## Calendar Questions

Go through Calendar Questions Page to get all the sample Calendar Questions for Practicing for Competitive Exam. Here you will also get some Formula that’ ll make Calendar Questions Solving easy. ### Formula or Rule for solving CALENDAR questions

• 100 years give us 5 odd days
• 200 years give us 5 x 2 = 10 – 7 (one week) = 3 odd days.
• 300 years have 5 x 3 = 15 – 14 (two weeks) =1 odd day.
• 400 years have [{5 x 4 + 1 (leap century)} – 21] (three weeks)= 0 odd days.
• January has 31 – 28 = 3 odd days.
• February has 28 – 28 = 0 odd day in a typical year and 1 odd day in a leap year and so on for all the other months.

### Points To Remember:

• Leap year rule:

• A year is a leap year if it is divisible by 4.
• However, if a year is divisible by 100, it is not a leap year unless it is also divisible by 400.
• For example, the year 2000 was a leap year because it is divisible by both 4 and 400.
• Odd days:

• The concept of odd days is used in the doomsday rule to find the day of the week for a given date within the same year.
• In a normal (non-leap) year, there are 365 days, which is equivalent to 52 weeks and 1 day (1 odd day).
• In a leap year, there are 366 days, which is equivalent to 52 weeks and 2 days (2 odd days).
• Doomsday rule:

• The doomsday rule is a mental calculation technique to find the day of the week for any date in a given year.
• It is based on a set of anchor days for each century (e.g., 1800, 1900, 2000) and applying simple arithmetic to find the day of the week.
• With practice, this rule can be used to quickly determine the day of the week for any date.

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## Practice Calendar Questions and Answers

1. If it is Sunday on Jan 1, 2006, determine the day on Jan 1, 2010? Friday

Friday

60.87%

Monday

Monday

17.39%

Thursday

Thursday

17.39%

None of the above

None of the above

4.35%

It was Saturday on 31st December 2005.

Odd days from the year 2006 to 2009 = (1+1+2+1) = 5 days

Therefore, it was Thursday on 31st December 2009, and the next day, 1st January 2010 will be Friday. 2. Find out the day for 28th of May 2006? sunday

sunday

44.44%

tuesday

tuesday

22.22%

saturday

saturday

22.22%

None of the above

None of the above

11.11%

28 May 2006

= (2005 + time from 1st Jan 2006 to 28th May 2006)

Number of odd days in 1600 years =0

Number of odd days in 400 years =0

Calculation for 4 years = (4 ordinary years + 1 leap year)

= (4 x 1 + 1x 2)

= 6 odd days

Also, total days =

Jan = 31

Feb = 28

March = 31

April = 30

May= 28

Sum of all these days = 148

Number of add days in these 148 days = (21 weeks + 1 day) = 1 odd day

Total odd days = 0+0+6+7 = 7 (which leaves no odd day)

Therefore, the required day is Sunday. 3. Find out the day after 61 days if today is Monday? Monday

Monday

9.52%

Tuesday

Tuesday

9.52%

Saturday

Saturday

76.19%

None of the above

None of the above

4.76%

It takes seven days to repeat each day in a week.

Therefore, after 63 days, it will be on Monday.

And after 61 days, it will be on Saturday. 4. If it is Monday on 6th March 2005, what is the day of 6th March 2004? Tuesday

Tuesday

10%

Sunday

Sunday

75%

Wednesday

Wednesday

5%

None of the above

None of the above

10%

2004 is a leap year and has 2 odd days

The month of Feb 2004 will not be calculated and hence, it has only 1 odd day.

Therefore, the day of 6th March 2005 will be one day after the day on 6th March 2004.

It is also given that 6th March 2005 is a Monday.

Hence, 6th March 2004 is a Sunday. 5. For April 2001, on what all dates will it be Wednesday? 5, 7, 9, 10

5, 7, 9, 10

10.53%

2, 3, 5, and 7

2, 3, 5, and 7

10.53%

4, 11, 18, and 25

4, 11, 18, and 25

68.42%

1, 2, 3, and 4

1, 2, 3, and 4

10.53%

Determining the day on 1st April 2001 = (2000 + time from 1.1.2001 to 1.4.2001)

Number of odd days in 1600 years = 0

Number of odd days in 400 years = 0

Calculation for Jan, Feb, March, and April= 31 +28 + 31 +1 = 91 days and 0 odd days

Total odd days = (0+0+0) =0

Day on 1st April 2001 = Sunday

Therefore, dates on which Wednesday will fall = 4, 11, 18, and 25 as counted by the interval of 7 days each. 6. Calculate the number of days in p weeks and p days. 8p

8p

55.56%

p^2

p^2

22.22%

5 + p

5 + p

11.11%

6p

6p

11.11%

p weeks and p days = (7p + p) days

= 8p days 7. From the options provided, which day is not a century’s last day? Tuesday

Tuesday

52.38%

Monday

Monday

19.05%

Wednesday

Wednesday

19.05%

None of the above

None of the above

9.52%

Odd days in 100 years = 5

Therefore, we can say that the last day of the first century is Friday.

Odd days in 200 years = (2 x 5) Ξ 3 odd days

Therefore, the last day of the second century is Wednesday

Odd days in 300 years = (5 x 3) = 15  Ξ 1 odd days

Therefore, we can say that the last day of the third century is Monday.

Odd days in 400 years = 0

Therefore, we can say that the last day of the 3rd centaury is Sunday.

The above cycle is repeated.

We can say that a century’s last day cannot be Tuesday, Thursday, or Saturday. 8. The day of the week on 7th February 2005 was a Monday. Determine the day of the week on 7th February 2004? Saturday

Saturday

41.18%

Friday

Friday

5.88%

Tuesday

Tuesday

23.53%

None of the above

None of the above

29.41%

2004 is a leap year and hence, number of odd days = 2

Therefore, the day of 8th Feb 2004 will be 2 days before the day on 8th Feb 2005.

Therefore, the day is Saturday. 9. Among the following, which of these is not a leap year? 300

300

66.67%

400

400

14.29%

1200

1200

4.76%

None of the above

None of the above

14.29%

The number which is divisible by 400 is a leap year.

Therefore, the year 300 is not a leap year. 10. It is on Friday, 7th December 2007. Which day of the week will fall on 7th December 2006? Thursday

Thursday

76.47%

Monday

Monday

11.76%

Tuesday

Tuesday

5.88%

None of the above

None of the above

5.88%

2006 is an ordinary year. Hence, the number of odd days = 1.

Also, the day of 7th December 2007 will be one day after the day on 7th December 2006.

Day on 7th December 2007= Friday.

Hence, 8th December 2006 = Thursday.  ×