# Calendar Questions and Answers – Quantitative Aptitude

Calendar Questions and AnswersFormulas & Tricks

Odd Days:

We are supposed to find the day of the week on a given date. So, We use the concept “Odd Days”

The No.of.days more than the complete weeks are called Odd Days.

Leap Year:

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

Every 4th century is a leap year and no other century is a leap year.

Note: A Leap year has 366 days.

Example:

• Each of the years 1948, 2004, 1676 etc., is a leap year.
• Each of the Years 400, 800, 1200, 1600, 2000 etc, is a leap year.
• None of the Years 2001, 2002, 2003, 2005, 1800, 2100 is a leap year.

Counting of Odd Days

• 1 Ordinary Year = 365 days =(52 weeks + 1 day)

Therefore, 1 Ordinary Year has 1 odd day.

• 1 Leap Year = 366 days = (52 weeks + 2 days)

Therefore 1 Leap Year has 2 Odd Days

• 100 Years = 76 Ordinary Years + 24 Leap Years

= (76 * 1 + 24 * 2) odd days = 124 odd days

= (17 weeks + days) = 5 odd days

Therefore, No.of. Odd days in 100 Years = 5.

No. of. Odd days in 200 Years = (5 * 2) = 3 odd days

No. of. Odd days in 300 Years = (5 * 3) = 1 odd days

No. of. Odd days in 400 Years = (5 * 4 + 1) = 0 odd days

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

• Day of the Week Related to Odd Days:

Calendar Questions and Answers With Detailed ExplanationQuantitative Aptitude

## Calendar

 Question 1
The First Republic Day of India was celebrated on January 26, 1950. What day of the Week was it?
 A Wednesday B Friday C Thursday D Tuesday
Question 1 Explanation:
Explanation:
Number of Odd days in 400/800/1200/1600/2000 Years are 0
Number of odd days in 300 Years = 1
49 Years = (12 Leap Years + 37 Years)
= (12 * 2, odd days + 37*1, odd days)
=24 + 37 = 61 odd days
On dividing 61 by 7, We get remainder 5,
Total Number of odd days in 1949 Years = 1+5 = 6odd days
Now, look at the Year 1950
Jan 26 = 26 days = 3 weeks + 5 days = 5 odd days
Total Number of Odd days = 6 + 5 = 11
= 4 Odd days

Odd days
0 = Sunday
1 = Monday
2 = Tuesday
3 = Wednesday
4 = Thursday
5 = Friday
6 = Saturday
Therefore, January 26th 1950 was Thursday
 Question 2
It was Sunday on Jan 1, 2006. What was the day of the Week Jan 1, 2010?
 A Monday B Friday C Sunday D Tuesday
Question 2 Explanation:
Explanation:
On 31st Dec, 2005 it was Saturday.
No. of. Odd days from 2006 to 2009 = (1 + 1 + 2 + 1) = 5days.

On 31st Dec 2009, it was Thursday.
Therefore, On 1st Jan, 2010 it is Friday
 Question 3
Given that on 9th Aug 2016 is Saturday. What was the day on 9th Aug 1616?
 A Saturday B Sunday C Friday D Monday
Question 3 Explanation:
Explanation:
We know that, After every 400 Years, the same day Occurs.
Thus, if 9th Aug 2016 is Saturday, Before 400 Years
i.e., On 9th Aug 1616 has to be Saturday.
 Question 4
What was the day of the week on 28th May, 2006?
 A Thursday B Friday C Saturday D Sunday
Question 4 Explanation:
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 * 1 + 1 * 2) = 6 odd days
Jan Feb Mar April May
(31 + 28 + 31 + 30 + 28 ) = 148days
148days = (21 weeks + 1days) = 1 odd days
Total No. of Odd Days = (0 + 0 + 6 + 1) = 7 = 0 Odd Days
Given day is Sunday
 Question 5
Today is Wednesday, after 68 days, it will be
 A Monday B Wednesday C Friday D Sunday
Question 5 Explanation:
Explanation:
Each day of a week is repeated after 7 days, So after 70 days, it will be Wednesday.
Therefore, after 68 days, it will be Monday.
 Question 6
The Calendar for the year 1993 will be same for the year
 A 1999 B 2004 C 2010 D 2021
Question 6 Explanation:
Explanation:
Note
Repetition of Leap Year => Add +28 to the Given Year
Repetition of non leap year
Step 1: Add +11 to the Given Year. If Result is a leap Year, Go to Step 2.
Step 2: Add +6 to the Given Year.

Solution:
Given Year is 1993, Which is a non leap year.
Step 1: Add +11 to the given year (i.e 1993 + 11) = 2004, Which is a leap year.
Step 2: Add +6 to the given year (i.e 1993 + 6) =1999
Therefore, The calendar for the year 1993 will be same for the year 1999
 Question 7
If Every seconds Saturday and all Sundays are holidays in a 30 days month beginning on Saturday, then how many working days are there in that month? (Month starts from Saturday)
 A 25 B 24 C 23 D 22
Question 7 Explanation:
Explanation:
As the month beings on Saturday, So 2nd, 9th, 16th, 23rd, 30th days will be Sunday. While 8th and 22nd days are the second Saturday. Thus, there are 7 holidays in all.
Hence, No.of working days = 30-7 = 23
 Question 8
The Last day of a century Cannot be
 A Monday B Wednesday C Tuesday D Friday
Question 8 Explanation:
Explanation:
100 years contain 5 odd days
Therefore, Last day of 1st centure is Friday
200 Years contain (5 *2) = 3 odd days.
Therefore, Last day of 2nd centure is Wednesday
300 Years contain (5 *3) = 15 = 1 odd days.
Therefore, Last day of 3rd centure is Monday
400 Years contain 0 odd days.
Therefore, Last day of 4th centure is Sunday
This Cycle is repeated.
Therefore, Last day of a century cannot be Tuesday or Thursday or Saturday.
 Question 9
Prove that any date in March of a Year is the same day of the week corresponding date in November that year.
 A Same day B Not Same day C Next Day D Previous Day
Question 9 Explanation:
Explanation:
We will show that the number of odd days between last day of February and last day of October is Zero. . .
March April May June July Aug Sept Oct
31 + 30 + 31 + 30 + 31 + 31 + 30 + 31
= 241 days = 35 Weeks = 0 Odd day.
No. of Odd days during this period = 0
Therefore, 1st March of an Year will be the same day as 1st November of that year.
Hence the result follows.
 Question 10
Today is Monday. After 61 days it will be:
 A Saturday B Sunday C Monday D Tuesday
Question 10 Explanation: