All questions of Calendar for UGC NET Exam

**Explanation:**

To determine the day of the week for a given date, we can use the concept of the Gregorian calendar and some basic calculations.

**Step 1: Determining the Reference Day**

- To find the day of the week for a specific date, we need to determine a reference day. In this case, we can choose a known day and its corresponding date that falls within the same year as the given date. Let's choose the reference day as January 1, 2010, which was a Friday.

**Step 2: Counting the Number of Days**

- The next step is to count the number of days between the reference day and the given date. For this, we need to consider both the number of days within the same year and the number of days in the intervening years.

- From January 1, 2010, to August 15, 2010, there are 226 days.
- Additionally, we need to consider the number of days in the intervening years (2011-2019) between the reference year and the given year. There are 9 years, and each year has 365 days, so the total number of intervening days is 9 * 365 = 3285.

- Therefore, the total number of days between the reference day and August 15, 2010, is 226 + 3285 = 3511.

**Step 3: Determining the Day of the Week**

- Now, we need to find the remainder when the total number of days is divided by 7. This remainder will give us the day of the week.

- 3511 divided by 7 equals 501 remainder 4.

- Since the reference day was Friday (which corresponds to 0), we can count 4 days forward to determine the day of the week for August 15, 2010.

- Friday (0) -> Saturday (1) -> Sunday (2) -> Monday (3) -> **Tuesday (4)**.

Therefore, the day of the week for August 15, 2010, was **Tuesday**.

Explanation:

Given Dates:
09/12/2001 (DD/MM/YYYY) = Sunday
09/12/1971 (DD/MM/YYYY)

Identifying the Day of the Week:
To determine the day of the week for 09/12/1971, we need to consider the number of days between 09/12/1971 and 09/12/2001.

Calculating the Difference:
The difference between 09/12/1971 and 09/12/2001 is 30 years.
Now, we need to consider the number of days in these 30 years, including leap years.

Leap Years:
From 1971 to 2001, there are 7 leap years (1972, 1976, 1980, 1984, 1988, 1992, 1996).

Calculating Total Days:
Number of days = 30 years * 365 days/year + 7 leap years = 10957 days

Day of the Week:
Now, when we divide the total days by 7 (since there are 7 days in a week), the remainder will give us the day of the week for 09/12/1971.

Calculation:
Sunday (09/12/2001) + 10957 days = Thursday

Therefore, if 09/12/2001 was a Sunday, then 09/12/1971 would have been a Thursday.

Explanation:

A Leap Year in 1895:
- A leap year is a year that is evenly divisible by 4, except for years that are evenly divisible by 100 but not by 400.
- In 1895, the calendar followed the standard leap year rules, so it had 366 days.

Calendar Matching with Year X:
- To find a possible value of X that has the same calendar as 1895, we need to look for a year that is also a leap year.
- The leap years are generally every 4 years, so we need to find a year that is 4 years away from 1895.

Possible Value of X:
- The year that is 4 years after 1895 is 1899 (1895 + 4 = 1899).
- Since 1899 is a leap year (divisible by 4), it will have the same calendar as 1895.
- Therefore, a possible value of X is 1899.

Conclusion:
- Option 'a) 1901' is incorrect because it is not a leap year.
- Option 'b) 1900' is incorrect because it is not 4 years after 1895.
- Option 'c) 1902' is incorrect because it is not a leap year.
- Option 'd) 1903' is incorrect because it is not 4 years after 1895.
- The correct answer is option 'a) 1901' as it is 4 years after 1895 and a leap year.