CALENDAR PROBLEMS




About the topic Calendar problems

Calendar problems play a major role quantitative aptitude test. There is no competitive exam without the questions from this topic. We have already learned this topic in our lower classes.Even though we have been already taught this topic in our lower classes, we need to learn some more short cuts which are being used to solve the problems in the above topic.

The only thing we have to do is, we need to apply the appropriate short cut and solve the problems in a limited time. This limited time will be one minute or less than one minute in most of the competitive exams.

Points to remember

Let the day of the week to be found on a given date. To find that, we are going to use the concept of odd days.
Odd Days :
                  In a given period, the number of days more than the complete weeks are called odd days.
Leap Year :
(i)Every year is divisible by 4 is a leap year, if it is not a century
(ii)Every 4th century is a leap year and no other century is a leap year.

Note:A leap years has 366 days and the month February has 29 days
Ordinary Year:
                  The year which is not a leap year is called an ordinary year.An ordinary year has 365 days and the month February has 28 days.
Counting of Odd Days :
(i)1 leap year has 366 days = 52 weeks + 2 days.
     So, 1 leap year has 2 odd days

(ii) 1 ordinary year = 365 days = 52 weeks + 1 day.
     So, 1 ordinary year has 1 odd day

100 years = 76 ordinary years + 24 leap years
                  = 76X1 + 24X2 = 76 + 48 = 124 days
                  = 17 weeks + 5 days = 5 odd days
SO, 100 years has 5 odd days

Number of odd days in 100 years = 5
Number of odd days in 200 years = 3
Number of odd days in 300 years = 1
Number of odd days in 400 years = 0

Similarly, each one of 800 years, 1200 years, 1600 years and 2000 years etc has 0 odd days.
Days of the week related to odd days:
                                         0 odd days ---> Sunday
                                         1 odd day   ---> Monday
                                         2 odd days ---> Tuesday
                                         3 odd days ---> Wednesday
                                         4 odd days ---> Thursday
                                         5 odd days ---> Friday
                                         6 odd days ---> Saturday
Shortcut to find the no. of leap years in the given no. of years :
                   Divide the given number of years by 4 and take the integer part. That is the number of leap years available in the given number of years.

Example : Let us find the number of leap years in 75 years. Divide 75
                  by 4. That is 75/4 = 18.75. In this, integer part is 18. There are
                  18 leap years in 75 years.

Why do students have to study this topic?

Students who are preparing to improve their aptitude skills and those who are preparing for this type of competitive test must prepare this topic in order to have better score. Because, today there is no competitive exam without questions from the topic Calendar problems. Whether a person is going to write placement exam to get placed or a students is going to write a competitive exam in order to get admission in university, they must be prepared to solve Calendar problems. This is the reason for why people must study this topic.

Benefit of studying this topic

As we mentioned in the above paragraph, a person who wants to get placed in a company and a students who wants to get admission in university for higher studies must write competitive exams like placement test and entrance exam. To meet the above requirement, it is very important to score more marks in the above mentioned competitive exams. To score more marks, they have to prepare this topic. Preparing this topic would definitely improve their marks in the above exams. Preparing this topic is not difficult task. We are just going to remember the stuff that we have already learned in our lower classes

How can students do Calendar problems?

Students have to learn few basic operations in this topic Calendar problems and some additional tricks. Already we are much clear with the four basic operations which we often use in math. They are addition, subtraction, multiplication and division. Even though we are much clear with these four basic operations, we have to be knowing some more stuff to do the problems which are being asked from this topic in competitive exams. The stuff which I have mentioned above is nothing but the tricks and shortcuts which need to solve the problems in a very short time. 

Shortcuts we use to solve the problems

Short cut is nothing but the easiest way to solve problems related to Calendar. In competitive exams, we will have very limited time to solve each problem. Then only we will be able to attend all the questions. If we do problems in competitive exams in perfect manner with all the steps, it will definitely take much time and we may not able to attend the other questions. So we need some other way in which the problems can be solved in a very short time. The way we need to solve the problem quickly is called as shortcut.

Here, we are going to have some Calendar problems . You can check your answer online and see step by step solution.

