Calendar: Definition, Practice Questions, Dates, Months, Types

Calendar is a topic in Logical Reasoning that appears in various exams. This topic is one of the most important topics and is commonly found in many Government exams and entrance exams such as VITEEE, CUET, UPSC, RBI Grade B, SSC, Railway, Defence, and other competitive exams. The questions related to this topic can be tricky, but they are generally easy and scoring, once the concept is fully understood. One should have complete knowledge of a calendar reasoning formula to solve the questions related to this topic.

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This Story also Contains

1. What is a Calendar?
2. Types of Calendar Reasoning Questions Asked in Various Exams
3. Practice and Resources on Calendar
4. Question Weightage of Calendar in Competitive Exams
5. Practice Calendar Reasoning Questions For Odd Days in a Calendar
6. Practice Calendar Reasoning Questions For Repeating Year in a Calendar
7. Practice Calendar Reasoning Questions For Reference Calendar
8. Practice Calendar Reasoning Questions For Finding the Day/Date of the Week in a Calendar
9. Calendar Reasoning Questions with Answers for VITEEE/ CUET
10. Calendar Reasoning Questions with Answers for CAT/ APICET/ SUAT/ JIPMAT
11. Calendar Reasoning Questions with Answers for SSC/ RRB exams

What is a Calendar?

A calendar is a chart showing a systematic arrangement of days, months, and weeks of a particular year. With the help of a calendar, we can easily determine the date, month, year, and weekday of a particular day. Let’s discuss some of the major points based on the structure of a calendar.

1. A calendar shows the arrangement of the days and months of a year.

2. A year has either 365 days or 366 days.

3. Any year which has 365 days is known as an ordinary calendar and any year which has 366 days is known as a leap year.

4. To determine if a year is a leap year or an ordinary year, check if the last two digits of the year are divisible by 4. If they are divisible by 4, then it is a leap year; otherwise, it is an ordinary year.

5. A year can be classified into one more category, i.e., century year. A century year is a block of 100 years. For example - 1600, 1700, 1800, 1900, 2000, etc.

6. If a year is a century year (i.e., the last two digits are 00), and is divisible by 4, it does not make every century year a leap year. To determine if a century is a leap year, check the divisibility of the complete year by 400. If it is divisible by 400, then it is a century leap year; otherwise, it is an ordinary century year.

7. When we divide the total number of days (365 or 366 days) in a year by 7, we get a quotient and a remainder as a result.

8. On dividing 365 days by 7, we get 52 as a quotient and 1 as a remainder. This indicates that a normal year has 52 weeks and 1 extra day and this extra day is referred to as an odd day.

9. On dividing 366 days by 7, we get 52 as a quotient and 2 as a remainder. This indicates that a leap year has 52 weeks and 2 extra days and this extra day is referred to as an odd day.

10. An ordinary year has 1 odd day, whereas a leap year has 2 odd days.

Types of Calendar Reasoning Questions Asked in Various Exams

There are different types of questions that are asked from the calendar reasoning topic in exams and they are as follows -

1. Find the odd days in a calendar

2. Find the repeating year in a calendar

3. Reference calendar

4. Find the day/date of a week in a calendar

Let’s discuss each type in detail with the help of examples -

1. Find the Odd Days in a Calendar

In these types of questions, candidates are asked to determine the number of odd days in a specific year or within a particular period. The number of odd days is calculated by dividing the total number of days by 7, and the remainder is the number of odd days. It's important to note that odd days are always less than 7.

For example, January has 31 days, and dividing 31 by 7 gives the remainder 3. Thus, January has 3 odd days. So, this means that any month with 31 days, has 3 odd days. Also, any month with 30 days, has 2 odd days. But, as usual, February will be an exceptional case. In an ordinary year, February has 28 days, and the division of 28 by 7 gives 0 as the remainder. Thus, in an ordinary year, February has 0 odd days. In a leap year, February has 29 days, and the division of 29 by 7 gives 1 as the remainder. Thus, in a leap year, February has 1 odd day. The following table shows the number of odd days in different months.

In the exams, questions based on this topic are not asked directly. The questions asked are based on the concept of odd days. The candidate has to find out the day of the week of a given date. In this type, no reference date or day will be provided. For these types of questions, we have to use the concept of odd days and need to remember a few codes to make the question easier.

The formula used to calculate the day is as follows:
The number of odd days = {Date + (Month code) + (Century code) + (last two digit of the year) + ((last two digit of the year/4)'s Quotient)}/7

Code for months -

Code for days -

Code for Centuries -

Example: What was the day of the week on the 10th of May 2007?

Answer: For 10th of May, 2007 -
Date = 10
Code for the month (May) = 1
Code for the century (2000) = 6
(2007 = 2000 + 7; For 2000 it will be 6 and for 7 it will be 7)
And, 7 ÷ 4 = 1(Quotient)
Now, on adding them we get→ 10 + 1 + 6 + 7 + 1 = 25
On dividing 25 by 7, we get 4 as a remainder. So, the number of odd days = 4
4 is the code for Thursday.
So, 10th of May 2007 will be Thursday.

