Calendar – Aptitude Questions and Answers
‘Calendars’ is one of the most important topics for government sector entrance exams. The topic “Calendar” falls under the category of Logical Reasoning as it involves a lot of logical discussion and analysis. One can definitely expect 2 to 4 problems in the question papers of various Govt and Bank Exams. In Calendar, questions are mainly based on finding the day of the week if we are given a date. For example, we may be asked to find the day on 2 February 1981.
The concepts as well as formulas used for Calendars in the aptitude section are explained below. Students must go through the sample questions and practice them to understand the topic.
Practice Quiz:
Calendars Formulas and Concepts:
What is Calendar?
A calendar is a system used to organize time into days, weeks, and months throughout the year. It typically includes important dates, such as holidays or special events. There are various types of calendars, depending on cultural or religious practices, but many follow the same basic structure. This includes evaluating leap years, decoding the days of the week, finding the day when another day is given or not given, and matching calendars for a particular month. Understanding these concepts can help make planning and keeping track of important dates much easier.
1. Odd Days:
To determine the day of the week for a specific date, we use the concept of “odd days”. Odd days refer to the extra or remaining days in a given period that exceeded complete weeks. For example, if a month has 30 days, there are two odd days because two days exceed four complete weeks. It is important to understand this concept when working with calendars and scheduling events on specific dates.
- Finding days from dates is based on calculating the number of odd days. By odd days, we mean a number of days more than a complete number of weeks.
- For example, the number of days in a non-leap year = 365 365 mod 7 = 1 So, the number of odd days in a non-leap year = 1
- Number of days in a leap year = 366 => Number of odd days in a leap year = 366 mod 7 = 2
- Number of odd days in 100 years (76 non-leap years + 24 leap years) = [(76 x 1) + (24 x 2)] mod 7 = (76 + 48) mod 7 = 124 mod 7 = 5 days
- Number of odd days in 200 years = (2 x Number of odd days in 100 years) mod 7 = 10 mod 7 = 3
- Number of odd days in 300 years = (3 x 5) mod 7 = 1
- Number of odd days in 400 years = (4 x 5 + 1) mod 7 = 21 mod 7 = 0 Note that here, we have added 1 day extra because the 400th year would itself be a leap year.
Month |
Number of odd days |
January |
3 |
February(ordinary/leap) |
(0/1) |
March |
3 |
April |
2 |
May |
3 |
June |
2 |
July |
3 |
August |
3 |
September |
2 |
October |
3 |
November |
2 |
December |
3 |
2. Leap Year:
- To check if a non – centennial year is a leap year, we divide it by 4. If the remainder is 0, the year is a leap year. For example, 2016 mod 4 = 0. Thus, we can safely deduce that 2016 is a leap year.
- To check if a centennial year is a leap year, we divide it by 400. If the remainder is 0, the year is a leap year. For example, 1700 mod 400 = 100. So, it was not a leap year. But 1600 mod 400 = 0. Thus, we can safely deduce that 1600 was a leap year.
3. Day of the Week Related to Odd Days:
No. of days: |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
Day: |
Sun. |
Mon. |
Tues. |
Wed. |
Thurs. |
Fri. |
Sat. |
Sample Questions on Calendars:
Type 1 Problems: Finding the day when another day is given
Q1. If January 1st, 2023 is a Sunday, what day of the week will be October 31st, 2023?
Solution:
To solve this problem, we need to count the number of days between January 1st, 2023, and October 31st, 2023, and then find out what day of the week October 31st, 2023 falls on. Total number of days between January 1st, 2023, and October 31st, 2023 = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 = 304 Now, we can find out what day of the week October 31st, 2023 falls on by adding 304 days to Sunday, which is the day of the week on January 1st, 2023. 304 divided by 7 leaves a remainder of 2, which means that 304 days after Sunday is two days after Sunday, which is Tuesday. Therefore, October 31st, 2023 is on a Tuesday.
Q2. If March 1st, 2024 is a Saturday, what day of the week will be September 1st, 2024?
Solution:
To solve this problem, we need to count the number of days between March 1st, 2024, and September 1st, 2024, and then find out what day of the week September 1st, 2024 falls on. A total number of days = 31 + 30 + 31 + 30 + 31 + 31 + 1 = 185 Now, we can find out what day of the week September 1st, 2024 falls on by adding 185 days to Saturday, which is the day of the week on March 1st, 2024. 185 divided by 7 leaves a remainder of 6, which means that 185 days after Saturday is six days after Saturday, which is Friday. Therefore, September 1st, 2024 is on a Friday.
Type 2 Problems: Finding the day when another day is not given
Q3. What was the day on 14 April 2000?
Solution :
1600 will have 0 odd days. 300 years will have 1 odd day. Now, in the next 99 years, we would be having 75 non-leap years and 24 leap years. => Number of odd days = (75 x 1) + (24 x 2) = 75 + 48 = 123 mod 7 = 4 odd days Total odd days till now = 1 + 4 = 5 Number of odd days in January = 31 mod 7 = 3 Number of odd days in February (2000 is a leap year) = 29 mod 7 = 1 Number of odd days in March = 31 mod 7 = 3 Number of odd days till 14 April 2000 in the month of April= 14 mod 7 = 0 So, the total number of odd days = 5 + 3 + 1 + 3 = 12 mod 7 = 5 Thus, 14 April 2000 was Friday (odd days = 5 => Friday)
Q4. What was the day on 16 August 1947?
Solution:
1600 will have 0 odd days. 300 years will have 1 odd day. Now, in the next 46 years, we would be having 35 non-leap years and 11 leap years. => Number of odd days = (35 x 1) + (11 x 2) = 35 + 22 = 57 mod 7 = 1 odd days Total odd days till now = 1 + 1 = 2 Number of odd days in January = 31 mod 7 = 3 Number of odd days in February (1947 is a non – leap year) = 28 mod 7 = 0 Number of odd days in March = 31 mod 7 = 3 Number of odd days in April = 30 mod 7 = 2 Number of odd days in May = 31 mod 7 = 3 Number of odd days in June = 30 mod 7 = 2 Number of odd days in July = 31 mod 7 = 3 Number of odd days till 16 August 1947 = 16 mod 7 = 2 So, the total number of odd days = 2 + 3 + 0 + 3 + 2 + 3 + 2 + 3 + 2 = 20 mod 7 = 6 Thus, 16 August 1947 was Saturday (odd days = 6 => Saturday)
Related Resources:
Test your knowledge of Calendar in Quantitative Aptitude with the quiz linked below, containing numerous practice questions to help you master the topic:-
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