How to master calendar problems in aptitude tests

How to master calendar problems in aptitude tests

Calendar questions assess the test-taker’s familiarity with the Gregorian calendar and thereby the ability to ascertain the day of the week of any given date.



How to Master Calendar Problems



How to Master Calendar Problems



 





Fundamentals of ascertaining the day of the week of a given date


A year is called a leap year if it has 366 days, and the year which has 365 days is called a non-leap year. Every year that is divisible by 4 is called a leap year. Any century (100 years) is a leap year only if it is a multiple of 400. The fundamental concept of odd days is used to find the day of the week of any given date. The number of odd days in a given period is the number of days more than the complete number of weeks during that period.


 

Counting of Odd days

To find out the number of odd days in a given period, divide the number of days in that period by 7. The remainder is the number of odd days.

For example,

1 Non-leap year = 365 days

Number of odd days = Remainder when divided by 7 = 1 day

 

Number of Odd days

The number of odd days for different periods is tabulated below:

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When odd days are calculated relative to the beginning of the calendar, the following codes can be used.

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Alternative Approach


The last day of February of any year is the same as the following dates – 4th April, 6th June, 8th August, 10th October and 12th December. It can be easily remembered as 4/4, 6/6, 8/8, 10/10, 12/12. It also falls on the same day as 5/9, 9/5, 7/11 and 11/7. This information can be used to determine the day of the week for a date in any month.



Solved Examples


Q.1. It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?


Solution:

On 31st December, 2005 it was Saturday.

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

 On 31st December 2009, it was Thursday.

Thus, on 1st Jan, 2010 it is Friday.


Q.2. What was the day of the week on 28th May, 2006?


Solution:

28 May, 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 n(31 +  28  +  31   +   30   +   28 ) = 148 daysn

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

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

Given day is Sunday.


Q.3. What was the day of the week on 17th June, 1998?


Solution:

17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)

Odd days in 1600 years = 0

Odd days in 300 years = (5 x 3)  1

97 years has 24 leap years + 73 ordinary years.

Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.

Jan.  Feb.   March    April    May      June n(31 +  28  +  31   +   30   +   31   +   17) = 168 daysn

 168 days = 24 weeks = 0 odd day.

Total number of odd days = (0 + 1 + 2 + 0) = 3.

Given day is Wednesday.


Q.4. What will be the day of the week 15th August, 2010?

Solution:

15th August, 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. n(31 +  28  +  31   +   30   +  31  + 30  + 31  + 15) = 227 daysn

 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.


Q.5. Today is Monday. After 61 days, it will be:


Solution:

Each day of the week is repeated after 7 days.

So, after 63 days, it will be Monday.

 After 61 days, it will be Saturday.




Q.6. If 6th March, 2005 is Monday, what was the day of the week on 6th March, 2004?


Solution:

The year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.

 The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.

Given that, 6th March, 2005 is Monday.

 6th March, 2004 is Sunday (1 day before to 6th March, 2005).


Q.7. On what dates of April, 2001 did Wednesday fall?


Solution:

We shall find the day on 1st April, 2001.

1st April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)

Odd days in 1600 years = 0

Odd days in 400 years = 0

Jan. Feb. March April

(31 + 28 + 31 + 1)  = 91 days  0 odd days.

Total number of odd days = (0 + 0 + 0) = 0

On 1st April, 2001 it was Sunday.

In April, 2001 Wednesday falls on 4th, 11th, 18th and 25th.


Q.8. The last day of a century cannot be


Solution:

100 years contain 5 odd days.

 Last day of 1st century is Friday.

200 years contain (5 x 2)  3 odd days.

 Last day of 2nd century is Wednesday.

300 years contain (5 x 3) = 15  1 odd day.

 Last day of 3rd century is Monday.

400 years contain 0 odd day.

 Last day of 4th century is Sunday.

This cycle is repeated.

 Last day of a century cannot be Tuesday or Thursday or Saturday.


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