{"id":5803567,"date":"2020-07-10T19:50:59","date_gmt":"2020-07-10T14:20:59","guid":{"rendered":"https:\/\/entri.app\/blog\/?p=5803567"},"modified":"2020-07-10T19:50:59","modified_gmt":"2020-07-10T14:20:59","slug":"tips-and-tricks-to-solve-calendar-problems","status":"publish","type":"post","link":"https:\/\/entri.app\/blog\/tips-and-tricks-to-solve-calendar-problems\/","title":{"rendered":"Tips and Tricks to Solve Calendar Problems"},"content":{"rendered":"

In competitive exams, Reasoning is one of the most scoring section. Some aspirants who have already gone through these exams consider it a more confusing part. From this <\/span>confusing section, there is one topic of \u201cTime And Calendar\u201d. Let’s discuss some tricks to Solve Calendar Problems for PSC, Bank, and SSC Exam here.<\/span><\/p>\n

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Basic Structure of a Calendar<\/strong><\/h3>\n
    \n
  1. Ordinary year:<\/b> Any year which <\/span>365<\/b> days is called an ordinary year.Ex: 1879, 2009, 2019, etc.<\/span><\/li>\n
  2. Leap year:<\/b> Any year which has <\/span>366<\/b> days is called a leap year.Ex: 2012, 2016 2020 etc.<\/span><\/li>\n
  3. The division of the number 365 by 7 gives the quotient 52 and remainder 1 which indicates that an ordinary year has 52 weeks and one extra day. This extra day is referred to as an \u201c<\/span>odd day<\/b>\u201d throughout the calendar topics.<\/span><\/li>\n
  4. A leap year has 366 days, the division of the number 366 by 7 gives the quotient 52 and remainder 2. This indicates that a leap year has 52 weeks and 2 extra days. These two extra days are also referred to as \u201c<\/span>odd days<\/b>\u201d.<\/span><\/li>\n<\/ol>\n

    An ordinary year has one odd day, whereas a leap year has two odd days.<\/b><\/p>\n

    Concept of an Odd Day<\/strong><\/h3>\n

    January has 31 days irrespective of whether it\u2019s an ordinary year or leap year. The division of the number 31 by 7 provides the remainder 3 hence the January has 3 odd days. On generalizing, any month which has 31 days has 3 odd days and any month which has 30 days has 2 odd days.<\/span><\/p>\n

    The only exception happens is in the case of February. The February month of an ordinary year has 28 days, division of 28 by 7 provides zero as remainder. Hence the number of odd days in February of an ordinary year will have 0 odd days and that of leap year will have 1 odd day as February in a leap year has 29 days.<\/span><\/p>\n

    The below table depicts the number of odd days in different months of a calendar year:<\/span><\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    Month<\/b><\/td>\nNumber of odd days<\/b><\/td>\n<\/tr>\n
    January<\/span><\/td>\n3<\/span><\/td>\n<\/tr>\n
    February(ordinary\/leap)<\/span><\/td>\n(0\/1)<\/span><\/td>\n<\/tr>\n
    March<\/span><\/td>\n3<\/span><\/td>\n<\/tr>\n
    April<\/span><\/td>\n2<\/span><\/td>\n<\/tr>\n
    May<\/span><\/td>\n3<\/span><\/td>\n<\/tr>\n
    June<\/span><\/td>\n2<\/span><\/td>\n<\/tr>\n
    July<\/span><\/td>\n3<\/span><\/td>\n<\/tr>\n
    August<\/span><\/td>\n3<\/span><\/td>\n<\/tr>\n
    September<\/span><\/td>\n2<\/span><\/td>\n<\/tr>\n
    October<\/span><\/td>\n3<\/span><\/td>\n<\/tr>\n
    November<\/span><\/td>\n2<\/span><\/td>\n<\/tr>\n
    December<\/span><\/td>\n3<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Click here to read the article on tricks to solve quadratic equations in bank exams<\/strong><\/a><\/p>\n