2. Find the Repeating Year in a Calendar

In these types of questions, candidates need to find the year or month which will have the same calendar as the given year or month. To determine the repeat year of the given year, the sum of the odd days of all the years between the given year and the resulting year must be 7 or 0.

Example: Which year has the same calendar year as the year 2009, (2014 or 2015)?

Answer: Let’s calculate the number of odd days -

The number of odd days between 2009 and 2014 = 5 + 1 = 6 (On dividing 6 by 7, the remainder is 6)

The number of odd days between 2009 and 2015 = 6 + 1 = 7 (On dividing 7 by 7, the remainder is 0)

The year which has the same calendar as the year 2009 will be 2015.

3. Reference Calendar

There are again a few types in the Reference Calendar. Let’s understand these types in detail -

a) Dates and Months are the Same but the Year is Different

In these types of questions, the candidate has to find out the day of the week for a given date using the day of the week provided for another date. In this case, the reference date and the date for which the day of the week needs to be determined will have the same date and month but different years. So, here, the concept used is the number of odd days in an ordinary year and a leap year and we know, that an ordinary year has 1 odd day, whereas a leap year has 2 odd days. There are 2 ways to solve these types of questions. Either calculate the days for each year respectively by adding the odd number of days to the day of each previous year or calculate the total years between both the given dates and the respective total odd number of days and then add it to the given day to get the day of the week on the particular given date.

In these types of questions, a candidate needs to determine the day of the week for a given date using the day of the week provided for another date. The reference date and the date for which the day of the week needs to be determined will have the same date and month but different years. The concept used here involves the number of odd days in an ordinary year and a leap year. An ordinary year has 1 odd day, whereas a leap year has 2 odd days.

There are 2 ways to solve these types of questions. You can either calculate the days for each year respectively by adding the odd number of days to the day of each previous year, or you can calculate the total years between both the given dates and the respective total odd number of days, and then add it to the given day to get the day of the week on the particular given date.

Example: If on 15 August 2003, it was Friday, then what day would it be on 15 August 2009?

1st Method: If 15 August 2003 → Friday, then 15 August 2004 → Friday + 2 = Sunday (2004 is a leap year)

Similarly, 15 August 2005 → Sunday + 1 = Monday

15 August 2006 → Monday + 1 = Tuesday

15 August 2007 → Tuesday + 1 = Wednesday

15 August 2008 → Wednesday + 2 = Friday

15 August 2009 → Friday + 1 = Saturday

2nd Method: The 15th of August 2003 was Friday.

So, the total number of days from the 15th of August 2003 to the 15th of August 2009 -

(6 × 365) + 2 = 2190 + 2 = 2192

(2 days are being added as 2004, and 2008 are leap years.)

Now, on dividing 2192 by 7, we get 1 as a remainder.

So, the number of odd days = 1

So, Friday + 1 day = Saturday

The 15th of August 2009 was Saturday.

b) Dates and Months are Different but the Year is the Same

In these types of questions, the candidate has to find out the day of the week for a given date using the day of the week provided for another date. In this case, the reference date and the date for which the day of the week needs to be determined will have the same year but different dates and months.

To solve these questions, the first step is to calculate the total number of days between both the given dates and then divide the total number of days by 7 to get the required quotient and remainder. The obtained remainder will be the number of odd days. Based on the number of odd days, we will get the day of the week.

Example: If on 15 March 2005, it was Friday, then what day would it be on 29 April 2005?

Answer: 15 March 2005 → Friday, 29 April 2005 → (?)

The number of days from 15 March 2005 to 29 April 2005 = 16 + 29 = 45

Divide 45 by 7 = 45/7 = 6 (Quotient), and 3 (Remainder)

Total number of odd days = 3

So, on 29 April 2005, it will be Friday + 3 = Monday

c) Dates, Months and Years are Different

In these types of questions, the candidate has to find out the day of the week for a given date using the day of the week provided for another date. In this case, the reference date and the date for which the day of the week needs to be determined will have different dates, months and years.

To solve these questions, the first step is to calculate the total number of days between both the given dates and then divide the total number of days by 7 to get the required quotient and remainder. The obtained remainder will be the number of odd days. Based on the number of odd days, we will get the day of the week.

Example: If on 15 August 2003, it was Sunday, then what day would it be on 27 May 2015?

Answer: 15 August 2003 → Sunday, 27 May 2015 → (?)

