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

What is Calendar in Reasoning? Definition, Concepts, Leap Year & Odd Days Explained

Understanding the calendar is the foundation of solving calendar reasoning questions, especially topics like day and date problems, odd days concept, and calendar tricks for competitive exams.

Calendar Definition in Reasoning with Easy Explanation

A calendar is a systematic chart that shows the arrangement of days, weeks, and months in a year. It helps us determine the exact date, day, month, and year, which is essential for solving calendar questions in exams like SSC, Banking, and other competitive tests.

Basic Structure of Calendar (Days, Weeks & Months)

• A calendar represents the proper arrangement of days and months in a year

• It helps in solving calendar reasoning questions related to dates and weekdays

• Widely used in day and date questions in competitive exams

Number of Days in a Year (Ordinary vs Leap Year)

• A year has either 365 days or 366 days

• A 365-day year is called an ordinary year

• A 366-day year is known as a leap year

• These concepts are important for solving calendar reasoning questions with answers

Leap Year Rules in Calendar Reasoning

• If the last two digits of a year are divisible by 4, it is generally a leap year

• This rule is commonly used in calendar questions and shortcut methods

Century Year Concept in Calendar Questions

• A century year is a year ending with 00 (e.g., 1600, 1700, 1800, 2000)

• These are frequently asked in calendar reasoning questions for SSC and Banking exams

Leap Year Rule for Century Years (Important Trick)

• Not all century years are leap years

• A century year is a leap year only if it is divisible by 400

• This is a key concept in calendar tricks and formulas

Odd Days Concept in Calendar Reasoning

• When total days in a year are divided by 7, the remainder is called odd days

• Odd days help in solving day and date questions quickly

Odd Days in Ordinary Year

• $365 \div 7 = 52$ weeks and 1 odd day

• So, an ordinary year has 1 odd day

Odd Days in Leap Year

• $366 \div 7 = 52$ weeks and 2 odd days

• So, a leap year has 2 odd days

Types of Calendar Reasoning Questions Asked in SSC, Banking & Competitive Exams

Understanding the different types of calendar reasoning questions helps you solve day and date problems quickly using the right concepts, formulas, and calendar tricks.

Find Odd Days in Calendar Reasoning Questions

Questions based on odd days concept in calendar require you to calculate the number of odd days in a given period. These are the most common calendar questions and form the base for solving advanced problems.

Repeating Year Questions in Calendar Reasoning

In these calendar reasoning questions, you need to identify which year has the same calendar as a given year. These are also known as repeating calendar questions and are frequently asked in competitive exams.

Reference Calendar Based Questions

These types of calendar questions provide a reference day or date (e.g., 1st January is Monday), and you need to find the day of another date using logical calculation and calendar tricks.

Day and Date Problems in Calendar Reasoning

These are the most important day and date questions, where you are asked to find the exact day for a given date or vice versa. They are commonly seen in SSC, Banking, and other reasoning exams.

Overview of Calendar Question Types with Examples

Each type of calendar reasoning question follows a specific approach and shortcut. In the next sections, we will discuss these types in detail with examples and calendar reasoning questions with answers to help you master the topic easily.

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 questions, the candidate must find the day of the week for a given date using the day of the week of another date with the same day and month but different years. The concept relies on counting odd days: an ordinary year has 1 odd day, and a leap year has 2 odd days.

There are two methods to solve this: either add the odd days year by year to the given day, or calculate the total number of years between the dates, find the total odd days, and add that to the known day to determine the day of the week for the required 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

Top 5 Tips to Solve the Calendar Questions

The candidates must follow the tips below to solve calendar reasoning questions that come in various competitive exams and entrance exams.

Tip 1: The candidates must remember that in an ordinary year, the first day of the month is the same for:

(1) January and October

(2) April and July

(3) February, March and November

(4) September and December.

Tip 2: The candidates must remember that in a leap year, the first day of the month is the same for:

(1) January, April, July.

(2) February and August.

(3) March and November.

(4) September and December.

Tip 3: The candidates must remember that the first day of the century must be either Monday, Wednesday, Friday or Saturday whereas the last day of the century cannot be Tuesday, Thursday or Saturday.

