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# Calendar – Aptitude MCQ Questions and Solutions with Explanations

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# Calendar – Aptitude MCQ Questions and Solutions with Explanations

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

## Answer & Solution

Answer: Option C
Solution:

On 31st December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
∴ On 31st December 2009, it was Thursday.
Thus, on 1st Jan, 2010 it is Friday.

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

## Answer & Solution

Answer: Option D
Solution:

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 17th June, 1998?

## Answer & Solution

Answer: Option C
Solution:

17th 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

Therefore 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 15th August, 2010?

## Answer & Solution

Answer: Option A
Solution:

15th 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     Mayb   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:

## Answer & Solution

Answer: Option B
Solution:

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

## Answer & Solution

Answer: Option A
Solution:

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 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.
Given that, 6th March, 2005 is Monday.
∴ 6th March, 2004 is Sunday (1 day before to 6th March, 2005).

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

## Answer & Solution

Answer: Option D
Solution:

We shall find the day on 1st April, 2001.
1st 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 1st April, 2001 it was Sunday.
In April, 2001 Wednesday falls on 4th, 11th, 18th and 25th.

8. How many days are there in x weeks x days?

## Answer & Solution

Answer: Option B
Solution:

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

9. The last day of a century cannot be

## Answer & Solution

Answer: Option C
Solution:

100 years contain 5 odd days.
∴ Last day of 1st century is Friday.
200 years contain (5 x 2) ≡ 3 odd days.
∴ Last day of 2nd century is Wednesday.
300 years contain (5 x 3) = 15 ≡ 1 odd day.
∴ Last day of 3rd century is Monday.
400 years contain 0 odd day.
∴ Last day of 4th century is Sunday.
This cycle is repeated.
∴ Last day of a century cannot be Tuesday or Thursday or Saturday.

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

## Answer & Solution

Answer: Option C
Solution:

The year 2004 is a leap year. It has 2 odd days.
∴ The day on 8th Feb, 2004 is 2 days before the day on 8th Feb, 2005.
Hence, this day is Sunday.

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