Remaining number of days in 2003 = 16 + 30 + 31 + 30 + 31 = 138

Divide 138 by 7 = 138/7 = 19 (Quotient), and 5 (Remainder)

Odd days in 2003 = 5

Similarly, odd days from 2004 to 2014 -

2004 → 2; 2005 → 1; 2006 → 1; 2007 → 1; 2008 → 2; 2009 → 1; 2010 → 1; 2011 → 1; 2012 → 2; 2013 → 1; 2014 → 1

Odd days from 2004 to 2014 = 14

Now, the number of odd days in 2015 till 27 May → 3 + 0 + 3 + 2 + 6 = 14

Total number of odd days = 5 + 14 + 14 = 33

Divide 33 by 7; Remainder = 5, i.e., the required number of odd days = 5

So, on 27 May 2015, it will be Sunday + 5 = Friday

4. Find the Day/Date of the Week in the Calendar

In these types of questions, instead of a direct date, some general information will be provided and the candidates are asked to find the day or date of the week in the calendar based on the given information. The problems related to this topic use the concept of odd days, some general facts or calendar reasoning tricks related to the calendar.

Example: If today is Sunday, then what will be the 24th day from today?

Answer: The 24th day from today means the total number of days is 23.

On dividing 23 by 7, the remainder obtained is 2.

So, 24th day from today = Sunday + 2 = Tuesday

Practice and Resources on Calendar

1. A Modern Approach to Verbal & Non-Verbal Reasoning by RS Aggarwal

2. Perfect Verbal Reasoning By Ajay Chauhan

3. SSC Reasoning by Rakesh Yadav

4. Logical and Analytical Reasoning by AK Gupta

Question Weightage of Calendar in Competitive Exams

The number of questions based on the calendar varies from exam to exam -
1) Questions asked in SSC exams, i.e., SSC MTS, SSC CGL, SSC CHSL, SSC CPO, and Steno - 1 to 3 questions.

2) Questions asked in Bank exams i.e., IBPS PO, IBPS Clerk, RRB PO, RRB Clerk - 2 to 3 questions.

3) Questions asked in the Railways exam i.e. Group D, NTPC, JE, ALP etc - 1 to 2 questions.

4) In various exams clock and calendar reasoning questions are asked together.

You can click below for clock and calendar reasoning questions with solutions and other important verbal reasoning topics.

Practice Calendar Reasoning Questions For Odd Days in a Calendar

Q1. Directions: What was the day of the week on the 24th of August 1923?

A) Monday

B) Sunday

C) Thursday

D) Friday (Correct)

Solution: Use the month, century, and day code for the calendar and apply the formula.

The formula used to calculate the day is as follows:
{Date + (Month code) + (Century code) + (last two digit of the year) + ((last two digit of the year/4)'s Quotient)}/7

Code for months -

Code for days -

Code for Centuries -

For 24th of August, 1923 -
Date = 24
Code for the month (August) = 2
Code for the century (1900) = 0
(1923 = 1900 + 23; For 1900 it will be 0 and for 23 it will be 23)
And, 23 ÷ 4 = 5(Quotient)
Now, on adding them we get→24 + 2 + 0 + 23 + 5 = 54
On dividing 54 by 7, we get 5 as a remainder. So, the number of odd days = 5
5 is the code for Friday.
So, the 24th of August 1923 will be Friday. Hence, the fourth option is correct.

Q2. Directions: What was the day of the week on the 18th of July 1850?

A) Wednesday

B) Monday

C) Saturday

D) Thursday (Correct)

Solution: Use the month, century, and day code for the calendar and apply the formula.

The formula used to calculate the day is as follows:
{Date + (Month code) + (Century code) + (last two digit of the year) + ((last two digit of the year/4)'s Quotient)}/7

Code for months -

Code for days -

Code for Centuries -

For 18th of July, 1850 -
Date = 18
Code for the month (July) = 6
Code for the century (1800) = 2
(1850 = 1800 + 50; For 1800 it will be 2 and for 50 it will be 50)
And, 50 ÷ 4 = 12(Quotient)
Now, on adding them we get→18 + 6 + 2 + 50 + 12 = 88
On dividing 88 by 7, we get 4 as a remainder. So, the number of odd days = 4
4 is the code for Thursday.
So, the 18th of July 1850 will be Thursday. Hence, the fourth option is correct.

Q3. Directions: What will be the day of the week on the 20th of January 2030?

A) Friday

B) Sunday (Correct)

C) Saturday

D) Monday

Solution: Use the month, century, and day code for the calendar and apply the formula.

The formula used to calculate the day is as follows:
{Date + (Month code) + (Century code) + (last two digit of the year) + ((last two digit of the year/4)'s Quotient)}/7

Code for months -

Code for days -

Code for Centuries -

For 20th of January 2030 -
Date = 20
Code for the month (January) = 0
Code for the century (2000) = 6
(2030 = 2000 + 30; For 2000 it will be 6 and for 30 it will be 30)
And, 30 ÷ 4 = 7(Quotient)
Now, on adding them we get→20 + 0 + 6 + 30 + 7 = 63
On dividing 63 by 7, we get 0 as a remainder. So, the number of odd days = 0
0 is the code for Sunday.
So, the 20th of January 2030 will be Sunday. Hence, the second option is correct.

