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# Calendar Test - CSEET

## Calendar Test

One of such topic that needs the immense amount of practice is calendar test related problems. Although the frequency of this question is less in exams they are of equal importance. If you have learned the basics of these questions then it will be easy for you to solve it.

Important concepts and ideas are given under.

### Leap and Ordinary Year

Leap year − Every year which is exactly divisible by 4 such as 1992, 1996 etc. is called leap year.

Every 4th century is also called as leap year. For a century to be a leap year, it should be exactly divisible by 400.

Example − 400, 800, 1200 are leap years because these are divisible by 400.

### Number of Odd Days

Apart from the complete number of weeks in a particular month, the extra days are called odd days.

### Calculation of Odd Days

1. An ordinary year has 365 days. When we divide 365 by 7, we get 52 as quotient and 1 as remainder. So that year has 52 weeks and single day. As the remainder is odd, we call it Odd day.

2. A leap year has 366 days that is 52 weeks and 2 days. So leap year has two odd days. A century has 100 years. Out of these years 76 years are ordinary years and 24 leaps years. So, 100 years contain 5 odd days, Similarly, 400 years contain 5 × 4 + 1 = 21 (no odd days)

NOTE

1. 5 × 3 = 15 days = 2 weeks + 1 odd day 5 × 1 = 5 days = 5 odd days

2. 400th year is a leap year therefore, one additional day is added.

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### Types of Problems

Type I

To find the day of a week by the help of number of odd days, when reference day is given.

Working Rule

1. Find the net number of odd days for period between the reference date and given date. The day of the week on the particular date is equal number of net odd days ahead of reference day but behind reference day.

Example 1 − January 5, 1991 was a Saturday. What day of the week was on March 3, 1992?

Solution − 1991 is ordinary year, so it has only 1 odd day. Thus January 5, 1992 was a day beyond Saturday. That is Sunday.

Now, in January 1992 there are 26days left. That is 5 odd days. In February 1992 there are 29 days that is 1 odd day. In March 1992 there 31 days, i.e. 3 odd days. So total number of days after January 5, 1992 = (5 + 1 + 3) = 9 days, i.e. 2 odd days.

Therefore, 3 March 1992 will be 2 days beyond Sunday.

Example 2 − Today is 21st August. The day of the week is Monday. This is a leap year. What will be the day of the week on this day after three years?

Solution − Since this is a leap year, so none of the next 3 years is a leap year. Hence the number of odd days = 3. So, the day of the week will be 3 days beyond Monday i.e. it will be Thursday.

Type II

To find the day of a week by the help of the number of odd days, when no reference day is given.

### Working Rule

1. On an assigned date, calculate the number of odd days.

2. In that case we count days according to number of the odd days.

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