Problems on Calendar - Solved Examples
| 11. The last day of a century cannot be | |
| A. Monday | B. Wednesday |
| C. Tuesday | D. Friday |
answer with explanation
Answer: Option C
Explanation:
We know that number of odd days in 100 years = 5
Hence last day of the first century is Friday
Number of odd days in 200 years = 5 x 2 = 10 = 3 (As we can reduce multiples of 7 from odd days which will not change anything)
Hence last day of the 2nd century is Wednesday
Number of odd days in 300 years = 5 x 3 = 15 = 1
Hence last day of the 3rd century is Monday
We know that umber of odd days in 400 years = 0. (∵ 5 x 4 + 1 = 21 = 0)
Hence last day of the 4th century is Sunday
Now this cycle will be repeated. Hence last day of a century will not be Tuesday or Thursday or Saturday.
its better to learn this by heart which will be helpful to save time in objective type exams
Explanation:
We know that number of odd days in 100 years = 5
Hence last day of the first century is Friday
Number of odd days in 200 years = 5 x 2 = 10 = 3 (As we can reduce multiples of 7 from odd days which will not change anything)
Hence last day of the 2nd century is Wednesday
Number of odd days in 300 years = 5 x 3 = 15 = 1
Hence last day of the 3rd century is Monday
We know that umber of odd days in 400 years = 0. (∵ 5 x 4 + 1 = 21 = 0)
Hence last day of the 4th century is Sunday
Now this cycle will be repeated. Hence last day of a century will not be Tuesday or Thursday or Saturday.
its better to learn this by heart which will be helpful to save time in objective type exams
| 12. January 1, 2008 is Tuesday. What day of the week lies on Jan 1, 2009? | |
| A. Saturday | B. Wednesday |
| C. Thursday | D. Saturday |
answer with explanation
Answer: Option C
Explanation:
Number of odd days in 2008 = 2 (since it is a leap year)
(we have taken the complete year 2008 because we need to find out the odd days from 01-Jan-2008 to 31-Dec-2008, that is the whole year 2008)
Given that January 1, 2008 is Tuesday.
Hence January 1, 2009 = (Tuesday + 2 odd days) = Thursday
Explanation:
Number of odd days in 2008 = 2 (since it is a leap year)
(we have taken the complete year 2008 because we need to find out the odd days from 01-Jan-2008 to 31-Dec-2008, that is the whole year 2008)
Given that January 1, 2008 is Tuesday.
Hence January 1, 2009 = (Tuesday + 2 odd days) = Thursday
| 13. If Jan 1, 2006 was a Sunday, What was the day of the week Jan 1, 2010? | |
| A. Friday | B. Thursday |
| C. Tuesday | D. Saturday |
answer with explanation
Answer: Option A
Explanation:
Given that Jan 1 2006 was a Sunday
Number of odd days in the period 2006-2009
= 3 normal years + 1 leap year
= 3 x 1 + 1 x 2 = 5 (note that we have taken the complete year 2006 because the period in 2006 is from 01-Jan-2006 to 31-Dec-2006, which is the whole year 2006. Then the complete years 2007, 2008 and 2009 are also involved)
Hence Jan 1 2010 = (Sunday + 5 odd days) = Friday
Explanation:
Given that Jan 1 2006 was a Sunday
Number of odd days in the period 2006-2009
= 3 normal years + 1 leap year
= 3 x 1 + 1 x 2 = 5 (note that we have taken the complete year 2006 because the period in 2006 is from 01-Jan-2006 to 31-Dec-2006, which is the whole year 2006. Then the complete years 2007, 2008 and 2009 are also involved)
Hence Jan 1 2010 = (Sunday + 5 odd days) = Friday
| 14. What was the day of the week on 17th June 1998? | |
| A. Monday | B. Sunday |
| C. Wednesday | D. Friday |
answer with explanation
Answer: Option C
Explanation:
17 Jun 1998 = (1997 years + period from 1-Jan-1998 to 17-Jun-1998)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)
Number of odd days in the period 1601-1900
= Number of odd days in 300 years
= 5 x 3 = 15 = 1
(As we can reduce perfect multiples of 7 from odd days without affecting anything)
Number of odd days in the period 1901-1997
= 73 normal years + 24 leap year
= 73 x 1 + 24 x 2 = 73 + 48 = 121 = (121 - 7 x 17) = 2 odd days
Number of days from 1-Jan-1998 to 17-Jun-1998
= 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 31(may) + 17(Jun)
= 168 = 0 odd day
Total number of odd days = (0 + 1 + 2 + 0) = 3
3 odd days = Wednesday
Hence 17th June 1998 is Wednesday.