Q4. Directions: What was the day of the week on the 10th of May 1713?

A) Monday

B) Sunday

C) Wednesday (Correct)

D) Friday

Solution: Use the month, century, and day code for the calendar and apply the formula.

The formula used to calculate the day is as follows:
{Date + (Month code) + (Century code) + (last two digit of the year) + ((last two digit of the year/4)'s Quotient)}/7

Code for months -

Code for days -

Code for Centuries -

For 10th of May 1713 -
Date = 10
Code for the month (May) = 1
Code for the century (1700) = 4
(1713 = 1700 + 13; For 1700 it will be 4 and for 13 it will be 13)
And, 13 ÷ 4 = 3(Quotient)
Now, on adding them we get→10 + 1 + 4 + 13 + 3 = 31
On dividing 31 by 7, we get 3 as a remainder. So, the number of odd days = 3
3 is the code for Wednesday.
So, the 10th of May 1713 will be Wednesday. Hence, the third option is correct.

Q5. Directions: What was the day of the week on the 5th of December 2015?

A) Saturday (Correct)

B) Friday

C) Thursday

D) Wednesday

Solution: Use the month, century, and day code for the calendar and apply the formula.

The formula used to calculate the day is as follows:
{Date + (Month code) + (Century code) + (last two digit of the year) + ((last two digit of the year/4)'s Quotient)}/7

Code for months -

Code for days -

Code for Centuries -

For 5th of December 2015 -
Date = 5
Code for the month (December) = 5
Code for the century (2000) = 6
(2015 = 2000 + 15; For 2000 it will be 6 and for 15 it will be 15)
And, 15 ÷ 4 = 3(Quotient)
Now, on adding them we get→5 + 5 + 6 + 15 + 3 = 34
On dividing 34 by 7, we get 6 as a remainder. So, the number of odd days = 6
6 is the code for Saturday.
So, the 5th of December 2015 will be Saturday. Hence, the first option is correct.

Practice Calendar Reasoning Questions For Repeating Year in a Calendar

Q1. The calendar of the year 1990 will be equal to which of the following years?

A) 1997

B) 1996

C) 2000

D) 2001 (Correct)

Solution: The given year 1990 is an ordinary year, so, the year which will be equal to the year 1990 will also be an ordinary year.

From the options, the years 1996 and 2000 are leap years. Thus, the remaining years from the options are 1997 and 2001. We will now calculate the odd number of days between 1990 and these years respectively and check the divisibility by 7.

The number of odd days between 1990 and 1997 = 7 + 2 = 9 (On dividing 9 by 7, we will get the remainder of 2.) This is not the equivalent year.

The number of odd days between 1990 and 2001 = 11 + 3 = 14 (On dividing 14 by 7, we will get the remainder of 0.) This is the correct answer.

So, the year that is equal to the year 1990 is 2001. Hence, the fourth option is correct.

Q2. The calendar of the year 2019 will be equal to which of the following years?

A) 2024

B) 2025

C) 2028

D) 2030 (Correct)

Solution: The given year 2019 is an ordinary year, so, the year which will be equal to the year 2019 will also be an ordinary year.

From the options, the years 2024 and 2028 are leap years. Thus, the remaining years from the options are 2025 and 2030. So, the year equal to 2019 will be either 2025 or 2030. We will now calculate the odd number of days between 2019 and these years respectively and check the divisibility by 7.

The number of odd days between 2019 and 2025 = 6 + 2 = 8 (On dividing 8 by 7, we will get the remainder of 1.) This is not the equivalent year.

The number of odd days between 2019 and 2030 = 11 + 3 = 14 (On dividing 14 by 7, we will get the remainder of 0.) This is the correct answer.

So, the year that is equal to the year 2019 is 2030. Hence, the fourth option is correct.

Q3. The calendar of the year 2005 will be equal to which of the following years?

A) 2016

B) 2011 (Correct)

C) 2010

D) 2008

Solution: The given year 2005 is an ordinary year, so, the year which will be equal to the year 2005 will also be an ordinary year.

From the options, the years 2016 and 2008 are leap years. Thus, the remaining years from the options are 2010 and 2011. So, the year equal to 2005 will be either 2011 or 2010. We will now calculate the odd number of days between 2005 and these years respectively and check the divisibility by 7.

The number of odd days between 2005 and 2011 = 6 + 1 = 7 (On dividing 7 by 7, we will get the remainder of 0.) This is the correct answer.

The number of odd days between 2005 and 2010 = 5 + 1 = 6 (On dividing 6 by 7, we will get the remainder of 6.) This is not the equivalent year.

So, the year that is equal to the year 2005 is 2011. Hence, the second option is correct.

Q4. The calendar of the year 2017 will be equal to which of the following years?

A) 2028

B) 2024

C) 2023 (Correct)

D) 2025

Solution: The given year 2017 is an ordinary year, so, the year which will be equal to the year 2017 will also be an ordinary year.

