**Quantitative Aptitude Calendars Study Material**

If someone asks you what day it was on IOth May 1575 or what day it would be on 12th September 2340, you may call him crazy for asking such silly questions. If you don’t know the rule how to find it, it may look like a Herculean task for you. But, truly speaking, it is not so difficult. In this chapter we will concentrate our discussion on finding its answer i.e; on what day of the week a particular date falls. The clue to the process of finding it lies in calculating the number of odd days, which is quite different from the odd numbers.

**See Also: **

The number of days more than the complete number of weeks in a given period are called odd days. In other words it is the remainder left when the given number of days is converted into weeks on dividing it by 7.

In ancient times many civilizations used calendars based on movement of the moon. These lunar calendars were not accurate and corrections had to be made frequently. Later, based on the fact that in the solar system all planets including Earth revolved around the Sun, solar calendars were developed. These solar calendars proved to be more accurate. 1

A solar year consists of 365 days, 5 hours, 48 minutes and 48 seconds. In Julian calendar, arranged in 47 BC by Julius Caesar, the year was taken as 365— days. In order to make up for the odd quarter of a day, an extra or intercalary day was added once in every fourth year and this was called ‘a Leap year. Thus, an ordinary year consists of 365 days and a leap year has 366 days. In a leap year, February has 29. days instead of 28 days for ordinary year. The calendar based on this system is known as the Old Style Calendar. But, as can be noticed on comparison, since the solar year is 11 minutes 12 seconds less than a quarter of a day, in due course of several years, Julian Calendar too became inaccurate by several days. It again called for a further correction to be made. To rectify this discrepancy Pope Gregory XIII devised another calendar known as Gregorian Calendar. According to it, not all century years are leap years, although all of them are divisible by 4. He made centurial years leap years only once in 4 centuries. Accordingly, only those century years which are divisible by 400 arc leap years, while other century years are ordinary years. For example; 1300, 1400 and 1500 are ordinary years but 1600 is a leap year. With this modification, the Gregorian Calendar came in close exactitude with the solar year and the difference between the two is only 26 seconds which amounts to a day, in 3323 years. These calendars are called as the New Style Calendars.

In India, Vikrami and many other calendars were used earlier. Now the Government of India has adopted the National Calendar based on Sakaera with Chaitra as its first month. The days of this national calendar have a direct permanent correspondence with the days of Gregorian —

Chaitra 1 falls on March 22 in an ordinary year and on March 21 in a leap year.

**Points to be Remembered:**

1. An ordinary year contains 365 days i.e., 52 weeks and 1 odd day.

2. A leap year contains 366 days i.e., 52 weeks and 2 odd days.

Note : For an year to be a leap year, both the following conditions should be satisfied:

(i) An year divisible by 4 is a leap year. For example, 1984, 1988,1992, 1996,2000etc.

(ii) In case of century years, only those divisible by 400 are leap years, while other century years arc not leap years. For example, 400,800,1200,1600,2000 etc. are leap years.

500, 600,700,900, 1000 etc. are not leap years.

3.100 years contains 24 leap years and 76 ordinary years.

Therefore 100 years :

= [(24 x 52) weeks + (24 x 2) add days) + [(76 x 52) weeks + (76 x 1) odd days)

= (24 + 76) x 52 weeks + (48 + 76) odd days.

= 5200 weeks + 124 odd days.

= 5200 weeks + 17 weeks + 5 odd days.

‘ = 5217 weeks + 5 odd days. • ”

i.e., 100 years contains 5 odd days.

200 years contains (5×2)= 10 =1 week + 3 odd days, i.e., 3 odd days.

300 years contains (5 x 3) = 15 = 2 weeks + 1 odd day, i.e, 1 odd day.

400 years is a leap year and hence it will contain (5 x 4) + 1 = 21 days which equals 3 weeks and and hence no odd day.

Similarly, 800, 1200, 1600, 2000 years each contain no odd day.

To find the day of the week on a particular date when no reference day is given :

(i) Count the net number of odd days on the given date.

(ii) Write

Sunday for 0 odd day

Monday for 1 odd day

Tuesday for 2 odd days

:

:

:

Saturday for 6 odd days

Sunday for 7 odd days which is same as 0 odd day.

Assumption : First January 1 A.D. was Monday.

**Ex. 1. What day of the week was on May 3, 1999?**

**Sol.** May 3, 1999 means 1998 complete years + first 4 months upto April of 1999 + 3 days of May.

1600 years have 0 odd day

300 years have 1 odd day

98 years have 24 leap years + 74 ordinary years

= (24 x 2) + (74 x 1)

= 122 days = 17 weeks + 3 odd days

1998 years have (0 + 1 + 3) = 4 odd days January has 31 days i.e., 3 odd days

1999 is an ordinary year.

