1st 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.
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
Solution:
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 daysTake 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 daysTake 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
34.
Today is Thursday. What will be the day of the week after 94 days?
(A) Sunday
(B) Monday
(C) Tuesday
(D) Wednesday
Solution:
94 days = (13 × 7) + 3 = 3 odd days. The required day is 3 days beyond Thursday i.e., Sunday
35.
If today is Monday, what will be the day 350 days from now?
(A) Monday
(B) Tuesday
(C) Wednesday
(D) Thursday
Solution:
350 days, $$\frac{{350}}{7}$$ = 50, no odd days, so it will be a Monday.
36.
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
37.
What was the day of the week on 28th May, 2006?
(A) Thursday
(B) Friday
(C) Saturday
(D) Sunday
Solution:
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) ≡ 6 odd days Jan. Feb. March April May (31 + 28 + 31 + 30 + 28 ) = 148 days ∴ 148 days = (21 weeks + 1 day) ≡ 1 odd day. Total number of odd days = (0 + 0 + 6 + 1) = 7 ≡ 0 odd day. Given day is Sunday.
38.
If Feb 12th,1986 falls on Wednesday then Jan 1st,1987 falls on which day?
(A) Wednesday
(B) Tuesday
(C) Thursday
(D) Friday
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.
39.
What will be the day of the week 15th August, 2010?
(A) Sunday
(B) Monday
(C) Tuesday
(D) Friday
Solution:
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 ≡ 4 odd days. Jan. Feb. March April Mayb June July Aug. (31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days ∴ 227 days = (32 weeks + 3 days) ≡ 3 odd days. Total number of odd days = (0 + 0 + 4 + 3) = 7 ≡ 0 odd days. Given day is Sunday.
40.
Today is 3rd November. The day of the week is Monday. This is a leap year. What will be the day of the week on this date after 3 years?
(A) Monday
(B) Tuesday
(C) Wednesday
(D) Thursday
Solution:
This is a leap year. So, none of the next 3 years will be leap years. Each year will give one odd day so the day of the week will be 3 odd days beyond Monday i.e. it will be Thursday.