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51. How many days are there from 3rd February, 2012 to 18th April 2012 (both inclusive)?
The given year is leap year, So February month has 29 days
From 3rd to 29th February = 27 days
In March = 31 days
From 1st to 18th April = 18 days
Total number of days = 27 + 31 + 18 = 76 days
Solution:
Here we have to count the number days from 3rd February, 2012 to 18th April 2012 ( both inclusive)The given year is leap year, So February month has 29 days
From 3rd to 29th February = 27 days
In March = 31 days
From 1st to 18th April = 18 days
Total number of days = 27 + 31 + 18 = 76 days
52. What was the day on 15th august 1947 ?
Odd days in 1600 years = 0
Odd days in 300 years = 1
46 years
= (35 ordinary years + 11 leap years)
= (35 x 1 + 11 x 2)
= 57 (8 weeks + 1 day)
= 1 odd day
Jan. Feb. Mar. Apr. May. Jun. Jul. Aug
( 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 )
= 227 days
= (32 weeks + 3 days)
= 3 odd days
Total number of odd days = (0 + 1 + 1 + 3) = 5 odd days
Hence, as the number of odd days = 5, given day is Friday.
Solution:
15th Aug, 1947 = (1946 years + Period from 1.1.1947 to 15.8.1947)Odd days in 1600 years = 0
Odd days in 300 years = 1
46 years
= (35 ordinary years + 11 leap years)
= (35 x 1 + 11 x 2)
= 57 (8 weeks + 1 day)
= 1 odd day
Jan. Feb. Mar. Apr. May. Jun. Jul. Aug
( 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 )
= 227 days
= (32 weeks + 3 days)
= 3 odd days
Total number of odd days = (0 + 1 + 1 + 3) = 5 odd days
Hence, as the number of odd days = 5, given day is Friday.
53. What was the day of the week on, 16th July, 1776?
Counting of odd days :
1600 years have 0 odd day
100 years have 5 odd days
75 years = (18 leap years + 57 ordinary years)
= [(18 x 2) + (57 x 1)]
= 93 (13 weeks + 2 days)
= 2 odd days
1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day
Jan Feb Mar Apr May Jun Jul
31 + 29 + 31 + 30 + 31 + 30 + 16
= 198 days
= (28 weeks + 2 days)
Total number of odd days = (0 + 2) = 2
Required day was 'Tuesday'.
Solution:
16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to 16th July, 1776)Counting of odd days :
1600 years have 0 odd day
100 years have 5 odd days
75 years = (18 leap years + 57 ordinary years)
= [(18 x 2) + (57 x 1)]
= 93 (13 weeks + 2 days)
= 2 odd days
1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day
Jan Feb Mar Apr May Jun Jul
31 + 29 + 31 + 30 + 31 + 30 + 16
= 198 days
= (28 weeks + 2 days)
Total number of odd days = (0 + 2) = 2
Required day was 'Tuesday'.
54. The maximum gap between two successive leap year is?
Ex: 1896 is a leap year.The next leap year comes in 1904 (1900 is not a leap year).
Explanation : The length of the solar year, however, is slightly less than 365 days-by about 11 minutes. To compensate for this discrepancy, the leap year is omitted three times every four hundred years.
In other words, a century year cannot be a leap year unless it is divisible by 400. Thus 1700, 1800, and 1900 were not leap years, but 1600, 2000, and 2400 are leap years.
Solution:
This can be illustrated with an example.Ex: 1896 is a leap year.The next leap year comes in 1904 (1900 is not a leap year).
Explanation : The length of the solar year, however, is slightly less than 365 days-by about 11 minutes. To compensate for this discrepancy, the leap year is omitted three times every four hundred years.
In other words, a century year cannot be a leap year unless it is divisible by 400. Thus 1700, 1800, and 1900 were not leap years, but 1600, 2000, and 2400 are leap years.
55. How many leap years does 100 years have?
Here, 100 ÷ 4 = 25
But, as 100 is not a leap year ⇒ 25 - 1 = 24 leap years.
Solution:
Given year is divided by 4, and the quotient gives the number of leap years.Here, 100 ÷ 4 = 25
But, as 100 is not a leap year ⇒ 25 - 1 = 24 leap years.