Explanation:
17 Jun 1998 = (1997 years + period from 1-Jan-1998 to 17-Jun-1998)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)
Number of odd days in the period 1601-1900
= Number of odd days in 300 years
= 5 x 3 = 15 = 1
(As we can reduce perfect multiples of 7 from odd days without affecting anything)
Number of odd days in the period 1901-1997
= 73 normal years + 24 leap year
= 73 x 1 + 24 x 2 = 73 + 48 = 121 = (121 - 7 x 17) = 2 odd days
Number of days from 1-Jan-1998 to 17-Jun-1998
= 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 31(may) + 17(Jun)
= 168 = 0 odd day
Total number of odd days = (0 + 1 + 2 + 0) = 3
3 odd days = Wednesday
Hence 17th June 1998 is Wednesday.
| 15. 6th March, 2005 is Monday, what was the day of the week on 6th March, 2004? | |
| A. Friday | B. Saturday |
| C. Wednesday | D. Sunday |
answer with explanation
Answer: Option D
Explanation:
Number of days from 6th March, 2004 to 5th March 2005 = 365 days
(Though Feb 2004 has 29 days as it is a leap year, it will not come in the required period)
365 days = 1 odd day
Given that 6th March, 2005 is Monday
Hence 6th March, 2004 = (Monday - 1 odd day) = Sunday
Explanation:
Number of days from 6th March, 2004 to 5th March 2005 = 365 days
(Though Feb 2004 has 29 days as it is a leap year, it will not come in the required period)
365 days = 1 odd day
Given that 6th March, 2005 is Monday
Hence 6th March, 2004 = (Monday - 1 odd day) = Sunday
| 16. What day of the week was 1 January 1901 | |
| A. Monday | B. Tuesday |
| C. Saturday | D. Friday |
answer with explanation
Answer: Option B
Explanation:
1 Jan 1901 = (1900 years + 1st Jan 1901)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)
Number of odd days in the period 1601-1900
= Number of odd days in 300 years
= 5 x 3 = 15 = 1
(As we can reduce perfect multiples of 7 from odd days without affecting anything)
1st Jan 1901 = 1 odd day
Total number of odd days = (0 + 1 + 1) = 2
2 odd days = Tuesday
Hence 1 January 1901 is Tuesday.
Explanation:
1 Jan 1901 = (1900 years + 1st Jan 1901)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)
Number of odd days in the period 1601-1900
= Number of odd days in 300 years
= 5 x 3 = 15 = 1
(As we can reduce perfect multiples of 7 from odd days without affecting anything)
1st Jan 1901 = 1 odd day
Total number of odd days = (0 + 1 + 1) = 2
2 odd days = Tuesday
Hence 1 January 1901 is Tuesday.
| 17. What day of the week will 22 Apr 2222 be? | |
| A. Monday | B. Tuesday |
| C. Sunday | D. Thursday |
answer with explanation
Answer: Option A
Explanation:
22 Apr 2222 = (2221 years + period from 1-Jan-2222 to 22-Apr-2222)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)
Number of odd days in the period 2001-2200
= Number of odd days in 200 years
= 5 x 2 = 10 = 3
(As we can reduce perfect multiples of 7 from odd days without affecting anything)
Number of odd days in the period 2201-2221
= 16 normal years + 5 leap years
= 16 x 1 + 5 x 2 = 16 + 10 = 26 = 5 odd days
Number of days from 1-Jan-2222 to 22 Apr 2222
= 31 (Jan) + 28 (Feb) + 31 (Mar) + 22(Apr) = 112
112 days = 0 odd day
Total number of odd days = (0 + 3 + 5 + 0) = 8 = 1 odd day
1 odd days = Monday
Hence 22 Apr 2222 is Monday.