    Evaluation of leap year<\/strong><\/h3>\n

    The Earth actually completes 1 orbit around the Sun in more than 365 days, i.e in 365 Days 5 Hours 48 minutes and 45 seconds or takes approximately 6 hours more. A leap year occurs every 4 years to adjust for the 1\/4th day, 6 x 4= 24 hours, so every 4th year has 366 days (or 2 odd days). And as far as the few odd minutes of the orbit time are concerned, well every 100 years starting 1 AD, the year is declared to be a non\u2013leap year, but every 4th century is a leap year. So any year divisible by 400 will be a leap year e.g.: 1200, 1600 and 2000. The years 1800, 1900 will be non leap years.<\/span><\/p>\n

    Note:<\/b><\/p>\n

    A year 700 is completely divisible by 4, but it is not considered as a leap year. For a century year, the logic follows that it should always be divisible by 400 not by 4. Even though the year 700 is divisible by 4 but not by 400. Hence, the year 700 cannot be considered as a leap year.<\/span><\/p>\n

    Ex: 400, 800, 1200, etc. are leap years as they are divisible by 400 and years 300, 700, 100, etc. are not leap years as they are not divisible by 400.<\/span><\/p>\n

    Based on this above evaluation of odd years, we can solve the majority of questions. Let\u2019s discuss a question here:<\/span><\/p>\n

    \u00a0Question:<\/strong> What day of the week was year 100 A.D December 31<\/span>st<\/span>?<\/span><\/p>\n

    Solution: <\/b>Let\u2019s consider the first 100 years i.e. Year 1.A.D to year 100 A.D<\/span><\/p>\n

    Dividing the first 100 by 4 we get that first 100 years had 76 ordinary years and 24 ordinary years. (The quotient when 100 is divided by 4 gives 25 but the year 100 itself is not a leap year as it is not divisible by 400 hence 24 is considered instead of 25)<\/span><\/p>\n

    Step 1:<\/b> 100 years = 76 ordinary years + 24 leap years<\/span><\/p>\n

    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 24*2 = 48 odd days. Adding both the results we get 76+48 = 124 odd days in total.<\/span><\/p>\n

    Step 2:<\/b> 100 years = (76 x 1 + 24 x 2) odd days = 124 odd days.<\/span><\/p>\n

    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.<\/span><\/p>\n

    Step 3:<\/b> 100 years = (17 weeks + 5 odd days)<\/span><\/p>\n

    The number of odd days in 100 years = 5.<\/span><\/p>\n

    Decoded day of the week<\/strong><\/h3>\n

    The week always begins with Monday and hence the Saturday and Sunday are referred to as weekends. In order to make the calculation easier and reduce its time during the exams.<\/span><\/p>\n

    The days of the week are coded as follows:<\/span><\/p>\n\n\n\n\n\n\n\n\n\n\n
    Code of the day<\/b><\/td>\nDay<\/b><\/td>\n<\/tr>\n
    0<\/span><\/i><\/td>\nSunday<\/span><\/i><\/td>\n<\/tr>\n
    1<\/span><\/i><\/td>\nMonday<\/span><\/i><\/td>\n<\/tr>\n
    2<\/span><\/i><\/td>\nTuesday<\/span><\/i><\/td>\n<\/tr>\n
    3<\/span><\/i><\/td>\nWednesday<\/span><\/i><\/td>\n<\/tr>\n
    4<\/span><\/i><\/td>\nThursday<\/span><\/i><\/td>\n<\/tr>\n
    5<\/span><\/i><\/td>\nFriday<\/span><\/i><\/td>\n<\/tr>\n
    6<\/span><\/i><\/td>\nSaturday<\/span><\/i><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    The week always begins with Monday and hence counting that as 1, 5 stands for Friday.<\/span><\/p>\n