From the options, the years 2028 and 2024 are leap years. Thus, the remaining years from the options are 2023 and 2025. So, the year equal to 2017 will be either 2023 or 2025. We will now calculate the odd number of days between 2017 and these years respectively and check the divisibility by 7.

The number of odd days between 2017 and 2023 = 6 + 1 = 7 (On dividing 7 by 7, we will get the remainder of 0.) This is the correct answer.

The number of odd days between 2017 and 2025 = 8 + 2 = 10 (On dividing 10 by 7, we will get the remainder of 3.) This is not the equivalent year.

So, the year that is equal to the year 2017 is 2023. Hence, the third option is correct.

Q5. The calendar of the year 1997 will be equal to which of the following years?

A) 2008

B) 2003 (Correct)

C) 2004

D) 2005

Solution: The given year 1997 is an ordinary year, so, the year which will be equal to the year 1997 will also be an ordinary year.

From the options, the years 2008 and 2004 are leap years. Thus, the remaining years from the options are 2003 and 2005. So, the year equal to 1997 will be either 2003 or 2005. We will now calculate the odd number of days between 1997 and these years respectively and check the divisibility by 7.

The number of odd days between 1997 and 2003 = 6 + 1 = 7 (On dividing 7 by 7, we will get the remainder of 0.) This is the correct answer.

The number of odd days between 1997 and 2005 = 8 + 2 = 10 (On dividing 10 by 7, we will get the remainder of 3.) This is not the equivalent year.

So, the year that is equal to the year 1997 is 2003. Hence, the second option is correct.

Practice Calendar Reasoning Questions For Reference Calendar

Q1. Directions: If the 2nd of March 2006 was Thursday, what day of the week was it on the 2nd of March 2022?

A) Friday

B) Wednesday (Correct)

C) Thursday

D) Monday

Solution: The 2nd of March 2006 was Thursday.

So, the total number of days from 2nd of March 2006 to 2nd of March 2022 -

(16 × 365) + 4 = 5840 + 4 = 5844

(4 days are being added as 2008, 2012, 2016, and 2020 are leap years.)

Now, on dividing 5844 by 7, we get 6 as a remainder.

So, the number of odd days = 6

So, Thursday + 6 days = Wednesday

The 2nd of March 2022 will be Wednesday. Hence, the second option is correct.

Q2. Directions: If the 9th of June 2010 was Wednesday, what day of the week was it on the 9th of June 2018?

A) Saturday (Correct)

B) Monday

C) Friday

D) Thursday

Solution: The 9th of June 2010 was Wednesday.

So, the total number of days from the 9th of June 2010 to the 9th of June 2018 -

(8 × 365) + 2 = 2920 + 2 = 2922

(2 days are being added as 2012, and 2016 are leap years.)

Now, on dividing 2922 by 7, we get 3 as a remainder.

So, the number of odd days = 3

So, Wednesday + 3 days = Saturday

The 9th of June 2018 will be Saturday. Hence, the first option is correct.

Q3. Directions: If the 17th of October 2011 was Monday, what day of the week was it on the 17th of October 2022?

A) Friday

B) Sunday

C) Monday (Correct)

D) Saturday

Solution: The 17th of October 2011 was Monday.

So, the total number of days from the 17th of October 2011 to the 17th of October 2022 -

(11 × 365) + 3 = 4015 + 3 = 4018

(3 days are being added as 2012, 2016, and 2020 are leap years.)

Now, on dividing 4018 by 7, we get 0 as a remainder.

So, the number of odd days = 0

So, Monday + 0 days = Monday

The 17th of October 2022 will be Monday. Hence, the third option is correct.

Q4. Directions: If the 30th of March 2014 was Sunday, what day of the week was it on the 31st of December 2014?

A) Friday

B) Sunday

C) Tuesday

D) Wednesday (Correct)

Solution: The 30th of March 2014 was Sunday.

The total number of days from the 30th of March 2014 to the 31st of December 2014 = 1 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31 = 276

Divide 276 by 7 = 276/7 = 39 (Quotient), and 3 (Remainder)

Total number of odd days = 3

So, on 31 December 2014, it will be Sunday + 3 = Wednesday

The 31st of December 2014 will be Wednesday. Hence, the fourth option is correct.

Q5. Directions: If the 15th of September 2022 was Thursday, what day of the week was it on the 25th of November 2022?

A) Friday (Correct)

B) Monday

C) Tuesday

D) Saturday

Solution: The 15th of September 2022 was Thursday.

The total number of days from the 15th of September 2022 to the 25th of November 2022 = 15 + 31 + 25 = 71

Divide 71 by 7 = 71/7 = 10 (Quotient), and 1 (Remainder)

Total number of odd days = 1

So, on 25 November 2022, it will be Thursday + 1 = Friday

The 25th of November 2022 will be Friday. Hence, the first option is correct.

