Calendar — Part 2 MCQs with Explanations

Formula : (Date + Month code + No.of years + No.of leap year + Century code)/7 $$\eqalign{ & = \frac{{26 + 1 + 50 + 12 + 0}}{7} \cr & = \frac{{89}}{7} \cr & = 5 \cr & = {\text{Thursday}} \cr} $$

30 days = 4 x 7 + 2 = 2 odd days The day is 2 days before Thursday i.e Tuesday

Given that January 1, 2007 was Monday. Odd days in 2007 = 1 (we have taken the complete year 2007 because we need to find out the odd days from 01-Jan-2007 to 31-Dec-2007, that is the whole year 2007) Hence January 1, 2008 = (Monday + 1 Odd day) = Tuesday

Given that 1.12.91 is the first Sunday Hence we can assume that 3.12.91 is the first Tuesday If we add 7 days to 3.12.91, we will get second Tuesday If we add 14 days to 3.12.91, we will get third Tuesday If we add 21 days to 3.12.91, we will get fourth Tuesday ⇒ Fourth Tuesday = (3.12.91 + 21 days) = 24.12.91

59 days = 8 weeks 3 days = 3 odd days Hence if today is Thursday, After 59 days, it will be = (Thursday + 3 odd days) = Sunday

Formula : (Date + Month code + No.of years + No.of leap year + Century code)/7 $$\eqalign{ & = \frac{{21 + 6 + 87 + 21 + 0}}{7} \cr & = \frac{{135}}{7} \cr & = 2 \cr & = {\text{Monday}} \cr} $$

Here we have to count the number days from 13 th May, 2005 to 18 rd August 2005 ( both inclusive) From 13 th to 31 st May = 19 days In June = 30 days In July = 31 days From 1st to 19th April = 19 days Total number of days = 19 + 30 + 31 + 19 = 99 days The number of odd days are = 14 x 7 + 1 = 99 So there is 1 odd day in the given period

For a year to have the same calendar with 1990 ,total odd days from 1990 should be 0. Take the year 1992 from the given choices. Total odd days in the period 1990-1991 = 2 normal years ⇒ 2 x 1 = 2 odd daysTake the year 1995 from the given choices. Number of odd days in the period 1990-1994 = 4 normal years + 1 leap year ⇒ 4 x 1 + 1 x 2 = 6 odd daysTake the year 1996 from the given choices. Number of odd days in the period 1990-1995 = 5 normal years + 1 leap year ⇒ 5 x 1 + 1 x 2 = 7 odd days = 0 odd days (As we can reduce multiples of 7 from odd days which will not change anything) Though number of odd days in the period 1990-1995 is 0, there is a catch here. 1990 is not a leap year whereas 1996 is a leap year. Hence calendar for 1990 and 1996 will never be the same.Take the year 2001 from the given choices. Number of odd days in the period 1990-2000 = 8 normal years + 3 leap years ⇒ 8 x 1 + 3 x 2 = 14 odd days = 0 odd days Also, both 1990 and 2001 are normal years. Hence 1990 will have the same calendar as that of 2001

A period of 100 years has 5 odd days . In 26 years , 4 are leap, remaining are ordinary years 125 years = 100 years + 26 years = 100 years + 6 leap years + 20 ordinary years = 5 odd days + 12 odd days + 20 odd days = 37 odd days = 5 x 7 +2 = 2 odd days

Day before yesterday was Thursday ⇒ Yesterday was a Friday ⇒ Today is a Saturday ⇒ Tomorrow is a Sunday

Here we have to count the number days from 3 rd February, 2012 to 18 th April 2012 ( both inclusive) The given year is leap year, So February month has 29 days From 3 rd to 29 th February = 27 days In March = 31 days From 1 st to 18 th April = 18 days Total number of days = 27 + 31 + 18 = 76 days

15 th Aug, 1947 = (1946 years + Period from 1.1.1947 to 15.8.1947) Odd days in 1600 years = 0 Odd days in 300 years = 1 46 years = (35 ordinary years + 11 leap years) = (35 x 1 + 11 x 2) = 57 (8 weeks + 1 day) = 1 odd day Jan. Feb. Mar. Apr. May. Jun. Jul. Aug ( 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 ) = 227 days = (32 weeks + 3 days) = 3 odd days Total number of odd days = (0 + 1 + 1 + 3) = 5 odd days Hence, as the number of odd days = 5, given day is Friday.

