Answer: C 14th march 1993 means (1992, 2 months and 14 days) Now 1900 years have 1 odd day. 92 years have 23 leap years and 69 ordinary years => 46+63 = 115 days = 3 odd days. 2 months + 14 days = 31+28+14 =73 = 3 odd days. Total number of odd days = (1+3+3) = 7 odd days. Therefore it will be Sunday on 14th March 1993.
Answer: D 1st December 1984 => 1983 years 11 months and 1 day Now, 1990 years have 1 odd day 83 year have (20 leap years and 63 ordinary years) => 40 +63 = 103 odd days => 5 odd days 11 month and 1 day. 0=336 days => 0 odd days. Total number of odd days = 1+5+0 =6. Thus, its was Saturday on 1st December 1984 and 1st Sunday was on 2nd December 1984. Subsequently Sunday's of the month were on 2nd, 9th, 16th, 23rd and 30th.
Answer: C 2nd july 1984 means => 1983 years 6 months and 2 days. 1900 years have 1 odd day 83 year have 20 leap years and 63 ordinary years. Thus, 40+63 = 103 odd days = 5 odd days. 6 month and 2 days = 184 days = 2 odd days Total number of odd days = (1+5+2) = 8 odd days = 1 odd day. Therefore it was Monday on 2nd july 1984.
Q. No. 4:
The year next to 1990 will have the same calender as that of the year 1990
Answer: B The year 1990 has 365 days i.e 1 odd day, year 1991 has 365 day i.e 1 odd day, year 1992 has 366 days i.e 2 odd days. Likewise year 1993,1994,1995 has 1 odd days. Total number of odd days (1990 - 1995) = (1+1+2+1+1+1) = 7 odd days = 0 odd days. Hence, year 1996 will have the same calender as that of year 1990.
Q. No. 5:
An application was received by inward clerk in the afternoon of a week day. Next day he forwarded it to the table of the senior clerk, who was on leave that day. The senior clerk put up the application to the desk officer next day in the evening. The desk officer studied the application and disposed off the matter on the same day i.e Friday. Which day was the application received by the inward clerk?
Answer: B 100 years contain 5 odd days.
Last day of 1st century is Friday 200 years contain (5 x 2) =>3 odd days.
Last day of 2nd century is Wednesday.
300 years contain (5 x 3) = 15=> 1 odd day.
Last day of 3rd century is Monday.
400 years contain 0 odd day.
Last day of 4th century is Sunday.
This cycle is repeated.
Last day of a century cannot be Tuesday or Thursday or Saturday.