Clock and calender
 If 2^{nd} June 2013 is Sunday then which day was on 2^{nd} June 2010?

View Hint View Answer Discuss in Forum
Consider from 2^{nd} June 2010 to 2^{nd} June 2013 we have total 2 non leap year and one leap year so number of odd days are 1 + 1 + 2 = 4 so 2^{nd} June 2010 must be 4 days back from Sunday and that day is Wednesday.
From Zeller's Formula:
f = k + [13 x m  1/ 5 ] + D + [D/4] + [C/4]  2 x C.
In this case k = 2 (since 2^{nd} June)
Month m = 4 (As march = 1, April = 2, May = 3, June = 4 )
D is the last two digit of year here D = 10 (As year is 2010)
C is 1 st two digit of century here C = 20 (As year is 2010)
f = 2 + [13 x 4  1 / 5] + 10 + [10/ 4] + [ 20/4]  2 x 20.
f = 2 + [51/5] + 10 +[2.5] + [5]  40.
f = 2 + 10 + 10 + 2 + 5  40 = 11
This  ve value of f can be made positive by adding multiple of 7
So f =  11 + 14 = 3
When divided by 7 we will get remainder 3, hence number of odd days is 3,
So 2^{nd} June 2010 is 3 days more than Monday, i.e Wednesday.Correct Option: A
Consider from 2^{nd} June 2010 to 2^{nd} June 2013 we have total 2 non leap year and one leap year so number of odd days are 1 + 1 + 2 = 4 so 2^{nd} June 2010 must be 4 days back from Sunday and that day is Wednesday.
From Zeller's Formula:
f = k + [13 x m  1/ 5 ] + D + [D/4] + [C/4]  2 x C.
In this case k = 2 (since 2^{nd} June)
Month m = 4 (As march = 1, April = 2, May = 3, June = 4 )
D is the last two digit of year here D = 10 (As year is 2010)
C is 1 st two digit of century here C = 20 (As year is 2010)
f = 2 + [13 x 4  1 / 5] + 10 + [10/ 4] + [ 20/4]  2 x 20.
f = 2 + [51/5] + 10 +[2.5] + [5]  40.
f = 2 + 10 + 10 + 2 + 5  40 = 11
This  ve value of f can be made positive by adding multiple of 7
So f =  11 + 14 = 3
When divided by 7 we will get remainder 3, hence number of odd days is 3,
So 2^{nd} June 2010 is 3 days more than Monday, i.e Wednesday.
 If 5^{th} march of a particular year is Friday then which day of the week will be on 5^{th} November.

View Hint View Answer Discuss in Forum
Here we have to find the number of odd days between, 5^{th} march and 5^{th} November,
Number of days in March is 26 or 5 odd days
(Here we have not included 5^{th} march)
Number of days in April is 30 or 2 odd days
Number of days in May is 31 or 3 odd days
Number of days in June is 30 or 2 odd days
Number of days in July is 31 or 3 odd days
Number of days in August is 31 or 3 odd days
Number of days in September is 30 or 2 odd days
Number of days in October is 31 or 3 odd days
Number of days in November is 5 or 5 odd days
(Here 5^{th} November is included)
So total number of odd days = 5 + 2 + 3 + 2 + 3 + 3 +2 + 3 + 5 = 28 when divided by 7 gives remainder 0 hence 5^{th} November will be same as that of 5^{th} march.Correct Option: B
Here we have to find the number of odd days between, 5^{th} march and 5^{th} November,
Number of days in March is 26 or 5 odd days
(Here we have not included 5^{th} march)
Number of days in April is 30 or 2 odd days
Number of days in May is 31 or 3 odd days
Number of days in June is 30 or 2 odd days
Number of days in July is 31 or 3 odd days
Number of days in August is 31 or 3 odd days
Number of days in September is 30 or 2 odd days
Number of days in October is 31 or 3 odd days
Number of days in November is 5 or 5 odd days
(Here 5^{th} November is included)
So total number of odd days = 5 + 2 + 3 + 2 + 3 + 3 +2 + 3 + 5 = 28 when divided by 7 gives remainder 0 hence 5^{th} November will be same as that of 5^{th} march.
 Which of the following day could be the 18^{th} October 2050?

