Aptitude - Calendar - Discussion

Muthaiyan good way of explanation.

Good explanation rajesh.

Step 1:[1997+1.1.1998to17.6.1998]
step2: 1997=1600+300+97=> 1600 is 0 odd days, 300 is 1 odd day
step3: calculate 97
step4:97/4= 24,=> [24-97=73]
step5:24leap year +73 ordinary year=>(24*2)+(73*1)=127
step6:127/7=17,=>17week+2 odd days
step7:1998 is an ordinary year therefore [31+28+31+30+31+17=168]
ste8:168/7=24weeks+0 odd days
step9:add all odd days we get 3 is an wednesday

How we gat an idea to take 1600 years and 300 years in above problem.

Ishan , Arya and Raviteja

Thanks to all u for ur such a explanation

How to calculate odd days for 300 years? I don't understand this only.

Here is complete solution, the year is 1998
16+3=19
See from 0 A.D
1600yr+300yr =1900yr
We are calling the odd days from the 0 A.D. So 1600 yr has 0 odd days
But the remaing 300 year has 15 odd days and rest you can do.

How you got ordinary days? could you explain?

For those who did not understand this problem,let me try to explain you.

The first thing to be cleared is that for every 400 years there are 0 odd days.

This is because,in 400 years,there are 97 leap years and 303 ordinary years(remember that 100,200,300 are not leap years but 400 is a leap year.Because any year ending with 00,that year should be divisible by 400 in order to become a leap year)

So the number of odd days at the end of 400 years will be [(97*2)+(303*1)]=497 days.Convert this into weeks,we get 71weeks + 0odd days=totally 0 odd days

Now coming to the problem,
1997=1600+397
for 1600 years 0 odd days
therefore 397 years can be classified as
1601-1700-->24 leap years + 76 ordinary years=[(24*2)+(76*1)]=
124 odd days
1701-1800-->24 leap years + 76 ordinary years=124 odd days
1801-1900-->24 leap years + 76 ordinary years=124 odd days
1901-1997-->24 leap years + 73 ordinary years=121 odd days
totally we have 124*3 + 121=493 odd days
convert this into weeks=70weeks +3 odd days
Hence at the end of the year 1997,we have 3 odd days

Now coming to the year 1998,
the number of odd days=31+28+31+30+31+17=168 odd days(jan,feb,...jun17th)
Convert this into weeks=24weeks +0 odd days
Hence the total odd days=3+0=3 odd days
therefore 17th june 1998=Wednesday

There is another way...

Remember the code FOR ALL PROBLEMS ABOUT CALENDER


MONTH : J F M A M J J A S O N D
CODE . : 0 3 3 6 1 4 6 2 5 0 3 5

and

CODE FOR THE YEAR
1700 -- 1799 = 4
1800 -- 1899 = 2
1900 -- 1999 = 0
2000 -- 2099 = 6
2100 -- 2199 = 4

AT FINAL EX: FOR 17 JUNE 1998

take the last two numbers in the year for ex. In 1998 take 98 and divide them by 4 and take only the quotient and leave the reminder

ex: 4)98(24
8
----
18
16
----
2
leave that reminder 2 and take only quotient 24.

add the following

month code for JUNE 4
year code for 1998 0
left quotient while
divided by 4 24
finally add the last
two numbers in year
1998 and date17(98+17) 115

------------
143

divide the amount by 7 i.e., 143/7
here take only the reminder and left the quotient.

ex: 7)143(2
14
----
03
take the value of reminder 3.


WEEK S M T W T F S
CODE 0 1 2 3 4 5 6

IN THE ABOVE PROBLEM ANSWER IS 3 THEN DAY IS W-Wednesday


ANOTHER EXAMPLE: 15 AUGUST 1947

ADD

month code 2
year code 0
quotient divided by 4 11
(47+15) 62
----
75
75/7 take only reminder 5
Answer is 5-F-Friday.