1. What was the day of the week on 16th July,1776?

         (A) Monday                      (B) Tuesday
         (C) Wednesday              (D) Thursday

jQuery UI Dialog functionality
16th July,1776 = (1775 years + Period from 01.01.1776 to 16.07.1776)

1775 = 1600 + 100 + 75

No.of odd days in 1600 years = 0 ----(1)
No.of odd days in 100 years = 5 ----(2)
No.of odd days in 75 years = 18 leap years + 57 ordinary years = 18X2+57X1=93days=13weeks+2days=2 odd days ----(3)

From 01.01.1776 to 16.07.1776, we have,

Jan-31days+Feb-29days+Mar-31days+Apr-30days+May-31days+Jun-30days+July-16days = 198days = 28weeks + 2 days = 2 odd days ----(4)

Adding (1),(2),(3) & (4), we get,
Total no.of odd days = 0+5+2+2 = 9days = 1 week + 2days = 2 odd days. 2 odd days corresponding to Tuesday

Hence, the day of the week on 16th July,1776 is Tuesday

2. On what dates of March 2005 did Friday fall ?

      (A) 4,11,18&25               (B) 5,12,19&26
      (C) 6,13,20&27               (D) 7,14,21&28

jQuery UI Dialog functionality
First of all, we have to find the day on 01.03.2005

01.03.2005 = (2004 years + Period from 01.01.2005 to 01.03.2005)

Number of odd days in 2000 years = 0 ----(1)
Number of odd days in 4 years = 1 leap year + 3 ordinary year
= 1X2 + 3X1 = 5 days ----(2)

From 01.01.2005 to 01.03.2005, we have,
Jan-31days + Feb-28days + Mar-1day = 60days
= 8weeks + 4days = 4 odd days ----(3)

Adding (1),(2)&(3), we get,total no.of odd days = 0+5+4
= 9 days = 1 week + 2days = 2 odd days.
2 odd days corresponding to Tuesday.So 01.03.2005 was Tuesday. Therefore, Friday lies on 04.03.2005.

Hence, the dates of March,2015 on which Fridays fell are 4,11,18&25

3. The calendar for the year 2007 will be the same for the year?

                      (A) 2014               (B) 2016
                      (C) 2017               (D) 2018

jQuery UI Dialog functionality
For example,if our answer is 2018, we must have the same day on 01.01.07 & 01.01.18 and number of odd days from 01.01.07 to 31.12.17 must be zero. Let us check the same thing for each of the given options.

Option (a)2015: No.of odd days from 01.01.2007 to 31.12.2014 = 2 leap years + 6 ordinary years = 2X2+6X1 = 10days = 1week + 3days = 3 odd days[not correct]

Option (b)2016: No.of odd days from 01.01.2007 to 31.12.2015 = 2 leap years + 7 ordinary years = 2X2+7X1 = 11 days = 1 week + 4 days = 4 odd days [not correct]

Option (c)2017: No.of odd days from 01.01.2007 to 31.12.2016 = 3 leap years + 7 ordinary years = 3X2+7X1 = 13 days = 2 weeks + 1 day = 1 odd day [not correct]

Option (d)2018: No.of odd days from 01.01.2007 to 31.12.2017 = 3 leap years + 8 ordinary years = 3X2+8X1 = 14 days = 2 weeks + 0 day = 0 odd day [correct]

Hence, Calendar for the year 2018 will be the same as for the year 2007

4. How many days are there in "x" weeks "x" days?

                  (A)7x2                            (B) 7x
                  (C) 8x2                           (D) 8x

jQuery UI Dialog functionality
Number of days in "x" weeks = 7x
Number of days in "x" days = x

Number of days in "x"weeks"x"days = 7x+x = 8x

Hence, the number of days in "x"weeks"x"days = 8x days

5. The last day of a century can not be

         (A) Monday                      (B) Tuesday
         (C) Wednesday              (D) Thursday

jQuery UI Accordion - Default functionality
From the notes given on this web page, we have

Number of odd days in 100 years (1st century) = 5
So, the last day of 1st century is Friday

Number of odd days in 200 years (2nd century) = 3
So, the last day of 2nd century is Wednesday

Number of odd days in 300 years (3rd century) = 1
So, the last day of 3rd century is Monday

Number of odd days in 400 years (4th century) = 0
So, the last day of 4th century is Sunday

This pattern is repeated

Therefore, the last day of a century can not be Tuesday or Thursday or Saturday.

Hence, option (c)Tuesday is the correct answer.



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