**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 |

Same day | |

Not Same day | |

Next Day | |

Previous Day |

**Answer: Option A**

__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 2 |

25 | |

24 | |

23 | |

22 |

**Answer: Option C**

__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 3 |

1999 | |

2004 | |

2010 | |

2021 |

**Answer: Option A**

__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.

Repetition of non leap year

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 4 |

Saturday | |

Sunday | |

Friday | |

Monday |

**Answer: Option A**

__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 5 |

Saturday | |

Sunday | |

Monday | |

Tuesday |

**Answer: Option A**

__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

Question 6 |

Monday | |

Wednesday | |

Friday | |

Sunday |

**Answer: Option A**

__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 7 |

Monday | |

Friday | |

Sunday | |

Tuesday |

**Answer: Option B**

__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 8 |

Wednesday | |

Friday | |

Thursday | |

Tuesday |

**Answer: Option C**

__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 9 |

Monday | |

Wednesday | |

Tuesday | |

Friday |

**Answer: Option C**

__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 10 |

Thursday | |

Friday | |

Saturday | |

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 * 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**

**
There are 10 questions to complete.
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