    Hence, the last day (December 31<\/span>st<\/span>) of the year 100 A.D was Friday.<\/span><\/p>\n

    Extension of the logic<\/b><\/p>\n

    Similarly, one can find the last day of the other century years by extending the same logic.<\/span><\/p>\n

    If 100 years had 5 odd days, then logically 200 years should have 10 odd days. Since 10 is greater than 7, the division of 10 by 7 gives the remainder 3. Hence, the 200 years had 3 odd days, which means the last day of the year 200 was Wednesday.<\/span><\/p>\n

    Number of odd days in 200 years = (5 x 2) = 10 = (7+3) = 3 odd days.<\/span><\/p>\n

    If 100 years had 5 odd days and 200 years 10 odd days logically 300 years should have 15 odd days. The division of 15 by 7 indicates it has 1 odd day from the remainder which indicates it is Monday. Hence, the last day of the year 300 was Monday.<\/span><\/p>\n

    Number of odd days in 300 years = (5 x 3) = 15 = (14+1) = 1 odd day.<\/span><\/p>\n

    Logically, 400 years should have 20 odd days since 400<\/span>th<\/span> year is a leap year as it is divisible by 400. This year will have 20+1 = 21 odd days, which when divided by 7 gives the zero (0) as remainder. Hence, 400 years had 0 odd days and that was Sunday.<\/span><\/p>\n

    The logical approach for the next few years is shown in the table given below:<\/span><\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    Century<\/b><\/td>\nNumber of odd days<\/b><\/td>\nDay of the week<\/b><\/td>\n<\/tr>\n
    100<\/span><\/td>\n5<\/span><\/td>\nFriday<\/span><\/td>\n<\/tr>\n
    200<\/span><\/td>\n3<\/span><\/td>\nWednesday<\/span><\/td>\n<\/tr>\n
    300<\/span><\/td>\n1<\/span><\/td>\nMonday<\/span><\/td>\n<\/tr>\n
    400<\/span><\/td>\n0<\/span><\/td>\nSunday<\/span><\/td>\n<\/tr>\n
    500 = (100+400)<\/span><\/td>\n(5+0) = 5<\/span><\/td>\nFriday<\/span><\/td>\n<\/tr>\n
    600 =(200+400)<\/span><\/td>\n(3+0) =3<\/span><\/td>\nWednesday<\/span><\/td>\n<\/tr>\n
    700 = (300+400)<\/span><\/td>\n(1+0) = 1<\/span><\/td>\nMonday<\/span><\/td>\n<\/tr>\n
    800 = (400+400)<\/span><\/td>\n(0+0) = 0<\/span><\/td>\nSunday<\/span><\/td>\n<\/tr>\n
    900 = ( 400 + 500)<\/span><\/td>\n(0 + 5) = 5<\/span><\/td>\nFriday<\/span><\/td>\n<\/tr>\n
    1000 = (500 +500)<\/span><\/td>\n(5+5)= (7+3) = 3<\/span><\/td>\nWednesday<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Observations from the table:<\/b><\/p>\n

      \n
    1. The cycle of a number of days repeats after every four centuries and also hence the days at which it ends. The order will always be Friday, Wednesday Monday, and Sunday.<\/span><\/li>\n
    2. A century will always end on either Friday, Wednesday, Monday, or Sunday (Decoded values of these days are 5, 3, 1, and 0 respectively).<\/span><\/li>\n
    3. A century will never end on Tuesday, Thursday, and Saturday (Decoded values of these days are 2, 4, and 6 respectively).<\/span><\/li>\n<\/ol>\n

      Zeller\u2019s Formula<\/b><\/h3>\n

      Along with these tricks, we can also use Zeller\u2019s Formula. i.e.<\/span><\/p>\n

      \u00a0d=k+[(13m-1)\/5]+D+[D\/4]+[C\/4]-2C mod 7<\/i><\/strong><\/p>\n