Calendar ........[competitive. exam prep.... topic 29]

 Most of the students are having difficulties in solving the Calendar problems.

The number of days more than the complete number of weeks in a given period, are called odd days. For example :

(1) In an ordinary year (of 365 days) there are 52 weeks and one odd day.

(2) In a leap year (of 366 days) there are 52 weeks and two odd days.

What is the Leap and Ordinary year ?

Every year which is exactly divisible by 4 such as 1988, 1992, 1996 et. is called a leap year.

Also every 4th century is a leap year. The other centuries, although divisible by 4, are not leap years. Thus, for a century to be a leap year, it should be exactly divisible by 400. For example :

(1) 400, 800, 1200, etc are leap years since they are exactly divisible by 400.

(2) 700, 600, 500 etc are not leap years since they are not exactly divisible by 400.

How to find number of odd days :An ordinary year has 365 days. If we divide 365 by 7, we get, 52 as quotient and 1 as remainder. Thus, we may say that an ordinary year of 365 days has 52 weeks and 1 day. Since, the remainder day is left odd-out we call it odd day.

Therefore, an ordinary year has 1 odd day.

A leap year has 366 days, i.e. 52 weeks and 2 days.

Therefore, a leap year has 2 odd days.

A century, ie, 100 years has :

76 ordinary years and 24 leap years.

= [(76 X 1 day] + [(24 X 2 days]

= 124 days

if we devide 124 with 7 there are 17 full weeks and 5 odd days remains.

Therefore, 100 years contain 5 odd days.

Now, (i) 200 years contain 10 odd days, ie, 3 odd days.

(ii) 300 years contain 8 odd days, ie, 1 odd day.

(iii) 400 years contain 6 + 1 = 21, ie, no odd day.

(Nore: 400th year is a leap year therefore, one additional day is added, So number of odd days in year 301 to 400 = 6 instead of 5)

Similarly, 800, 1200 etc contain no odd day.

Already many logics are there to solve these kinds of problems, but all these logics are difficult to understand. So  here is the simple way to solve calendar problems.                

 Also Read: How to solve Logical reasoning part in Elitmus?

In order to solve these type of problems you must know some codes.

Year Code

Month Code

Day Code

Steps to solve:-

Step 1:  Add the day digit to last two digit of the year.

Step 2:   Divide the last two digits of the year by four.

Step 3:   Add the Quotient value in step 3 to result obtain in step 1.

Step 4:   Add Month Code and year codes to the result obtain in step3.

Step 5:   Divide the result of step4 by seven.

Step 6:  Obtain the remainder and match with the day code.

Example:-

Trick for solving Questions where is 1999 or less.

• First of all, take the last two digit of year and divide it by 4. For example- In case of year 1985, divide 85 by 4 and keep the result for future use.
• Then use the code of the month from calender shown above. If the year is a leap year, consider the code for leap year.
• Then write the date.
• At the end, add all these data and find the result.
• To find the day, divide the result by 7. You will get some remainder. 
• Match the code of remainder with above given code table and find the answer.

Trick for solving questions where Year is 2000 or more

• Consider the last 3 digit of the year. If it is less than 100, add 100 to it. For example- In year 2012, the last three digit is 012 which is less than 100 so add 100 to add and make it 112(100+012). Once done, divide it by 4 and keep the result.
• Then use the code of the month from calender shown above. If the year is a leap year, consider the code for leap year.
• Then write the date.
• At the end, add all these data and find the result.
• To find the day, divide the result by 7. You will get some remainder. 
• Match the code of remainder with above given code table and find the answer.

Examples with solution......................................................................

Q1.What was the day on 31st Oct 1984?

       (a)Friday              (b)Sunday

       (c)Wednesday     (d)Monday

Sol- In 1984, divide 84/4 = 21

       Code for Oct = 0 (Refer the above shown calender for codes)

       Mentioned Date= 31

Result = 84+21+0+31 =136

To find the day of week, divide the result by 7 and write the remainder = 136/7 = 3 (remainder)

3 is the code for Wednesday so the answer will be wednesday.

Q2.What was the day on 27th Dec 1985?

       (a)Friday                   (b)Monday

       (c)Tuesday                (d)Sunday

Sol-  Here the year is 1985 so divide 85/4= 21

 Code for Dec = 5

Mentioned Date= 27

Result = 85+21+5+27 =138

To find the day of week, divide the result by 7 and write the remainder = 138/7 = 5(remainder)

5 is the code for Friday so the answer will be friday.

Q3. Find the day of the week on 26th Jan 2012? 

         (a)Tuesday              (b)Thursday

         (c)Friday                 (d)Sunday

Sol- Here the year is 2012 so follow the second rule and consider the last 3 digit i.e 012. It is less than hundred so add 100 to it. After adding 100, it becomes 112. Now, divide it by 4. You will get 28. Also note that it is a leap year.VV

Code of Jan = 6 (due to leap year)

Mentioned Date= 26

Result = 112+28+6+26 = 172

To find the day of week, divide the result by 7 and write the remainder = 172/7= 4

4 is the code for Thursday so the answer will be Thursday.

ALL THE BEST...............