Exercise :: Calendar - General Questions
- Calendar - Important Formulas
- Calendar - General Questions
1. | It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010? |
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Answer: Option C Explanation: On 31st December, 2005 it was Saturday. Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
Thus, on 1st Jan, 2010 it is Friday. |
2. | What was the day of the week on 28th May, 2006? |
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Answer: Option D Explanation: 28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006) Odd days in 1600 years = 0 Odd days in 400 years = 0 5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2)
Jan. Feb. March April May (31 + 28 + 31 + 30 + 28 ) = 148 days
Total number of odd days = (0 + 0 + 6 + 1) = 7 Given day is Sunday. |
3. | What was the day of the week on 17th June, 1998? |
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Answer: Option C Explanation: 17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998) Odd days in 1600 years = 0 Odd days in 300 years = (5 x 3) 97 years has 24 leap years + 73 ordinary years. Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.
Jan. Feb. March April May June (31 + 28 + 31 + 30 + 31 + 17) = 168 days
Total number of odd days = (0 + 1 + 2 + 0) = 3. Given day is Wednesday. |
4. | What will be the day of the week 15th August, 2010? |
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Answer: Option A Explanation: 15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010) Odd days in 1600 years = 0 Odd days in 400 years = 0 9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days
Jan. Feb. March April May June July Aug. (31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days
Total number of odd days = (0 + 0 + 4 + 3) = 7 Given day is Sunday. |
5. | Today is Monday. After 61 days, it will be: |
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Answer: Option B Explanation: Each day of the week is repeated after 7 days. So, after 63 days, it will be Monday.
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