Q6. Directions: If the 13th of January 2019 was Sunday, what day of the week was it on the 23rd of May 2019?

A) Friday

B) Monday

C) Sunday

D) Thursday (Correct)

Solution: The 13th of January 2019 was Sunday.

The total number of days from the 13th of January 2019 to the 23rd of May 2019 = 18 + 28 + 31 + 30 + 23 = 130

Divide 130 by 7 = 130/7 = 18 (Quotient), and 4 (Remainder)

Total number of odd days = 4

So, on 23 May 2019, it will be Sunday + 4 = Thursday

The 23rd of May 2019 will be Thursday. Hence, the fourth option is correct.

Q7. Directions: If the 14th of April 2019 was Sunday, what day of the week was it on the 30th of March 2024?

A) Friday

B) Saturday (Correct)

C) Tuesday

D) Sunday

Solution: The 14th of April 2019 was Sunday.

Remaining number of days in 2019 = 16 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31 = 261

Divide 261 by 7 = 261/7 = 37 (Quotient), and 2 (Remainder)

Odd days in 2019 = 2

Similarly, odd days from 2020 to 2023 -

2020 → 2; 2021 → 1; 2022 → 1; 2023 → 1

Odd days from 2020 to 2023 = 5

Now, the number of odd days in 2024 till 30 March → 3 + 1 + 2 = 6

Total number of odd days = 2 + 5 + 6 = 13

Divide 13 by 7; Remainder = 6, i.e., the required number of odd days = 6

So, on 30 March 2024, it will be Sunday + 6 = Saturday

The 30th of March 2024 will be Saturday. Hence, the second option is correct.

Q8. Directions: If the 4th of February 2012 was Saturday, what day of the week was it on the 16th of April 2016?

A) Tuesday

B) Monday

C) Saturday (Correct)

D) Sunday

Solution: The 4th of February 2012 was Saturday.

Remaining number of days in 2012 = 25 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31 = 331

Divide 331 by 7 = 331/7 = 47 (Quotient), and 2 (Remainder)

Odd days in 2012 = 2

Similarly, odd days from 2013 to 2015 -

2013 → 1; 2014 → 1; 2015 → 1

Odd days from 2013 to 2016 = 3

Now, the number of odd days in 2016 till 16 April → 3 + 1 + 3 + 2 = 9

Total number of odd days = 2 + 3 + 9 = 14

Divide 14 by 7; Remainder = 0, i.e., the required number of odd days = 0

So, on 16 April 2016, it will be Saturday + 0 = Saturday

The 16th of April 2016 will be Saturday. Hence, the third option is correct.

Q9. Directions: If the 17th of May 2005 was Tuesday, what day of the week was it on the 23rd of July 2012?

A) Tuesday

B) Monday (Correct)

C) Saturday

D) Sunday

Solution: The 17th of May 2005 was Tuesday.

Remaining number of days in 2005 = 14 + 30 + 31 + 31 + 30 + 31 + 30 + 31 = 228

Divide 228 by 7 = 228/7 = 32 (Quotient), and 4 (Remainder)

Odd days in 2005 = 4

Similarly, odd days from 2006 to 2011 -

2006 → 1; 2007 → 1; 2008 → 2; 2009 → 1; 2010 → 1; 2011 → 1

Odd days from 2006 to 2011 = 7

Now, the number of odd days in 2012 till 23 July → 3 + 1 + 3 + 2 + 3 + 2 + 2 = 16

Total number of odd days = 4 + 7 + 16 = 27

Divide 27 by 7; Remainder = 6, i.e., the required number of odd days = 6

So, on 23 July 2012, it will be Tuesday + 6 = Monday

The 23rd of July 2012 will be Monday. Hence, the second option is correct.

Q10. Directions: If the 21st of June 2021 was Monday, what day of the week was it on the 11th of June 2023?

A) Tuesday

B) Monday

C) Saturday

D) Sunday (Correct)

Solution: The 21st of June 2021 was Monday.

Remaining number of days in 2021 = 9 + 31 + 31 + 30 + 31 + 30 + 31 = 193

Divide 193 by 7 = 193/7 = 27 (Quotient), and 4 (Remainder)

Odd days in 2021 = 4

Similarly, odd days in 2022 = 1

Now, the number of odd days in 2023 till 11 June → 3 + 0 + 3 + 2 + 3 + 4 = 15

Total number of odd days = 4 + 1 + 15 = 20

Divide 20 by 7; Remainder = 6, i.e., the required number of odd days = 6

So, on 11 June 2023, it will be Monday + 6 = Sunday

The 11th of June 2023 will be Sunday. Hence, the fourth option is correct.

Practice Calendar Reasoning Questions For Finding the Day/Date of the Week in a Calendar

Q1. Directions: Today, August 4th, is Tuesday. What day will it be on the 26th of this month?

A) Thursday

B) Tuesday

C) Wednesday (Correct)

D) Monday

Solution: 4th August → Tuesday, 26th August → (?)

