Introduction to Calendar | Quantitative Techniques for CLAT PDF Download

Calendar

A calendar is a system of organizing units of time for the purpose of reckoning time over extended periods. By convention, the day is the smallest calendrical unit of time; the measurement of fractions of a day is classified as timekeeping.

• Ordinary year: An year having 365 days is called ordinary years.
• Leap year: Year having 366 days is called a leap year. Every leap year is exactly divisible by 4 and ordinary years are not completely divisible by 4.
• Odd days: In a given period, the days apart from complete weeks are called odd days. An ordinary year has one odd day, i.e. 365/7 = 52 weeks + 1 day While the leap year has two odd days, i.e. 366/7 = 52 + 2 days.
  
  

To Find the Number of Odd Days

• In 100 years there are 24 leap years + 76 ordinary years
  = 24 x 52 weeks + 24 x 2 days + 76 x 52 weeks + 76 days
  = 6 days  + 6 days
  = 12 days = 1 week and 5 days
  So, in 100 years there are 5 odd days similarly in 200 years there are 3 odd days and in 300 years there is 1 odd day in 400 years there is 0 odd day similarly in 800 years, 1200 years and 1600 years there is 0 odd day.
• Odd days in Feb: In an ordinary year, Feb has no odd day, whereas in a leap year Feb has one odd day.
• 1^st day of the century must be Tuesday, Thursday or Saturday and Last day of a century cannot be Tuesday, Thursday or Saturday.

To Find a Particular Day When a Day and a Date is Given

Step I: Find out the number of odd days between the given date and the date for which the day is to be found one.

Step II: From the given day count the odd days in the forward direction to arrive at the day on the given date.

Example 1: If 10th January 1992 was Saturday, what day of the week was on 6^th March 1993.

Solution: 

Here’s a step-by-step way to find the day:

1. From 10 Jan 1992 to 10 Jan 1993 is one full year.
   – 1992 was a leap year, so that span is 366 days.

2. From 10 Jan 1993 to 6 Mar 1993:
   – Jan 11–31 Jan: 21 days
   – Feb 1993 (non-leap): 28 days
   – 1–6 Mar: 6 days
   Total = 21 + 28 + 6 = 55 days

3. Total days advanced from 10 Jan 1992 to 6 Mar 1993 = 366 + 55 = 421 days

4. Find 421 mod 7 to see how many weekdays we move forward:
   421 ÷ 7 = 60 weeks + 1 day ⇒ remainder 1

5. Starting from Saturday, moving ahead by 1 day gives Sunday.
   Answer: 6 March 1993 was a Sunday.

Example 2: On April 4, 1988, it was Monday. What day of the week was on 5^th Nov. 1987.

Solutions: No. of days between 5^th Nov. 1987 to 4^th April 1988 6^th Nov 1987 to 30 Nov = 25 days
 Dec 1987 = 31 days
Jan 1988 = 31 days
Feb 1988 = 29 days
March 1988 = 31 days
4^th April 1988 = 4 days
Total = 151 days
No. of odd days = 151 / 7 = 21 weeks – 4 days
So, since 5^th Nov. 1987 is prior to 4^th April 1988
We are to count 4 days backwards from Monday, the required day is Thursday.

To Find the Day on a Particular Date if Day and Date is not Given

• The procedure can be understood from the given example.

Example 3: Find the day of the week on 26^th Jan. 1960.

Solution: 

Here’s a step-by-step way to find the weekday of 26 January 1960:

1. Anchor date: 26 January 1950 was a Thursday (the first Republic Day)

2. Span to 1960: From 26 Jan 1950 up to 26 Jan 1960 is exactly 10 years.

3. Count leap years between: 1950–1959 includes the leap years 1952 and 1956 → 2 leap years

4. Total days advanced:

   • Non-leap years: 10 – 2 = 8 years × 365 days = 2920 days

   • Leap years: 2 × 366 = 732 days

   • Sum = 2 920 + 732 = 3652 days

5. Reduce mod 7:
   3652 ÷ 7 = 521 weeks + 5 days ⇒ remainder 5
   So from Thursday we move forward 5 weekdays.

6. Shift Thursday by 5 days:

   • +1 → Friday

   • +2 → Saturday

   • +3 → Sunday

   • +4 → Monday

   • +5 → Tuesday

Answer: 26 January 1960 fell on a Tuesday.

Miscellaneous Examples

1. On which day of the week did India celebrate its first Republic Day on January 26, 1950?

(a) Thursday

(b) Friday

(c) Sunday

(d) Tuesday

Answer : (a) Thursday

1. Split the year into century and year-within-century
• Century = 1900s → century‐code = 0 odd-days
• Year-within-century = 50

2. Compute odd-days contributed by the year-within-century

• Take the last two digits: 50

• Divide by 4 (to count leap years): ⌊50⁄4⌋ = 12

• Sum: 50 + 12 = 62 → 62 mod 7 = 62 – 56 = 6 odd-days

3. Month-code for January in a non-leap year
January → 0 odd-days

4. Date contribution
26 → 26 mod 7 = 26 – 21 = 5 odd-days

5. Total odd-days
century (0) + year (6) + month (0) + date (5) = 11
11 mod 7 = 11 – 7 = 4 odd-days

6. Map odd-days to weekdays (taking Sunday = 0)

0 → Sunday

1 → Monday

2 → Tuesday

3 → Wednesday

4 → Thursday

5 → Friday

6 → Saturday

4 ⇒ Thursday

2. What is the day on 15th August 2050?

(a) Saturday

(b) Monday

(c) Wednesday

(d) Tuesday

Answer: (b) Monday

15th August, 200 = (2049 years + Period from 1st January 2050 to 15th August 2050)

Odd days in 1600 years = 0
Odd days in 400 years = 0

49 years = (12 leap years + 37 ordinary years) = (12 × 2 + 37 × 1) = 61 odd days = 5 odd days

Odd days in the year 2050 =
January = 3, February = 0, March = 3, April = 2, May = 3, June = 2, July = 3, August = 1

Total odd days in 2050 till 15th August = 3 + 0 + 3 + 2 + 3 + 2 + 3 + 1 = 17 odd days or 3 odd days

Total odd days = 5 + 3 = 8 odd days or 1 odd day.

Days corresponding to odd-day counts:
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

Therefore, it will be Monday on 15th August 2050.

Hence, Monday is correct.

3. How many odd days are there in 100 years?

(a) 0

(b) 1

(c) 5

(d) 3

Answer 3: (c) 5

In 100 years there are 24 leap years and the 100th year itself is not a leap year.
So the number of ordinary years = 76 years.

Total odd days in a leap year = 2, so in 24 leap years = 48 odd days.
Total odd days in an ordinary year = 1, so in 76 ordinary years = 76 odd days.

Total odd days = 48 + 76 = 124 odd days = 17 weeks and 5 odd days.

Hence, the correct answer is 5.