Hence, February has 28 days, i.e, 0 odd day March has 31 days i.e., 3 odd days April has 30 days i.e. 2 add days May 3, gives 3 more odd days.

Total number of odd days on May 3, 1999 = (4 + 3 + 0 + 3 + 2 + 3) odd days.

= 15 odd days i.e., 1 odd day.

Hence, May 3, 1999 was Monday Ans.

To find the day of the week on a particular date when reference day is given :

(i) Find the net number of odd days for the period between the reference date and the given date.

Note : Exclude the reference day but count the given date for counting the number of net odd days.

(ii) The day of the week on the particular date is equal number of net odd days ahead of the reference day (if the reference day was before this date) but behind the reference day (if this date was behind the reference day).

This will become more clear from the examples illus-trated below.

**Ex. 2. January 16, 1997 was a Thursday. What day of the week was on January 4, 2000?**

**Sol**. First we look for the leap years during this period.

1997, 1998, 1999 are not leap years.

1998 and 1999 together have net 2 odd days.

Number of days remaining in 1997 = 365 – 16 = 349 days = 49 weeks 6 odd days.

January 4, 2000 gives 4 odd days.

.’. Total number of odd days = 2 + 6 + 4= 12 days = 1 week + 5 odd days.

Hence, January 4, 2000 would be 5 days beyond Thursday i.e, it was on Tuesday. Ans.

**Ex. 3. February 20, 1999 was Saturday. What day of the week was on December 30, 1997?**

**Sol.** The year during this interval was 1998 and it was not a leap year.

Number of odd days in 1999 upto February 19 :

January 1999 gives 3 odd days.

19 February 1999 gives 5 odd days.

1998, being an ordinary year, gives 1 odd day.

In 1997, December 30 and 31 give 2 odd days.

.•. Total number of odd days = 3 + 5+ 1 + 2 = 11 days 4 odd days.

Therefore, December 30, 1997 would fall 4 days before Saturday i.e., on Tuesday Ans.

**Ex. 4. March 5, 1999 was on Friday. What day of the week was on March 5, 2000?**

**Sol**. Year 2000 was a leap year.

Number of days remaining in 1999 = 365 – [31 days Jan. + 28 days Feb. + 5 days March]

= 301 days = 43 weeks i.e., 0 odd day

Number of days passed m 2000 ‘.

January 31 days give 3 odd days.

February 29 days (being leap year) give 1 odd day

March 5 days give 5 odd days.

.’. Total number of odd days = 0 + 3+ 1 + 5 = 9 days i.e.. 2 odd days.

Therefore, March 5, 2000 would be two days beyond Friday i.e., on Sunday Ans.

Note : (i) February : 28 days (ordinary year) give 0 odd days 29 days (leap year) give 1 odd day.

(ii) January, March, May, July, August, October, and December each has 31 days and therefore give 3 odd days.

(iii)April, June, September and November each has 30 days and therefore give 2 odd days.

(iv) Write Sunday for 0 odd day, Monday for 1 odd day, Tuesday for 2 odd days, Wednesday for 3 odd days, Thurs¬day for 4 odd days, Friday for 5 odd days, Saturday for 6 odd days and again Sunday for 7 or 0 odd days.

**Ex.5. What day of the week was on June 21, 1437?**

**Sol**. June 21, 1437 contains 1436 complete years + first 5 months of the year 1437 + 21 days of June

[ 36 years have 9 leap years and a leap year has one more odd day than ordinary year]

1200 years give 0 odd day

200 years give 3 odd days

36 years give (36 + 9) = 45 or 3 odd days

Total number, of odd days upto 21st June 21, 1437 = 6+4 = 10 days or 3 odd days. So, June 21, 1437 was a Wednesday Ans

**Ex. 7. August 15, 1992 was Saturday. What day of the week was on August 15, 1993?**

**Sol**. February and August 15, 1993. February 1993 contains 28 days because 1993 was an ordinary year.

The interval, August 15, 1992 to August 15, 1993, is one complete ordinary year of 365 days and hence has 1 odd day.

August 15, 1993 would be one day beyond Saturday i.e., on Sunday Ans.

**Ex. 8. August 15, 1992 was Saturday. What day of the week was on August 15, 1991?**

**Sol.** In the given interval of one year the month of February falls in 1992 which was a leap year. Therefore, this one year interval would be of 366 days which gives 2 odd days.

August 15. 1991 would be 2 days before Saturday i.e., on Thursday Ans.