The total number of days = 22

Divide 22 by 7, remainder obtained = 1

So, 26th August → Tuesday + 1 = Wednesday

Hence, the third option is correct.

Q2. Directions: If it was Sunday on the 30th of a month, then what day was it on the 1st of that month?

A) Saturday (Correct)

B) Friday

C) Sunday

D) Monday

Solution: 30th of a month → Sunday, 1st of the same month → (?)

The total number of days = 29

Divide 29 by 7, remainder obtained = 1

So, 1st of the month → Sunday - 1 = Saturday

Hence, the first option is correct.

Q3. Directions: If it was Tuesday on the 23rd of a month, then what day was it on the 7th of that month?

A) Tuesday

B) Wednesday

C) Sunday (Correct)

D) Thursday

Solution: 23rd of a month → Tuesday, 7th of the same month → (?)

The total number of days = 16

Divide 16 by 7, remainder obtained = 2

So, 7th of the month → Tuesday - 2 = Sunday

Hence, the third option is correct.

Q4. Directions: If today is Friday then what day will it be after 60 weeks?

A) Monday

B) Tuesday

C) Friday (Correct)

D) Saturday

Solution: Today is Friday.

60 weeks means (60 × 7) days

On dividing (60 × 7) by 7, the remainder obtained = 0

So, after 60 weeks. It will be Friday. Hence, the third option is correct.

Q5. Directions: If today is Saturday then what day will it be after 43 weeks?

A) Sunday

B) Saturday (Correct)

C) Monday

D) Friday

Solution: Today is Saturday.

43 weeks means (43 × 7) days

On dividing (43 × 7) by 7, the remainder obtained = 0

So, after 43 weeks, it will be Saturday. Hence, the second option is correct.

Q6. Directions: If 3 days before yesterday was Sunday then what will be 3 days after tomorrow?

A) Sunday

B) Monday (Correct)

C) Tuesday

D) Wednesday

Solution: 3 days before yesterday was Sunday.

Yesterday, it was (Sunday + 3) = Wednesday

Today, it is Thursday and tomorrow, it will be Friday.

So, 3 days after tomorrow, it will be (Friday + 3) = Monday

So, 3 days after tomorrow, it will be Monday. Hence, the second option is correct.

Q7. Directions: If 5 days before yesterday was Monday then what will be 8 days after tomorrow?

A) Tuesday (Correct)

B) Sunday

C) Monday

D) Wednesday

Solution: 5 days before yesterday was Monday.

Yesterday, it was (Monday + 5) = Saturday

Today, it is Sunday and tomorrow, it will be Monday.

So, 8 days after tomorrow, it will be (Monday + 8) = Tuesday

So, 8 days after tomorrow, it will be Tuesday. Hence, the first option is correct.

Q8. Directions: If it is Saturday after 6 days. What was 9 days before yesterday?

A) Monday

B) Friday

C) Wednesday

D) Thursday (Correct)

Solution: After 6 days, it is Saturday.

Today, it is (Saturday - 6) = Sunday and yesterday, it was Saturday.

So, 9 days before yesterday, it was (Saturday - 9) = Thursday

So, 9 days before yesterday, it was Thursday. Hence, the fourth option is correct.

Q9. Directions: Aniket celebrates his birthday on Sunday, 6 March 2004. When will he again celebrate his birthday on Sunday?

A) 2008

B) 2009

C) 2011

D) 2010 (Correct)

Solution: 6 March 2004 → Sunday

6 March 2005 → Monday

6 March 2006 → Tuesday

6 March 2007 → Wednesday

6 March 2008 → Friday (Leap year)

6 March 2009 → Saturday

6 March 2010 → Sunday

So, on 6 March 2010, he will again celebrate his birthday on Sunday. Hence, the fourth option is correct.

Q10. Directions: Diya remembers that her brother’s birthday comes after the 17th but before the 21st of February, but her brother remembers that it is after the 19th and before the 24th of February. When does her brother’s birthday come?

A) 22 February

B) 21 February

C) 18 February

D) 20 February (Correct)

Solution: According to Diya, her brother’s birthday is on one of the following dates:

18 February, 19 February, 20 February

But, according to Diya’s brother, her brother’s birthday is on one of the following dates:

20 February, 21 February, 22 February, 23 February

From the above, it appears that February 20th is a common date in both sets of information.

So, Diya’s brother’s birthday comes on 20 February. Hence, the fourth option is correct.

For practice you must attempt calendar reasoning mock test and download the e-book of calendar reasoning questions pdf below:
Calendar Questions with Solutions PDF

Calendar Reasoning Questions with Answers for VITEEE/ CUET

Q-1) Directions: If 31 January 2022 was a Monday, what was the day on the same date 100 years ago?

1) Wednesday

2) Sunday

3) Tuesday

4) Monday

Hint: Calculate the number of days and divide the number by 7 to get the answer.