Explanation:
22 Apr 2222 = (2221 years + period from 1-Jan-2222 to 22-Apr-2222)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)
Number of odd days in the period 2001-2200
= Number of odd days in 200 years
= 5 x 2 = 10 = 3
(As we can reduce perfect multiples of 7 from odd days without affecting anything)
Number of odd days in the period 2201-2221
= 16 normal years + 5 leap years
= 16 x 1 + 5 x 2 = 16 + 10 = 26 = 5 odd days
Number of days from 1-Jan-2222 to 22 Apr 2222
= 31 (Jan) + 28 (Feb) + 31 (Mar) + 22(Apr) = 112
112 days = 0 odd day
Total number of odd days = (0 + 3 + 5 + 0) = 8 = 1 odd day
1 odd days = Monday
Hence 22 Apr 2222 is Monday.
| 18. Today is Thursday. The day after 59 days will be? | |
| A. Monday | B. Tuesday |
| C. Saturday | D. Sunday |
answer with explanation
Answer: Option D
Explanation:
59 days = 8 weeks 3 days = 3 odd days
Hence if today is Thursday, After 59 days, it will be = (Thursday + 3 odd days)
= Sunday
Explanation:
59 days = 8 weeks 3 days = 3 odd days
Hence if today is Thursday, After 59 days, it will be = (Thursday + 3 odd days)
= Sunday
| 19. What is the year next to 1990 which will have the same calendar as that of the year 1990? | |
| A. 1992 | B. 2001 |
| C. 1995 | D. 1996 |
answer with explanation
Answer: Option B
Explanation:
For a year to have the same calendar with 1990 ,total odd days from 1990 should be 0.
Take the year 1992 from the given choices.
Total odd days in the period 1990-1991= 2 normal years
≡ 2 x 1 = 2 odd days
Take the year 1995 from the given choices.
Number of odd days in the period 1990-1994 = 4 normal years + 1 leap year
≡ 4 x 1 + 1 x 2 = 6 odd days
Take the year 1996 from the given choices.
Number of odd days in the period 1990-1995= 5 normal years + 1 leap year
≡ 5 x 1 + 1 x 2 = 7 odd days ≡ 0 odd days
(As we can reduce multiples of 7 from odd days which will not change anything)
Though number of odd days in the period 1990-1995 is 0, there is a catch here.
1990 is not a leap year whereas 1996 is a leap year.
Hence calendar for 1990 and 1996 will never be the same.
Take the year 2001 from the given choices.
Number of odd days in the period 1990-2000= 8 normal years + 3 leap years
≡ 8 x 1 + 3 x 2 = 14 odd days ≡ 0 odd days
Also, both 1990 and 2001 are normal years.
Hence 1990 will have the same calendar as that of 2001
Explanation:
For a year to have the same calendar with 1990 ,total odd days from 1990 should be 0.
Take the year 1992 from the given choices.
Total odd days in the period 1990-1991= 2 normal years
≡ 2 x 1 = 2 odd days
Take the year 1995 from the given choices.
Number of odd days in the period 1990-1994 = 4 normal years + 1 leap year
≡ 4 x 1 + 1 x 2 = 6 odd days
Take the year 1996 from the given choices.
Number of odd days in the period 1990-1995= 5 normal years + 1 leap year
≡ 5 x 1 + 1 x 2 = 7 odd days ≡ 0 odd days
(As we can reduce multiples of 7 from odd days which will not change anything)
Though number of odd days in the period 1990-1995 is 0, there is a catch here.
1990 is not a leap year whereas 1996 is a leap year.
Hence calendar for 1990 and 1996 will never be the same.
Take the year 2001 from the given choices.
Number of odd days in the period 1990-2000= 8 normal years + 3 leap years
≡ 8 x 1 + 3 x 2 = 14 odd days ≡ 0 odd days
Also, both 1990 and 2001 are normal years.
Hence 1990 will have the same calendar as that of 2001
| 20. January 1, 2004 was a Thursday, what day of the week lies on January 1 2005. | |
| A. Saturday | B. Monday |
| C. Saturday | D. Tuesday |
answer with explanation
Answer: Option C
Explanation:
Given that January 1, 2004 was Thursday.
Odd days in 2004 = 2 (because 2004 is a leap year)
(Also note that we have taken the complete year 2004 because we need to find out the odd days from 01-Jan-2004 to 31-Dec-2004, that is the whole year 2004)
Hence January 1, 2005 = (Thursday + 2 odd days) = Saturda
y
Explanation:
Given that January 1, 2004 was Thursday.
Odd days in 2004 = 2 (because 2004 is a leap year)
(Also note that we have taken the complete year 2004 because we need to find out the odd days from 01-Jan-2004 to 31-Dec-2004, that is the whole year 2004)
Hence January 1, 2005 = (Thursday + 2 odd days) = Saturda
y
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