Calendar Questions and Answers

Answer: C

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.

Answer: D

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.

Answer: C

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

 168 days = 24 weeks = 0 odd day.

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

Given day is Wednesday.

Answer: A

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 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.

Answer: B

Only the remainder on division by 7 matters - the odd days.

\[ 61 = 8 \times 7 + 5 \quad\Longrightarrow\quad 5 \text{ odd days} \]

Counting five days on from Monday:

\[ \text{Tue, Wed, Thu, Fri, } \mathbf{Sat} \]

The eight complete weeks return to Monday and contribute nothing.

Answer: A

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).

Answer: D

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.

Answer: B

Each week is seven days, so

\[ x \text{ weeks } + x \text{ days} = 7x + x = 8x \text{ days} \]

The two \( x \)'s are added, not multiplied - \( 7x^{2} \) would be the result of multiplying them.

Answer: C

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.

Answer: C

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.