Tip 4: The calendar of an ordinary year repeats after 6 or 11 years whereas the calendar of a leap year repeats in 28 years. Therefore while solving these types of questions the candidates must remember this tip and should eliminate the options to mark the answer.

Tip 5: The candidates must remember the following points:

(1) In 400 years there are 0 odd days.

(2) In 300 years there is 1 odd day.

(3) In 200 years there are 3 odd days.

(4) In 100 years there are 5 odd days.

Best Books for Calendar Reasoning & Logical Reasoning Preparation

This section lists the most recommended books for mastering calendar reasoning questions, along with their key features, exam relevance, and how they help improve your speed and accuracy in day and date problems.

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.

Verbal Reasoning Topics

Verbal reasoning tests your ability to understand and analyze information presented in words or numbers. Key topics include distance and direction, arithmetical problems, seating arrangement, blood relation, missing number, series, Venn diagrams, syllogism, binary logic, critical reasoning, clock reasoning, and mathematical operations. These are essential areas for competitive exams preparation. You can click below for clock and calendar reasoning questions with solutions and other important verbal reasoning topics.

Important Calendar Formulas and Tricks for Quick Solving (Day & Date Questions Made Easy)

Mastering the right calendar formulas and tricks is the key to solving calendar reasoning questions quickly and accurately. With concepts like odd days, leap year rules, and smart shortcuts, you can easily crack even tricky day and date questions in exams like SSC, Banking, and RRB.

Odd Days Formula in Calendar Reasoning (Core Concept for Day Calculation)

The odd days concept in calendar reasoning is the most important foundation for solving calendar questions.

• Odd days = Remainder when total number of days is divided by 7

• $1 \text{ odd day} = 1 \text{ day shift}$

• $7 \text{ odd days} = 0$ (complete week cycle)

Standard Values to Remember:

• Ordinary year (365 days) → $1$ odd day

• Leap year (366 days) → $2$ odd days

Usage:
This formula is widely used in day and date questions, helping you calculate the exact weekday by counting total odd days.

Leap Year Shortcut Tricks for Calendar Questions

Understanding leap years is crucial for solving calendar reasoning questions with answers.

Quick Rules:

• If a year is divisible by 4 → Leap year

• Century year (ending with 00) must be divisible by 400 → Leap year

Examples:

• 2024 → Leap year

• 1900 → Not a leap year

• 2000 → Leap year

Exam Tip:
Leap years contribute 2 odd days, making them important in calendar tricks to find day quickly.

Calendar Tricks to Find Day Quickly (Fast Calculation Techniques)

These calendar tricks help you solve day and date problems within seconds:

• Use reference day (e.g., 1 Jan as Monday)

• Count total odd days between given dates

• Use month codes or memorize odd days per month

• Break long durations into years + months + days

These tricks are highly useful in calendar reasoning questions for SSC CGL, Banking, and RRB exams.

Step-by-Step Approach to Solve Calendar Reasoning Questions

A structured method ensures accuracy while solving calendar questions.

Easy Method to Solve Day and Date Questions

Follow these steps for solving day and date questions:

1. Identify the reference date (if given)

2. Calculate total number of odd days

3. Add odd days to the reference day

4. Find the final day using modulo 7 logic

This is the most reliable approach for solving calendar reasoning questions step by step.

Shortcut Techniques for Calendar Questions (Time-Saving Tricks)

To improve speed in exams, use these shortcut techniques for calendar questions:

• Memorize odd days of months (30 → 2 odd days, 31 → 3 odd days)

• Use pattern repetition (calendar repeats after certain years)

• Skip unnecessary calculations using direct formulas

• Practice previous year calendar reasoning questions with answers

These shortcuts help you solve calendar questions quickly while maintaining accuracy in competitive exams.

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.

Practice Questions on Calendar PDF Download

For practice, you must attempt the calendar reasoning mock test and download the e-book of calendar reasoning questions PDF below:

Download Now: 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 the 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.

Non-Verbal Reasoning topics

Non-verbal reasoning focuses on solving problems using visual and spatial reasoning. Important topics include analogy, figure series, classification, embedded figures, cube and dice reasoning, figure counting, completion of figures, paper folding, and mirror and water images. These topics are commonly tested in competitive exams. For non-verbal reasoning, follow the topics below:

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.