Calender
Questions
Q1.
Today is 21st August . The day of the week is Monday . This is a leap year. What will be the day of the week on this day after 3
years ?
Solution.
Since this is a leap year, none of the next 3 years is a leap year.So, the day of the week will be 3 days beyond
Monday, i.e. it will be Thursday.
Solution
Q2.
It was Thursday on 2nd Jan 1993. What day of the week will be on 14th March 1993.
Total No. of days = 29(Jan) + 28(Feb) + 14 (March)
= 72 days = 10 weeks + 1 day
= 1 odd days
Thus the
given date will fall on one day beyond first day , So the day will be Friday .
Solution
Q3.
The first Republic Day of India was celebrated on 26th January 1950. What was the day of the week on that date ?
Solution.
Total no. of odd days = 1600 yrs have 0 odd day
+ 300 yrs have 1 odd day + 49 yrs (12 leap + 37 ordinary)have
5 odd days + 26 days of Jan have 5 odd days
= 0 + 1 + 5 + 5 = 4 odd days
So, the day was Thursday .
Solution
Q4.
200 years contains _______________ odd days .
(a) 1 (b) 2 (c) 3 (d) 4
Solution.
3 odd days
Note : As we know that 100 yrs has 76 ordinary years and 24 leap years .
Thus 100 years contains 5 odd
days
Solution
Q5.
On Jan 12, 1980, it was Saturday . The day of the week on Jan 12 1979 was :
(a) Saturday (b) Sunday (c) Thursday (d) Friday (e) Monday
Solution.
1979 is an ordinary year and Jan 12 , 1980 lies before the month of Feb, so the whole year has one odd day . Therefore , the day
one day before Saturday, i.e. Friday .
Solution
