Calendar - Solved Examples

Answer - B

Explanation

15th June 1776 = (1775 years + Period from 01.01.1776 to 15.06.1776)
Counting of odd days:
No of odd days in 1600 years = 0
No of odd days in 100 years = 5
75 years = 18 leap years + 57 ordinary years
= 18*2 + 57*1
= 36 + 57
= 93 odd days
= 13 weeks + 2 odd days = 2 odd days
∴ 1775 years have (0+5+2) = 7 odd days = 0 odd days.

Jan to May = (31+29+31+30+31)
= 152 days
Add 15 days of June.
= 152 + 15
= 167 days
= 23 weeks + 6 days
= 6 odd days.
∴ Total number of odd days = 0 + 6 = 6 odd days.
Hence 15.06.1776 was Saturday.

Answer - A

Explanation

1997, 1998 and 1999 are not leap years.
1998 and 1999 has 2 odd days.

No of days remaining in 1997 = 365 - 15 = 350
= 50 weeks of 0 odd days.

05.01.2000 = 5 odd days.
Total no of odd days = 2 + 0 + 5 = 7
7 days from Wednesday is Wednesday. 
∴ Jan 5, 2000 was also Wednesday.

Answer - A

Explanation

We will count the no of odd days from the year 2007 onwards to get the sum equal to 0 odd days.

Answer - B

Explanation

We must have same day on 1.1.2003 and 1.1.2014. 
Along these lines, number of odd days somewhere around 31.12.2002 and 
31.12.2013 must be 0. This period has 3 jump years and 8 common years. 
Number of odd days = (3*2+8*1) =14=0 odd days. 
∴ Calendar for the year 2003 will serve for the year 2014.

Answer - C

Explanation

fifteenth Aug.1947 =(1946 years +period from 1.1.1947 to 15.8.1947) 
Odd days in 1600 years =0 
Odd days in 300 years = (5*3) =15 =1946 years = (11 jump years+35 customary years) 
= (11*2 +35*1) odd days= 57 days 
= (8 weeks +1 day) = 1 odd day 

∴ odd days in 1946 years= (0+1+1) =2 

Jan + Feb. + March + April + May + June + July + Aug 
(31 + 28 +31+ 30 + 31 +30+31+15) = 227 days 

227 days = (32 weeks +3 days) = 3 odd days. 
Aggregate no. of odd days = (2+3) = 5 

Consequently the required day is Friday.