16 th July, 1776 = (1775 years + Period from 1 st Jan, 1776 to 16 th July, 1776) Counting of odd days : 1600 years have 0 odd day 100 years have 5 odd days 75 years = (18 leap years + 57 ordinary years) = [(18 x 2) + (57 x 1)] = 93 (13 weeks + 2 days) = 2 odd days 1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day Jan Feb Mar Apr May Jun Jul 31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days = (28 weeks + 2 days) Total number of odd days = (0 + 2) = 2 Required day was 'Tuesday'.

This can be illustrated with an example. Ex: 1896 is a leap year.The next leap year comes in 1904 (1900 is not a leap year). Explanation : The length of the solar year, however, is slightly less than 365$$\frac{1}{4}$$ days-by about 11 minutes. To compensate for this discrepancy, the leap year is omitted three times every four hundred years. In other words, a century year cannot be a leap year unless it is divisible by 400. Thus 1700, 1800, and 1900 were not leap years, but 1600, 2000, and 2400 are leap years.

Given year is divided by 4, and the quotient gives the number of leap years. Here, 100 ÷ 4 = 25 But, as 100 is not a leap year ⇒ 25 - 1 = 24 leap years.

If the period between the two months is divisible by 7, then that two months will have the same calendar. (a). Oct + Nov = 31 + 30 = 61 (not divisible by 7) (b). Apr + May + Jun + Jul + Aug + Sep + Oct = 30 + 31 + 30 + 31 + 31 + 30 + 31 = 214 (not divisible by 7) (c). Jun + July + Aug + Sep = 30 + 31 + 31 + 30 = 122 (not divisible by 7) (d). Apr + May + June = 30 + 31 + 30 = 91 (divisible by 7) Hence, April and July months will have the same calendar.

Given year is divided by 4, and the quotient gives the number of leap years. Here, 300 ÷ 4 = 75 But, as 100, 200 and 300 are not leap years ⇒ 75 - 3 = 72 leap years.

Let us find the day on 1 st July, 2004. 2000 years have 0 odd day. 3 ordinary years have 3 odd days. Jan. Feb. March April May June July 31 + 29 + 31 + 30 + 31 + 30 + 1 = 183 days = (26 weeks + 1 day) Total number of odd days = (0 + 3 + 1) odd days = 4 odd days. ∴ 1 st July 2004 was 'Thursday' Thus, 1 st Monday in July 2004 as on 5 th July. Hence, during July 2004, Monday fell on 5th, 12th, 19th and 26th.

NOTE : Repetition of leap year ⇒ Add + 28 to the Given Year. Repetition of non leap year Step 1 : Add + 11 to the Given Year. If Result is a leap year, Go to step 2. Step 2: Add + 6 to the Given Year. Solution : Given Year is 2005, Which is a non leap year. Step 1 : Add + 11 to the given year (i.e 2005 + 11) = 2016, Which is a leap year. Step 2 : Add + 6 to the given year (i.e 2005 + 6) = 2011 Therefore, The calendar for the year 2005 will be same for the year 2011

First,we count the number of odd days for the left over days in the given period. Here,given period is 12.2.1986 to 1.1.1987 Feb Mar Apr May June July Aug Sept Oct Nov Dec Jan 16 31 30 31 30 31 31 30 31 30 31 1 (left days) 2 + 3 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 1(odd days) = 1 odd day So, given day Wednesday + 1 = Thursday is the required result.

15 th August, 1947 = (1946 years + Period from 1 st Jan., 1947 to 15th ) Counting of odd days: 1600 years have 0 odd day. 300 years have 1 odd day. 47 years = (11 leap years + 36 ordinary years) = [(11 x 2) + (36 x 1) ] odd days = 58 odd days = 2 odd days Jan Feb Mar Apr May Jun Jul Aug = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 days = (32 weeks + 3 days) = 3, Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days. Hence, the required day was 'Saturday'.

We will show that the number of odd days between last day of February and last day of October is zero. March April May June July Aug. Sept. Oct. 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 = 241 days = 35 weeks = 0 odd day Number of odd days during this period = 0. Thus, 1 st March of an year will be the same day as 1 st November of that year. Hence, the result follows

350 days = $$\frac{{350}}{7}$$ = 50 weeks, i.e No odd days, So it will be a Saturday.

We already know that the calendar after a leap year repeats again after 28 years. Here 1988 is a Leap year, then the same calendar will be in the year = 1988 + 28 = 2016.

We know that, After every 400 years, the same day occurs. Thus, if 9 th August 2016 is Saturday, before 400 years i.e., on 9 th August 1616 has to be Saturday.

