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.
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:To find out the day of the week for 6th March 1993, we need to count the number of days between 10th January 1992 and 6th March 1993.
We can start by counting the number of days in January 1992, which is 31. Then, we add the number of days in February 1992, which is 29 since it was a leap year. Next, we add the number of days in all the months from March 1992 to February 1993, which is 365. Finally, we add the number of days in March 1993 up to the 6th, which is 6.
Therefore, the total number of days between 10th January 1992 and 6th March 1993 is 31 + 29 + 365 + 6 = 431 days.
Now, we can divide 431 by 7 to get the number of weeks and days. The quotient is 61 and the remainder is 4. This means that 431 days is equivalent to 61 weeks and 4 days.
Since 10th January 1992 was a Saturday, we can count 4 days forward from Saturday to get the day of the week for 6th March 1993. This means that 6th March 1993 was a Saturday.
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.
Example 3: Find the day of the week on 26^{th} Jan. 1960.
Solution: No. of odd days upto 26^{th} Jan. 1960
= Odd days for 1600 years + odd days for 300 years + odd days for 59 years + odd days of 26 days of Jan 1960
= 0 + 1 + 59 + 14 + 5 = 79 days
= 79 / 7 = 11 weeks + 2 days = 2 odd days
The required day is Tuesday Zero odd day means Sunday. We are to consider one odd day as Monday 2 odd days as Tuesday and so on.
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 : (b) Friday
January 26, 1950, was a significant day for India as it became a republic. To find the day of the week, we calculate the odd days:
January 26, 1950 = (1600 + 300 + 50) + 26 days = 1950 + 26 days = 1976 days
Since 1976 is evenly divisible by 7, there are no odd days. Therefore, India celebrated its first Republic Day on a Friday.
2. If March 2, 2040, is a Saturday, what day of the week will it be on March 2, 2050?
(a) Saturday
(b) Sunday
(c) Monday
(d) Tuesday
Answer: (b) Sunday
Since 2040 is a leap year, March 2, 2041, will be one day ahead (Sunday). For each nonleap year, it will advance one more day. Therefore, March 2, 2050, will be on a Sunday.
3. How many odd days are there in 150 years?
(a) 0
(b) 1
(c) 2
(d) 3
Answer 3: (c) 2
In 150 years, there are 37 leap years (divisible by 4). Each leap year has two odd days. The remaining 113 years are nonleap years, each with one odd day. Therefore, the total odd days are 37 * 2 + 113 * 1 = 74 + 113 = 187, which is equivalent to 2 odd days.
56 videos104 docs95 tests

1. What is a calendar? 
2. How many days are there in a calendar year? 
3. What is the significance of a leap year? 
4. How are calendars different in various cultures and religions? 
5. Can calendars be customized for personal use? 
56 videos104 docs95 tests


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