Calendar Questions and Answers

Answer: C

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

2005 is not divisible by 4, so it is an ordinary year.

\[ 365 \rightarrow 1 \text{ odd day} \]

\[ \text{Saturday} + 1 = \textbf{Sunday} \]

Answer: B

Count odd days backwards from 2006 until they reach a multiple of seven.

\[ 2005:1,\ 2004:2,\ 2003:1,\ 2002:1,\ 2001:1,\ 2000:2,\ 1999:1,\ 1998:1,\ 1997:1,\ 1996:2,\ 1995:1 \]

\[ \text{total} = 14 \quad\Longrightarrow\quad 0 \text{ odd days} \]

So the 2006 calendar had been used in 1995.

Answer: B

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: C

The year 2008 is a jump year.
So, it has 2 odd days.
First day of the year 2008 is Tuesday (Given).

In this way, first day of the year 2009 is
2 days past Tuesday.
Subsequently, it will be Thursday.

Answer: C

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

2008 is divisible by 4 and is not a century year, so it is a leap year with 366 days.

\[ 366 = 52 \times 7 + 2 \quad\Longrightarrow\quad 2 \text{ odd days} \]

\[ \text{Tuesday} + 2 = \textbf{Thursday} \]

An ordinary year advances the date by one weekday; a leap year by two.

Answer: C

Count the days from one date to the other and reduce modulo 7.

From 18 March 1994 to 18 March 1995 is a full ordinary year, 365 days, so

\[ 365 \equiv 1 \pmod 7 \]

Going back from 18 March 1995 to 25 February 1995 is 21 days, which is exactly 3 weeks and so leaves no odd day. Net shift from the given Friday:

\[ +1 - 0 = +1 \quad\Longrightarrow\quad \text{Friday} + 1 = \textbf{Saturday} \]

1995 is not a leap year and February 1995 has 28 days, so no extra day creeps in.

Answer: D

If the period between the two months is divisible by 7, then that two months will have the same calender .

 (a). Oct + Nov  = 31 + 30 = 61 (not divisible by 7)

(b). Apr + May + Jun + Jul + Aug + Sep + Oct = 30 + 31 + 30 + 31 + 31 + 30 + 31  = 214 (not divisible by 7)

(c). Jun + July + Aug + Sep = 30 + 31 + 31 + 30 = 122 (not divisible by 7)

(d). Apr + May + June = 30 + 31 + 30 = 91 (divisible by 7)

Hence, April and July months will have the same calendar.

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: D

Let us find the day on 1st July, 2004.

2000 years have 0 odd day. 3 ordinary years have 3 odd days.

Jan. Feb. March April May June July

31 + 29 + 31 + 30 + 31 + 30 + 1

= 183 days = (26 weeks + 1 day) = 1 t .

Total number of odd days = (0 + 3 + 1) odd days = 4 odd days. '

:. 1st July 2004 was 'Thursday',-,-

Thus, 1st Monday in July 2004 _as on 5th July.

Hence, during July 2004, Monday fell on 5th, 12th, 19th and 26th. .

Answer: A

A year is a leap year if it is divisible by 4, except that a century year must also be divisible by 400.

• \( 800 \div 400 = 2 \) ✓
• \( 1200 \div 400 = 3 \) ✓
• \( 2000 \div 400 = 5 \) ✓
• \( 700 \div 400 = 1.75 \) ✗

So 700 is not a leap year, even though it is divisible by 4. The 400-year rule exists because the tropical year is a little under \( 365.25 \) days.