Solution:

31 January 2022 is a Monday.

The number of odd days in a leap year = 2

The number of odd days in a non-leap year = 1

Number of leap years between 2022 and 1922→25

Number of non-leap years between 2022 and 1922→75

Let's calculate the number of odd days→(25 × 2) + (75 × 1) = 50 + 75 = 125

Dividing 125 by 7, the remainder is 6.

Thus, 31 Jan 1922 will be = Monday – 6 = Tuesday

So, 31 Jan 1922 is Tuesday. Hence, the third option is correct.

Q-2) Directions: Raman remembers that the examination is after the 15th of May but before the 18th of May, while Deep remembers that the examination is before the 21st of May but after the 16th of May. On which date of May is the examination?

1) 17th

2) 18th

3) 19th

4) 20th

Hint: Find the dates on which the two time frames overlap.

Solution:

Raman remembers that the examination is after 15th May but before 18th May. Hence, the possible dates are 16th May and 17th May.

Deep remembers that the examination is before 21st May but after 16th May. Hence, the possible dates are 17th May, 18th May, 19th May, and 20th May.

The only date which is common to both statements is the 17th of May. Hence, the first option is correct.

Calendar Reasoning Questions with Answers for CAT/ APICET/ SUAT/ JIPMAT

Q-1) Directions: If the 16th of January 2015 was Friday, then what was the day of the week on the 16th of January 2010?

1) Monday

2) Tuesday

3) Saturday

4) Sunday

Hint: Calculate the total number of days and divide it by 7.

Solution:

Let's calculate the total number of days from the 16th of January 2010 to the 16th of January 2015 –

5 × 365 = 1825 + 1 = 1826

(1 day is being added as 2012 is a leap year.)

Now, on dividing 1826 by 7, we get 6 as a remainder.

So, Friday – 6 days = Saturday

So, the 16th of January 2010 is Saturday. Hence, the third option is correct.

Q-2) Directions: What was the day of the week on 05th of April 1234?

1) Tuesday

2) Sunday

3) Monday

4) Wednesday

Hint: Use the month, century, and day code for the calendar and apply the formula.

Solution:

Code for months -

Code for days –

Code for centuries –

For, 5th of April 1234; –

Formula: {Date + (Month code) + (Century code) + ( last two digit of the year) + ((last two digit of the year / 4)'s Quotient)} / 7

Date = 5

Code for the month (April) = 6

Code for the century (1200) = 6

(1234 = 1200 + 34; For 1200 it will be 6 and for 34 it will be 34)

And, 34 ÷ 4 = 8(Quotient)

Now, on adding them we get→5 + 6 + 6 + 34 + 8 = 59

On dividing 59 by 7, we get 3 as a remainder.

3 is the code for Wednesday.

So, the 05th of April 1234 will be Wednesday. Hence, the fourth option is correct.

Calendar Reasoning Questions with Answers for SSC/ RRB exams

Q-1) Directions: If 30 April 1983 was a Saturday, then what was the day of the week on 13 August 1989?

1) Thursday

2) Sunday

3) Monday

4) Friday

Hint: Calculate the total number of days and divide it by 7.

Solution:

Calculate the number of days between April 30, 1983 and August 13, 1989.

Number of remaining days in 1983 ⇒ May = 31; June = 30; July =31; August = 31; September = 30; October = 31; November = 30; December = 31

⇒ 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31 = 245

From 1984 to 1988 ⇒ (5 × 365) + 2 = 1825 + 2 = 1827

(2 days are being added as there are two leap years 1984 and 1988)

From Jan 1, 1989 to August 13, 1989 ⇒ Jan = 31; Feb = 28; March = 31; April = 30; May = 31; June = 30; July = 31; August = 13

⇒ 31 + 28 + 31 + 30 + 31 + 30 + 31 + 13 = 225

Total number of days = 245 + 1827 + 225 = 2297

On dividing 2297 by 7, we get 1 as the remainder. So, add 1 day to Saturday, we will get Sunday.

So, 13th August 1989 will be Sunday. Hence, the second option is correct.

Q-2) Directions: If the day before yesterday was Thursday, when will Sunday be?

1) Tomorrow

2) Day after tomorrow

3) Today

4) Two days after today

Hint: If the day before yesterday was Thursday, think about what day yesterday would be, and then calculate how many days are left until you reach Sunday.

Solution:

If the day before yesterday was Thursday, then yesterday was Friday and today is Saturday.

So, tomorrow will be Sunday.

Therefore, Sunday will be tomorrow. Hence, the first option is correct.

For non verbal reasoning follow the below topics:

About the Faculty
Tanu Gupta, with over a decade of experience as a reasoning faculty, specializes in preparing students for various entrance examinations and career development. Her extensive work with multiple educational platforms and institutions has honed her expertise in logical and analytical thinking. Her dedication to innovative teaching methods ensures these articles provide practical insights and expert guidance.