56. Which two months in a year have the same calendar?
(a). Oct + Nov = 31 + 30 = 61 (not divisible by 7)
(b). Apr + May + Jun + Jul + Aug + Sep + Oct = 30 + 31 + 30 + 31 + 31 + 30 + 31 = 214 (not divisible by 7)
(c). Jun + July + Aug + Sep = 30 + 31 + 31 + 30 = 122 (not divisible by 7)
(d). Apr + May + June = 30 + 31 + 30 = 91 (divisible by 7)
Hence, April and July months will have the same calendar.
Solution:
If the period between the two months is divisible by 7, then that two months will have the same calendar.(a). Oct + Nov = 31 + 30 = 61 (not divisible by 7)
(b). Apr + May + Jun + Jul + Aug + Sep + Oct = 30 + 31 + 30 + 31 + 31 + 30 + 31 = 214 (not divisible by 7)
(c). Jun + July + Aug + Sep = 30 + 31 + 31 + 30 = 122 (not divisible by 7)
(d). Apr + May + June = 30 + 31 + 30 = 91 (divisible by 7)
Hence, April and July months will have the same calendar.
57. How many leap years do 300 years have?
Here, 300 ÷ 4 = 75
But, as 100, 200 and 300 are not leap years ⇒ 75 - 3 = 72 leap years.
Solution:
Given year is divided by 4, and the quotient gives the number of leap years.Here, 300 ÷ 4 = 75
But, as 100, 200 and 300 are not leap years ⇒ 75 - 3 = 72 leap years.
58. On what dates of July. 2004 did Monday fall?
2000 years have 0 odd day. 3 ordinary years have 3 odd days.
Jan. Feb. March April May June July
31 + 29 + 31 + 30 + 31 + 30 + 1
= 183 days
= (26 weeks + 1 day)
Total number of odd days = (0 + 3 + 1) odd days = 4 odd days.
∴ 1st July 2004 was 'Thursday'
Thus, 1st Monday in July 2004 as on 5th July.Hence, during July 2004, Monday fell on 5th, 12th, 19th and 26th.
Solution:
Let us find the day on 1st July, 2004.2000 years have 0 odd day. 3 ordinary years have 3 odd days.
Jan. Feb. March April May June July
31 + 29 + 31 + 30 + 31 + 30 + 1
= 183 days
= (26 weeks + 1 day)
Total number of odd days = (0 + 3 + 1) odd days = 4 odd days.
∴ 1st July 2004 was 'Thursday'
Thus, 1st Monday in July 2004 as on 5th July.Hence, during July 2004, Monday fell on 5th, 12th, 19th and 26th.
59. The year next to 2005 will have the same calendar as that of the year 2005?
Repetition of leap year ⇒ Add + 28 to the Given Year.
Repetition of non leap year
Step 1 : Add + 11 to the Given Year. If Result is a leap year, Go to step 2.
Step 2: Add + 6 to the Given Year.
Solution :
Given Year is 2005, Which is a non leap year.
Step 1 : Add + 11 to the given year (i.e 2005 + 11) = 2016, Which is a leap year.
Step 2 : Add + 6 to the given year (i.e 2005 + 6) = 2011
Therefore, The calendar for the year 2005 will be same for the year 2011
Solution:
NOTE :Repetition of leap year ⇒ Add + 28 to the Given Year.
Repetition of non leap year
Step 1 : Add + 11 to the Given Year. If Result is a leap year, Go to step 2.
Step 2: Add + 6 to the Given Year.
Solution :
Given Year is 2005, Which is a non leap year.
Step 1 : Add + 11 to the given year (i.e 2005 + 11) = 2016, Which is a leap year.
Step 2 : Add + 6 to the given year (i.e 2005 + 6) = 2011
Therefore, The calendar for the year 2005 will be same for the year 2011
60. If Feb 12th,1986 falls on Wednesday then Jan 1st,1987 falls on which day?
Here,given period is 12.2.1986 to 1.1.1987
Feb Mar Apr May June July Aug Sept Oct Nov Dec Jan
16 31 30 31 30 31 31 30 31 30 31 1 (left days)
2 + 3 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 1(odd days)
= 1 odd day
So, given day Wednesday + 1 = Thursday is the required result.
Solution:
First,we count the number of odd days for the left over days in the given period.Here,given period is 12.2.1986 to 1.1.1987
Feb Mar Apr May June July Aug Sept Oct Nov Dec Jan
16 31 30 31 30 31 31 30 31 30 31 1 (left days)
2 + 3 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 1(odd days)
= 1 odd day
So, given day Wednesday + 1 = Thursday is the required result.