Dates and calendar Question and answer

Dates and Calendars - Zeller's Rule Shortcut

Calculate the day of the week for any date. Zeller’s Rule can be used to find the day on any particular date in the calendar in the history. All you have to know is the formula given below and how to use it.

Zeller’s Rule Formula:

F = K + [(13xM - 1)/5] + D + [D/4] + [C/4] – 2C

where, K = Date, M = Month, C = The first two digits year and D = Last two digits of the year

*  In Zellers rule, months start from March. March = 1, April = 2, May = 3 and so on… till Dec = 10,

Jan = 11 Feb. = 12

*    Also remember that when you have to find day of the first or second month of any year, then Year=Given year-1 i.e., When you want to find Day of 15-2-1990., K=15, Month=12, D=Given Year-1=1990-1=1989=89

Ex: find the day of the 27-08-2014 ?

Sol: K=27,M=6,C=20 and D=14

Replacing the values in the formula, we get F = 27 + [{(13 x 6)- 1}/5] + 14 + 14/4 + 20/4 – (2 x 20)

Therefore, F = 27 + 77/5 + 14 + 14/4 + 20/4 –40 Which gives.. F =27 + 15.5 + 14 + 3.5 + 5 – 40

[ We have to Consider only the integral value and ignore the value after decimal. So, the equation changes a bit as shown below. We have just removed value after decimal ]

F = 27 + 15 + 14 + 3 + 5 – 40 Therefore, F = 3. Now that you have a numerical value for the day, divide the number by 7. We need the remainder only. For example, in this case, the remainder is 3.

Now, match the remainder with the chart below:

1 = Monday 2 = Tuesday 3 = Wednesday 4 = Thursday 5 = Friday 6 = Saturday 7 = Sunday

Here, 3 represents Wednesday .

So by Zeller’s rule, 27th of August, 2014 was on a Wednesday . So,Today is wednesday.

This formula will help you a lot in any Calendar question that you may encounter in Quant or DI. Remember that it is necessary to know the formula properly or else, even a little mistake can render the answer incorrect.

Above zeller rule is applicable to find exact day of given date. If you to find the day of the given date in a problem where a day is given on a specified date. then we have to follow the given below procedure:

MODULE OBJECTIVE:

The module calendar is used to find many problems related to odd days, leap year, and counting of odd days and many. The module clock is used to find many problems related to find angle between hour and minute hand of a clock, at what time the hands of clock will be together and many.

We are supposed to find the day of the week on a given date. Odd Days:For this, we use the concept of 'odd days'. In a given period, the number of days more than the complete weeks are called odd days.

Leap Year:

(i). Every year divisible by 4 is a leap year, if it is not a century.

(ii). Every 4th century is a leap year and no other century is a leap year. Note: A leap year has 366 days.

Examples:

i.  Each of the years 1948, 2004, 1676 etc. is a leap year.

ii.   Each of the years 400, 800, 1200, 1600, 2000 etc. is a leap year.

iii.   None of the years 2001, 2002, 2003, 2005, 1800, 2100 is a leap year. Ordinary Year: The year which is not a leap year is called an ordinary years. An ordinary year has 365 days.

Counting of Odd Days:

1.1 ordinary year = 365 days = (52 weeks + 1 day.)1 ordinary year has 1 odd day.

2.1 leap year = 366 days = (52 weeks + 2 days) so,1 leap year has 2 odd days.

3.100 years = 76 ordinary years + 24 leap years= (76 x 1 + 24 x 2) odd days = 124 odd days.

= (17 weeks + days)5 odd days. Number of odd days in 100 years = 5.

Number of odd days in 200 years = (5 x 2),3 odd days. Number of odd days in 300 years = (5 x 3),1 odd day. Number of odd days in 400 years = (5 x 4 + 1) 0 odd day.

Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc. has 0 odd days.

Some codes to remember the months and weeks of number of odd days :

a) Week
Day		-	Code
Sunday		–	1
Monday		–	2
Tuesday		–	3
Wednesday –			4
Thursday		–	5
Friday		–	6
Saturday		–	0
b) Month
Jan	–1
July	–0
feb	–4
Aug	–3
Mar	–4
Sep	–6
Apr	–0
Oct	–1
May	–2
Nov	–4
june	–5
Dec	–6

Dates and calendar question and answer:

1. It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?

A.	Sunday	B.	Saturday
C.	Friday	D.	Wednesday

Answer: Option C

Explanation:

On 31^st December, 2005 it was Saturday.

Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.

On 31^st December 2009, it was Thursday.

Thus, on 1^st Jan, 2010 it is Friday.

2.What was the day of the week on 28^th May, 2006?

A.	Thursday	B.	Friday
C.	Saturday	D.	Sunday

Answer: Option D

Explanation:

28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006) Odd days in 1600 years = 0

Odd days in 400 years = 0

5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2)  6 odd days

Jan.	Feb.	March	April	May
(31 +	28	+  31	+   30	+   28	) = 148 days
148 days = (21 weeks + 1 day)				1 odd day.

Total number of odd days = (0 + 0 + 6 + 1) = 7  0 odd day.

Given day is Sunday.

3.What was the day of the week on 17^th June, 1998?

A.	Monday	B.	Tuesday
C.	Wednesday	D.	Thursday

Answer: Option C

Explanation:

17^th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)

Odd days in 1600 years = 0

Odd days in 300 years = (5 x 3)  1

97 years has 24 leap years + 73 ordinary years.

Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.

Jan.	Feb.	March	April	May	June
(31 +	28	+  31	+   30	+   31	+   17)	= 168 days

 168 days = 24 weeks = 0 odd day.

Total number of odd days = (0 + 1 + 2 + 0) = 3.

Given day is Wednesday.

4.What will be the day of the week 15^th August, 2010?

A.	Sunday	B.	Monday
C.	Tuesday	D.	Friday

Answer: Option A

Explanation:

15^th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010) Odd days in 1600 years = 0

Odd days in 400 years = 0

9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days  4 odd days.

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.
Total number of odd days = (0 + 0 + 4 + 3) = 7						0 odd days.
Given day is Sunday.

5.Today is Monday. After 61 days, it will be:

A.	Wednesday	B.	Saturday
C.	Tuesday	D.	Thursday

Answer: Option B

Explanation:

Each day of the week is repeated after 7 days.

So, after 63 days, it will be Monday.

 After 61 days, it will be Saturday.

6. If 6^th March, 2005 is Monday, what was the day of the week on 6^th March, 2004?

A.	Sunday	B.	Saturday
C.	Tuesday	D.	Wednesday

Answer: Option A

Explanation:

The year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.

 The day on 6^th March, 2005 will be 1 day beyond the day on 6^th March, 2004.

Given that, 6^th March, 2005 is Monday.

 6^th March, 2004 is Sunday (1 day before to 6^th March, 2005).

7.On what dates of April, 2001 did Wednesday fall?

A.	1st, 8th, 15th, 22nd, 29th	B.	2nd, 9th, 16th, 23rd, 30th
C.	3rd, 10th, 17th, 24th	D.	4th, 11th, 18th, 25th

Answer: Option D

Explanation:

We shall find the day on 1^st April, 2001.

1^st April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001) Odd days in 1600 years = 0

Odd days in 400 years = 0

Jan. Feb. March April (31 + 28 + 31 + 1) = 91 days  0 odd days.

Total number of odd days = (0 + 0 + 0) = 0

On 1^st April, 2001 it was Sunday.

In April, 2001 Wednesday falls on 4^th, 11^th, 18^th and 25^th.

8.How many days are there in x weeks x days?
A.	7x2	B.	8x
C.	14x	D.	7

Answer: Option B

Explanation:

x weeks x days = (7x + x) days = 8x days.

9.The last day of a century cannot be
A.	Monday	B.	Wednesday
C.	Tuesday	D.	Friday

Answer: Option C

Explanation:

100 years contain 5 odd days.

 Last day of 1^st century is Friday.

200 years contain (5 x 2)  3 odd days.

 Last day of 2^nd century is Wednesday. 300 years contain (5 x 3) = 15  1 odd day.

 Last day of 3^rd century is Monday. 400 years contain 0 odd day.

 Last day of 4^th century is Sunday. This cycle is repeated.

 Last day of a century cannot be Tuesday or Thursday or Saturday.

10.On 8^th Feb, 2005 it was Tuesday. What was the day of the week on 8^th Feb, 2004?

A.	Tuesday	B.	Monday
C.	Sunday	D.	Wednesday

Answer: Option C

Explanation:

The year 2004 is a leap year. It has 2 odd days.

 The day on 8^th Feb, 2004 is 2 days before the day on 8^th Feb, 2005. Hence, this day is Sunday.


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