Practice Clocks and Calendars Questions and Answers

Calendar Question & Answer

Calendar is a part of Verbal reasoning and forms an important portion of the syllabus of various government and bank exams like SSC, SBI PO, IBPS SO, IBPS PO, RBI Grade B, SBI Clerk, RRB Assistant Loco Pilot among many others.

What is a Calendar?

A calendar is a systematic arrangement of day, week and month in a defined pattern with which we can easily recognise the required date, month or week of a particular day.

A fun fact about calendars is that, the most popular calendar in the world is the Gregorian calendar. It also happens to be the Indian National Calendar which was adopted by India on March 22, 1957.

The Indian National calendar consists of the following-

• Day
• Week
• Month
• Year
• Date

Structure of a Calendar

Day

A day is the seventh part of a week and has 24 hours. It is the smallest unit of the calendar.

Week

A week is the 52nd part of a year. It is grouped as 7 days.

• Sunday
• Monday
• Tuesday
• Wednesday
• Thursday
• Friday
• Saturday

Month

A month is the 12th part of a year. It has 28/29/30/31 days.

Year

A year is the 100th part of a century. It is the time taken by the Earth to make one revolution around the Sun. A year has 12 months.

• January
• February
• March
• April
• May
• June
• July
• August
• September
• October
• November
• December

Century

A block of 100 years is called a century. For example, each one of the year 1100,1800, 2000, 2100 and 2900 are century years.

Ordinary Year

• An ordinary year has 365 days (52 weeks + 1 odd day).
• Such years are not divisible by 4. For example, years like 2001, 2002, 2003, 2005, 2006, 2007, 2009, 2010 and 2011 etc.
• Ordinary years in the form of centuries are not exactly divisible by 400. For example, 100, 200, 500, 600, 700 and 900 etc.
• In an ordinary year, the first and the last days of the year are the same. For instance, in an ordinary year, if January 01 falls on Monday, then December 31 will also be on Monday.
• In any two consecutive ordinary years, the date of the next year will be one day ahead of the same date of the previous year. For instance, if March 02, 2010 is Tuesday, then March 02 2011 will be Wednesday.

Leap Year

• A leap year is a year which has 366 days (52 weeks + 2 days).
• Such years are exactly divisible by 4. For instance, 2004, 2008, 2012, 2016 etc.
• Leap years in the form of a century are also exactly divisible by 400. For example, 400, 800, 1200, 1600, 2000 etc.
• If a leap year comes immediately after an ordinary year, then the day on a certain date of the leap year will be-
  • 1 day ahead of the same date of the ordinary year upto February.
  • 2 days ahead of the same date of the ordinary year from March to December.
• If an ordinary year comes immediately after a leap year, then the day on a certain date of the ordinary year will be-
  • 2 days ahead of the same date of the leap year upto February.
  • 1 day ahead of the same date of the leap year from March to December.

Concept of Odd Day

Extra days apart from the complete weeks in a given period are called odd days.

Let’s understand this in detail-

• A period of 7 days= 7 + 0 extra day= 0 odd day
• A period of 8 days= 7 + 1 extra day= 1 odd day
• A period of 9 days= 7 + 2 extra days= 2 odd days
• A period of 10 days= 7 + 3 extra days= 3 odd days
• A period of 11 days= 7 + 4 extra days= 4 odd days
• A period of 12 days= 7 + 5 extra days= 5 odd days
• A period of 13 days= 7 + 6 extra days= 6 odd days
• A period of 14 days= 7 + 7= 2 complete weeks= 0 odd days

Formula to calculate odd number of days

Divide the period by 7, i.e (Period/7)

• If Period/7 does not leave any remainder, then there are zero odd days.
• However, if Period/7 leaves any remainder, then that remainder will be the number of odd days.

Let us look at an example-

Q1) Find the number of odd days for a 28 days period.

Solution:

As per the formula,

Number of odd days = Period/7= 28/7= 0

Explanation:

A period of 28 days= (4 x 7 + 0) days= 4 complete weeks + 0 odd day

As 28 is exactly divisible by 7, therefore, it leaves no remainder.

Q2) Find the number of odd days for a 30 days period.

Solution:

Number of odd days= 30/4= 2 odd days

Explanation:

Here, 30 days= (4 x 7 + 2) days= 4 complete weeks + 2 odd days

As 30 is not exactly divisible by 7, it leaves a remainder of 2.

Note-

• The months with 31 days contain (4 x 7 + 3) days= 3 odd days
• The months with 30 days contain (4 x 7 + 2) days= 2 odd days

The odd days for each month have been given in the table below-

Counting of Odd days

Ordinary year

Number of days in an Ordinary year= 365 days

Therefore, odd days in a year= 365/7= 1 odd day

Which means, 365 days= 52 x 7 + 1= 1 odd day

So, an ordinary year has 1 odd day.

