In competitive exams, it is important that you have mind clear while solving any questions. Because when your mind is clear you can solve every question and that too at a very brisk pace. So how do you clear your mind? There is no science in this, with the right amount of perseverance and effort you can become an expert in solving the questions in a competitive exam. Have enough practice so that while giving an exam you don’t have to think whether you have solved this question earlier or not.

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## Calendar Test

One such type of question 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 than it will become very easy for you to solve it. You can be asked whether the year mentioned in the question is a leap year or not! You will be given a date and a day and you will be asked to determine the day that will fall on the same date on next year and many more.

We will try to cover some of the questions of calendar test and will explain you in detail the tricks and methods to solve the questions. After that, there will be some practice questions which you can solve and make calendar test your strength. To understand the calendar you need to be aware of odd day concept.

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## Odd Days Concept

Suppose we have to calculate the number of odd days in 1200 years. In 1200 there are 3 years which are divided by 100 and 400. (400, 800, and 1200). And the remaining years are only divided by 100. (100, 200, 300, 500, 600, 700, 900, 1000, and 1100). Every 100 years have 76 ordinary years and 24 leap years.

Odd days in ordinary year = (52 weeks + 1) days. Odd days in a leap year = (52 weeks +2) days. So odd days in 100 years will be (76 x 1 + 24 x 2) which is 124 odd days. This can also be written as 17 weeks + 5 days. So every 100 years will have 5 odd days. Similarly, 200 years will have (5 x 2) = 3 odd days and 300 years will have 1 day.

The number of odd days in 400 years will be ( 5 x 4 + 1) because 400 is itself a leap year and that is why it has one odd day extra. Thus odd days in 400 will be 0. This is same for every year which is a multiple of 100 and 400. Thus 1200 will also have odd days. Now we will solve some examples to have a better understanding of this, concept.

## Solved Examples

Q. January 2, 2007, was Tuesday. What will be the day on January 2, 2008?

1. Monday 2. Tuesday

3. Wednesday 4. Thursday

Here you can see that 2007 is neither a multiple of 4. Thus 2007 is an ordinary year. So the odd day in 2007 will be 1. Now since the 2nd day of January 2007 was Tuesday, 2nd day of January 2008 will be one day beyond Tuesday. And that is why January 2, 2008, will be Wednesday. Thus the correct answer is the option (3).

Q. For what year will the calendar be the same as for the year 2009?

1. 2021 2. 2022

3. 2023 4. 2024

For the year to have the same calendar as 2009 you need the sum of the number of odd days. When this sum is divisible by 7 than that year will have the same calendar as 2009.

Year | 2010 | 2011 | 2012 | 2013 | 2014 | 2015 |

Odd days | 1 | 1 | 2 | 1 | 1 | 1 |

Here, the sum of odd days in these years is 7. Thus 2016 will have the same calendar as 2009.

Q. Mr. Khan has celebrated his birthday on Tuesday 30th, September 1997. When will he celebrate his birthday on the same day again?

1. 30 Sept 2004 2. 30 Sept 2002

3. 30 Sept 2001 4. 30 Sept 2003

The name day will be repeated after every 7 days. 1997 is an ordinary so it will have 1 odd day and thus 1998 will also gain 1 day.

30 Sept | 1998 | 1999 | 2000 | 2001 | 2002 | 2003 |

Day Gain | 1 | 1 | 2 | 1 | 1 | 1 |

Thus on 30th September 2003, Mr Khan will celebrate his birthday again on Tuesday.

### More Solved Examples

Q. What was the day on 10th August 1947?

1. Saturday 2. Sunday

3. Monday 4. Tuesday

In this type of question start by subtracting one year from the year given in the question. Thus 1947-1 = 1946. Now, remember the number of odd days in 100, 200, 300, and multiple of 400 years in 1946. Thus, it will be 1600 + 300 + 46 = 1946. 1600 will have 0 odd days, 300 will have one 1 odd day, and 46 will 46 odd days.

So the number of odd days will be, 0 + 1 + (46) For the number in the bracket, divide it by 4 and the quotient is to be added in the odd days. Here quotient will be 11, so odd days = 0 + 1 + 46 + 11 = 58. When you divide 58 by 7 you will have 2 as remainder. Thus December 31st, 1946 was Tuesday. From here count the number of days until 10th August 1947.

Month | Jan | Feb | Mar | Apr | May | June | July | Aug | Total |

Days | 31 | 28 | 31 | 30 | 31 | 30 | 31 | 10 | 222 |

A total number of days will be 222 days and add 2 from the previous year. Divide 224 by 7, you will have 0 remainders and thus 10th August 1947 was Sunday. So, your correct answer is (2).

## Practice Questions

1. What was the day on 30th May 2006?

1. Friday 2. Sunday

3. Tuesday 4. Monday

The answer is (3)

2. If it was Monday on the 4th of April, 1988 than what was the day on 6th November, 1989?

1. Monday 2.Wednesday

3. Friday 4. Saturday

The answer is (1)

3. On January 12, 1979, it was Friday, what was the day on 12th of January, 1980?

1. Friday 2. Saturday

3. Sunday 4. Monday

The answer is (2)

4. Suppose 1st March is Wednesday, then which month of the same year will also have Wednesday as the first day?

1. August 2. September

3. October 4. November

The answer is (4).

5. If the second-hand moves 480 then how much space (in terms of degrees) will the minute hand move?

1. 80 2. 60

3. 120 4. 40

The answer is (1).