View Hint View Answer Discuss in Forum
From Zeller's Formula
f = k + [13 x m  1 / 5] + D + [D/4] + [C/4]  2 x C.
In this case k = 18 (since 18^{th} October)
Month m = 8 (As march = 1, April = 2, May = 3, October = 8)
D is the last two digit of year here D = 50 (As year is 2050)
C is the 1^{st} two digit of century here C = 20 (As year is 1950)
f = 18 + [13 x 8  1 / 5] + 50 + [50/4] + [20/4]  2 x 20.
f = 18 + [103/5] + 50 + [12.5] + [5]  40.
f = 18 + 20 + 50 + 12 + 5  40 = 65.
When divided by 7 we will get remainder 2, hence number of odd days is 2,
So 18^{th} October 2050 is 2 days more than Sunday, i.e Tuesday.Correct Option: A
From Zeller's Formula
f = k + [13 x m  1 / 5] + D + [D/4] + [C/4]  2 x C.
In this case k = 18 (since 18^{th} October)
Month m = 8 (As march = 1, April = 2, May = 3, October = 8)
D is the last two digit of year here D = 50 (As year is 2050)
C is the 1^{st} two digit of century here C = 20 (As year is 1950)
f = 18 + [13 x 8  1 / 5] + 50 + [50/4] + [20/4]  2 x 20.
f = 18 + [103/5] + 50 + [12.5] + [5]  40.
f = 18 + 20 + 50 + 12 + 5  40 = 65.
When divided by 7 we will get remainder 2, hence number of odd days is 2,
So 18^{th} October 2050 is 2 days more than Sunday, i.e Tuesday.
 Which of the following day of the week was 15^{th} august 1947?

View Hint View Answer Discuss in Forum
From Zeller's Formula:
f = k + [13 x m  1 / 5] + D + [D/4] + [C/4]  2 x C.
In this case k = 15 (since 15^{th} August)
Month m = 6 (As march = 1, April = 2, May = 3, August = 6)
D is the last two digit of year here D = 47 (As year is 1947)
C is the 1^{st} two digit of century here C = 19 (As year is 1947)
f = 15 + [13 x 6  1 / 5] + 47 + [47/4] + [19/4]  2 x 19.
f = 15 + [77/5] + 47 + [11.75] + [4.75]  38.
f = 15 + 15 + 47 + 11 + 4  38 = 54.
When divided by 7 we will get remainder 5, hence number of odd days is 3,
A remainder of 0 corresponds to Sunday, 1 means Monday,
So 15^{th} August 1947 is 5 days more than Sunday, i.e Friday.Correct Option: B
From Zeller's Formula:
f = k + [13 x m  1 / 5] + D + [D/4] + [C/4]  2 x C.
In this case k = 15 (since 15^{th} August)
Month m = 6 (As march = 1, April = 2, May = 3, August = 6)
D is the last two digit of year here D = 47 (As year is 1947)
C is the 1^{st} two digit of century here C = 19 (As year is 1947)
f = 15 + [13 x 6  1 / 5] + 47 + [47/4] + [19/4]  2 x 19.
f = 15 + [77/5] + 47 + [11.75] + [4.75]  38.
f = 15 + 15 + 47 + 11 + 4  38 = 54.
When divided by 7 we will get remainder 5, hence number of odd days is 3,
A remainder of 0 corresponds to Sunday, 1 means Monday,
So 15^{th} August 1947 is 5 days more than Sunday, i.e Friday.
 How many weekends are there in March 2009?

View Hint View Answer Discuss in Forum
From Zeller's Formula we can find that 1^{st} March 2009 is Sunday so well have 5 Saturdays and 5 Sundays in total 10 weekends.
Correct Option: B
From Zeller's Formula we can find that 1^{st} March 2009 is Sunday so well have 5 Saturdays and 5 Sundays in total 10 weekends.