## Clock and calender

#### Clock and calender

1. If 2nd June 2013 is Sunday then which day was on 2nd June 2010?

1. Consider from 2nd June 2010 to 2nd June 2013 we have total 2 non leap year and one leap year so number of odd days are 1 + 1 + 2 = 4 so 2nd 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 2nd 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 2nd June 2010 is 3 days more than Monday, i.e Wednesday.

##### Correct Option: A

Consider from 2nd June 2010 to 2nd June 2013 we have total 2 non leap year and one leap year so number of odd days are 1 + 1 + 2 = 4 so 2nd 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 2nd 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 2nd June 2010 is 3 days more than Monday, i.e Wednesday.

1. If 5th march of a particular year is Friday then which day of the week will be on 5th November.

1. Here we have to find the number of odd days between, 5th march and 5th November,
Number of days in March is 26 or 5 odd days
(Here we have not included 5th 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 5th 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 5th November will be same as that of 5th march.

##### Correct Option: B

Here we have to find the number of odd days between, 5th march and 5th November,
Number of days in March is 26 or 5 odd days
(Here we have not included 5th 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 5th 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 5th November will be same as that of 5th march.

1. Which of the following day could be the 18th October 2050?

1. 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 18th 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 1st 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 18th 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 18th 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 1st 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 18th October 2050 is 2 days more than Sunday, i.e Tuesday.

1. Which of the following day of the week was 15th august 1947?

1. 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 15th 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 1st 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 15th 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 15th 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 1st 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 15th August 1947 is 5 days more than Sunday, i.e Friday.

1. How many weekends are there in March 2009?

1. From Zeller's Formula we can find that 1st 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 1st March 2009 is Sunday so well have 5 Saturdays and 5 Sundays in total 10 weekends.