Leap year

Number of days in a leap year= 366 days

So, odd days in a leap year= 366/7= 2 odd days

Which means, 366 days= 52 x 7 + 2= 2 odd days

Therefore, a leap year has 2 odd days.

Century

100 years= 76 ordinary years + 24 leap years

We know that an ordinary year has 1 odd day and a leap year has 2 odd days.

Hence, 76 ordinary years will have 76 odd days and 24 leap years will have 24 x 2 = 48 odd days. Adding both the results we get 76+48 = 124 odd days in total.

Therefore, number of odd days in 100 years= 76 odd days of 76 ordinary years + (24 x2) odd days of leap years= 76 odd days + 48 odd days= 124 odd days

Dividing the total odd days 124 by 7 gives the quotient as 17 and a remainder as 5. This indicates that 124 days had 17 weeks and 5 odd days.

Dividing 124/7= 5 odd days

124= 17 x 7 + 5 odd days

So, odd days in 100 years= 5 days.

Types of questions covered under this chapter

The following types of questions can be covered under this topic-

• Finding the day on a particular date when the reference day is given.
• Finding the day on a particular date when the reference day is not given.
• Finding a week on the basis of another week day.

Let’s look at all the types in detail-

Finding the day on a particular date when the reference day is given.

In these types of questions, a particular day and its date is given. On the basis of that the candidates have to find the day on a given day.

For example-

Q1) If March 17, 2008 was Monday, what was April 01, 2012?

Solution:

The total number of odd days from March 17, 2008 to March 17, 2012.

Since March 17, 2008 was Monday and March 17, 2012 is 5 days more than Monday.

Then, adding 5 odd days to Monday, we get Saturday.

Hence, March 17 to April 01 we have 15 days.

Saturday+15=Sunday.

Adding 15 days or (15 = 14+1) to Saturday, we get the answer as Sunday.

Finding the day on a particular date when the reference day is not given.

Here, the candidates have to find out the day of the week when a reference date/day is not given.

Here, we can use the concept of odd days to find the answer.

Let’s understand with an example-

Q1) Find the day of the week on January 26, 1950.

Solution:

Odd days for 1600 years= 0

Odd days for 300 years= 1

Odd days for 49 years= (12 x 2 + 37 x 1)= 61

Odd days for 26 days of January 1950= 5

Therefore, total odd days= 0 + 1 + 61 + 5= 67 days

So, 67/7= 4 odd days

i.e., 67 days= (9 x 7 + 4) days= 9 weeks + 4 days

Since the number of odd days= 4

The required day will be= Thursday

Finding a week day on the basis of another week day.

In such questions, a week day is to be found out on the basis of some reference day of the week.

Important Concepts Regarding Days

For example-

Q1) If today is Sunday, then what day of the week will be 3 days after tomorrow?

Solution:

Today= Sunday

Tomorrow= Sunday + 1 day= Monday

Therefore, 3 days after tomorrow= Monday + 3 odd days= Thursday

Sample questions on Calendar

Q1) It was Sunday on January 01, 2006. What was the day of the week January 01, 2010?

A. Sunday

B. Saturday

C. Friday

D. Wednesday

Solution: Option C.

Explanation:

On December 31, 2005 it was Saturday.

Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.

On December 31, 2009, it was Thursday.

Thus, on January 01, 2010 it is Friday.

Q2) What was the day of the week on May 28, 2006?

A. Thursday

B. Friday

C. Saturday

D. Sunday

Solution: Option D

Explanation:

May 28, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)

Odd days in 1600 years = 0

Odd days in 400 years = 0

5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) 6 odd days

Jan. Feb. March April May

(31 + 28 + 31 + 30 + 28 ) = 148 days

148 days = (21 weeks + 1 day) 1 odd day.

Total number of odd days = (0 + 0 + 6 + 1) = 7 0 odd days.

Given day is Sunday

Q3) what will be the day of the week on August 15, 2010?

A. Sunday

B. Monday

C. Tuesday

D. Friday

Solution: Option A

Explanation:

August 15, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)

Odd days in 1600 years = 0

Odd days in 400 years = 0

9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days 4 odd days.

Jan. Feb. March April May June July Aug.

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days

227 days = (32 weeks + 3 days) 3 odd days.

Total number of odd days = (0 + 0 + 4 + 3) = 7 0 odd days.

Given day is Sunday.

Conclusion

The chapter of Calendar is slightly tricky as it involves many concepts. However, you can improve your accuracy and speed by practicing as many sample questions as possible. Candidates can expect around 3 to 5 questions on this topic.