Given that the month begins on a Friday and has 31 days Sundays = 3 rd , 10 th , 17 th , 24 th , 31 st ⇒ Total Sundays = 5 Every second & fourth Saturday is holiday. 2 nd & 4 th Saturday in every month = 2 Total days in the month = 31 Total working days = 31 - (5 + 2) = 24 days.

The year 1996 is a leap year. So, it has 2 odd days. But 17th March comes after 29th February. So, the day on 17 th March, 1997 will be 1 day beyond the day on 17 th March,1996. Here 17 th March, 1997 is Monday. So, 17 th March, 1996 is Sunday.

When a century year leaves a remainder 0, when divided by 400 then it is a leap year (366 days). So, 1200 has 366 days.

Given Today is 20 th January 2017, Sunday In january, we have 31 days February - 28 days (Non leap year) March - 31 days April - 30 days ⇒ Remaining days = 31 - 20 = 11 in Jan 11 in Jan + 28 in Feb + 31 in Mar = 11 + 28 + 31 = 70 days More 20 days to complete 90 days ⇒ upto 20 th April Therefore, after 90 days from today i.e, 20 th Jan 2017 is 20 th Apr 2017. Now, the day of the week will be $$\frac{{90}}{7}$$ ⇒ Remainder '6' As the day starts with '0' on sunday 6 ⇒ Saturday. Required day is 20 th April, Saturday.

The year 2006 is an ordinary year. So, it has 1 odd day. So, the day on 24 th Nov, 2007 will be 1 day beyond the day on 24 th Nov, 2006. But, 24 th Nov, 2007 is Thursday. 24 - 10 = 14 days. Therefore, 2 weeks ago it is same day. Thus, 10 th Nov, 2006 is one day before 10 th Nov, 2007 i.e. it is Wednesday.

Given year 2003, when divided by 4 leaves a remainder of 3. NOTE: When remainder is 3, 11 is added to the given year to get the result. So, 2003 + 11 = 2014

The year 1985 is an ordinary year. So, it has 1 odd day. So, the day on 19 th June, 1985 will be 1 day after the day on 19 th June, 1984. But, 19 th June,1984 is Monday So, 19 th June, 1985 is Tuesday.

The day of the week repeats every 7 days. Given today is Friday. Again Friday is repeated on the 7 th day, 14 th ,... on 7 multiple days. Hence, Friday is on the 63 rd day, as 63 is multiple of 7. Now, the required day of the week on the 65 th day is Sunday.

We know that, 1 year = 365 days 1 day = 24 hours 1 hour = 60 minutes 1 minute = 60 seconds. Then, 1 year = 365 x 24 x 60 x 60 seconds. = 8760 x 3600 1 year = 31536000 seconds. Hence, 10 years = 31536000 x 10 = 315360000 seconds.

Given year 1939, when divided by 4 leaves a remainder of 3. NOTE: When remainder is 3, 11 is added to the given year to get the result. So, 1939 + 11 = 1950

We know that the day repeats every 7 days, 14 days, 21 days, ......... So if today is Monday, after 7 days it is again Monday, after 14 days again it is Monday. Hence, after 2 weeks i.e, 14 days the day repeats and is the same day.

Calculating Age has 2 conditions. Let your Birthday is on January 1 st . 1. If the month in which you are born is completed in the present year i.e, your birthday, then Your Age = Present year - Year you are born As of now, present year = 2018 i.e, Age = 2018 - 1995 = 23 years. 2. If the month in which you are born is not completed in the present year i.e, your birthday, then Your Age = Last year - Year you are born As of now, present year = 2018 i.e, Age = 2017 - 1995 = 22 years.

2 nd Jan 1901 means (1900 years and 2 day) Now, 1600 years have 0 odd day 300 years have 1 odd day 2 days has 2 odd day Total no. of odd days = 0 + 1 + 2 = 3 days Hence, the day on 2 nd Jan 1901 was Wednesday.

16 June, 1993 = (1992 years + Period from 1.1.1993 to 16.6.1993) Odd days in 1600 years = 0 Odd days in 300 years = 1 92 years = (69 ordinary years + 23 leap year) = (69 x 1 + 23 x 2)= 3 odd days Jan. Feb. March April May June (31 + 28 + 31 + 30 + 31 + 16 ) = 167 days 167 days = (23 weeks + 6 days) =6 odd days. Total number of odd days = (0 + 1 + 3 + 6) = 3 odd days. Given day is Wednesday.

For an century year,it is divided by 400,when leaves a remainder 0 then it is a leap year. So, 1600