Calendar Questions and Answers – Formulas & 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.
Month | No. Of. Odd Days |
Jan | 3 |
Feb (Normal / Leap) | 0 / 1 |
Mar | 3 |
Apr | 2 |
May | 3 |
June | 2 |
July | 3 |
Aug | 3 |
Sep | 2 |
Oct | 3 |
Nov | 2 |
Dec | 3 |
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:
No. Of. Days | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
Day | Sun | Mon | Tue | Wed | Thur | Fri | Sat |
Calendar Questions and Answers With Detailed Explanation – Quantitative Aptitude
Calendar
Question 1 |
Wednesday | |
Friday | |
Thursday | |
Tuesday |
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 |
Monday | |
Friday | |
Sunday | |
Tuesday |
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 |
Saturday | |
Sunday | |
Friday | |
Monday |
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 |
Thursday | |
Friday | |
Saturday | |
Sunday |
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 |
Monday | |
Wednesday | |
Friday | |
Sunday |
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 |
1999 | |
2004 | |
2010 | |
2021 |
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 |
25 | |
24 | |
23 | |
22 |
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 |
Monday | |
Wednesday | |
Tuesday | |
Friday |
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 |
Same day | |
Not Same day | |
Next Day | |
Previous Day |
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 |
Saturday | |
Sunday | |
Monday | |
Tuesday |
Explanation:
Each Day of the week is repeated after 7 days
So, After 63 days, it will be Monday.
Therefore, After 61 days, it will be Saturday