Calendar Month Code In Reasoning

Now this will be your last step you have written all the data and add them all after this you will find a result.

Calendar month code in reasoning. Take the last two digits of the year given i e. There are understandable simple solutions useful for rrb alp group d and bank jobs. Calendar calendar problem tricks calendar reasoning concept problems questions solutions clock and calendarinstall imo for free calendar pdf https.

The division of the number 31 by 7 provides the remainder 3 hence january has 3 odd days. Step 3 divide 47 by 4 i e 11 and add 11 to the result i e 47 0 15 11. At your third step you have to write the date.

Divide the last two digits of the year by four. Therefore 29th day of the month occurs 4400 97 or 4497 times ex given that on 10th november 1981 is tuesday what was the day on 10th november 1581. Secondly you have to use the code of the month as given in the calendar which is given above.

Add month code and year codes to the result obtain in step3. 31 5 n 36 n 36 38 40 or 42. 31 00 31.

Step 5 divide the sum by 7 i e 75 7 we get remainder 5. Quotient of 00 4 0 31 0 31. Add the day digit to last two digit of the year.

On generalising any month which has 31 days has 3 odd days and any month which has 30 days has 2 odd days. Add the quotient value in step 3 to result obtain in step 1. 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.