

= 16 weeks 1 day = 1 odd day.

Total 1991 yrs = 0+1+1 = 2 odd days.

jan feb mar april may
31 28 31 30 16 = 136 days.

= 19 weeks 3 days.

= 3 odd days.

Total till 16 may 1992 = 2+3 = 5.
5 = FRIDAY.
But in actual calendar it's SATURDAY. CAN I get the correct method to find it?

2000 to 2010 having 10 years.

In which 3 leap year (2000-01), (2004-05), (2008-09) & 7 normal years.

3 leap years have 6 odd days (3 years * 2 odd days) = 6 odd day.

Remaining 7 year having 7 odd day (individually 1) & for year (2010-11).

January February March April May June July Aug.

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days.

No of odd day = 227/7 = Remainder is 3.

So total odd days are 3+6+7 = 16.

Now 16/7, remainder is 2.

So Tuesday will come.

So how answer will come (Sunday)?

If you divide eery month by "7" I.E.
JAN = 31/7 R = 3
SIMILARLY
JAN = 3
FEB = 0
MARCH = 3
APRIL = 2
MAY = 3
JUNE = 2
JULY = 3
AUGUST = 15 (BECAUSE NO. OF DAYS COMPLETED )
SO, THE SUM IS;
3+0+3+2+3+2+3+15 = 31.

NOW DIVIDE 31/7, R = 3.
0 = SUNDAY
1 = MONDAY
2 = TUESDAY
3 = WEDNESDAY
4 = THURSDAY
5 = FRIDAY
6 = SATURDAY
7 = SUNDAY (= 0)

Then, R = 3.

Then the answer would be " WEDNESDAY".

If we divide 1600/400 we will get zero as remainder so it has zero odd days.
so in 2000 yrs we have 0 odd days.
In 9 yrs there r 2 leap years
leap yrs have 2 odd days(366/7 2 as remainder) and
non leapyr have 1 odd day totaling we have 2*2+7*1=11 odd days
Now cal no: of days in 2010 upto aug 15 then dividing by 7
will give no of odd days 2010 which is 3.

= 1600 + 400 + 9 + (jan+feb+mar+apr+may+jun+jul) + 15 days,
= 0 + 0 + ((2*2) + (7*1)) + (3+0+3+2+3+2+3) + 15 days,
= (4+7) + 16 + 15,
= 11 + 31,
= 42.

To find odd days?
= 42/7 = 6 weeks and 0(zero odd days)
Odd days:
0 -> sunday
1 = Mon,
2 = Tue,
3 = Wen,
4 = Thu,
5 = Fri,
6 = Sat.
0 = 7 = Sun.

So, Sunday is the answer.

@Sunita.

97 year == Divide 97 by 4:
= Quotient is 24.
= So 24 is leap year and in 1 leap year has 2 odd days.
= Than (97-24= 73): 73 is ordinary year and 1 ordinary year has 1 odd day.
= 24*2 + 73*1= 121 days.
=121/7 = than reminder equal to 2.
= So 2 is odd days.

Given that 15th august, 2010.

1) Add day and last two digits of the year 15+10 = 25.

2) Divide last two digit of the year by 4, so 10/4 = 2.

3) Add quotient value to step 1.

= 25+2+2(month code)+6(year code) = 35.

4) Divide 35/7 = 5 it means remainder is 0(day code) Sunday.

I have a question.

How in 100 years, 76 was ordinary years and 24 was leap years?

Because generally, every 4 year is a leap year, so 100 is completely divisible by 4 and we get 25, so in general, in 100 years, there were 25 leap years?

Please, anyone, explain this to me.

Why is it that the year's code is varying? Sometimes Jan is given code as 1, Feb as 4, Mar 4 April 0 and so on and again in another instance you find Jan given code 0, feb as 3 Mar 3 April 6, may 2. What brings the variation?

Anyone, please explain.