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
2.
The calendar of year 1939 is same as which year?
(A) 1943
(B) 1964
(C) 1950
(D) 1956
Solution:
Given year 1939, when divided by 4 leaves a remainder of 3. NOTE: When remainder is 3, 11 is added to the given year to get the result. So, 1939 + 11 = 1950
3.
The year next to 2005 will have the same calendar as that of the year 2005?
(A) 2016
(B) 2022
(C) 2011
(D) None
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
4.
On 8th Feb, 2005 it was Tuesday. What was the day of the week on 8th Feb, 2004?
(A) Tuesday
(B) Monday
(C) Sunday
(D) Wednesday
Solution:
The year 2004 is a leap year. It has 2 odd days. ∴ The day on 8th Feb, 2004 is 2 days before the day on 8th Feb, 2005. Hence, this day is Sunday.
5.
If the first day of a year (other than leap year) was Friday, then which was the last day of that year?
(A) Wednesday
(B) Thursday
(C) Friday
(D) Saturday
Solution:
Given that first day of a normal year was Friday Odd days of the mentioned year = 1 (Since it is an ordinary year) Hence First day of the next year = (Friday + 1 Odd day) = Saturday Therefore, last day of the mentioned year = Friday
6.
It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
(A) Sunday
(B) Saturday
(C) Friday
(D) Wednesday
Solution:
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. ∴ On 31st December 2009, it was Thursday. Thus, on 1st Jan, 2010 it is Friday.
7.
On what dates of July. 2004 did Monday fall?
(A) 6th, 10th, 21th, 30th
(B) 12th, 7th, 19th, 28th
(C) 5th, 10th, 24th, 17th
(D) 5th, 12th, 19th, 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.
8.
If 1st October is Sunday, then 1st November will be
(A) Monday
(B) Tuesday
(C) Wednesday
(D) Thursday
Solution:
Given that 1st October is Sunday Number of days in October = 31 31 days = 3 odd days (As we can reduce multiples of 7 from odd days which will not change anything) Hence 1st November = (Sunday + 3 odd days) = Wednesday
9.
The maximum gap between two successive leap year is?
(A) 4
(B) 8
(C) 2
(D) 1
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$$\frac{1}{4}$$ 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.
10.
Which calendar was used again in the year 1856?
(A) 1828
(B) 1850
(C) 1830
(D) 1852
Solution:
Given year 1856, when divided by 4 leaves a remainder of 0. NOTE: When remainder is 0, 28 is subtracted to the given year to get the result. So, 1856 - 28 = 1828