Calendar Questions and Answers

Answer: D

Count the days from 6 March to 11 November, then add the part-day hours.

\[ \text{March (after the 6th)} = 25, \ \text{Apr } 30, \ \text{May } 31, \ \text{Jun } 30 \]

\[ \text{Jul } 31, \ \text{Aug } 31, \ \text{Sep } 30, \ \text{Oct } 31, \ \text{Nov } 11 \]

\[ \text{total} = 250 \text{ days} \]

From 9 a.m. to 9 a.m. is \( 250 \times 24 = 6000 \) hours, and from 9 a.m. to 8 p.m. is another 11:

\[ 6000 + 11 = 6011 \text{ hours} \]

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: D

Here from 4th December 1999 to 3rd January 2000,
there is a difference of 30 days.
Number of odd days = 30/7 = 2 odd days

Then no. of odd days = 2 => tuesday and wednesday(from monday).

Hence on 3rd january it is wednesday.

Answer: B

1 Oct, 1994 = (1993 years + Period from 1.1.1993 to 1.10.1993)

Odd days in 1600 years = 0

Odd days in 300 years = 1

93 years = (70 ordinary years + 23 leap year) = (70 x 1 + 23 x 2)= 4 odd days

Jan  Feb  March  April  May  June  July  Aug  sept  Oct

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 1 ) = 274 days

274 days = (39 weeks + 1 day) =1 odd day.

Total number of odd days = (0 + 1 + 4 + 1) = 6 odd days.

Given day is Saturday.

Answer: A

Days in 4 years => 

Let the first year is Normal year i.e, its not Leap year. A Leap Years occurs once for every 4 years.

4 years => 365 + 365 + 365 + 366(Leap year)

4 years => 730 + 731 = 1461

Therefore, Number of Days in 4 Years = 1461 Days.

Answer: A

In a correct clock the hands coincide every

\[ \frac{720}{11} = 65\tfrac{5}{11} \text{ minutes} \]

Here they take 65 minutes, slightly less, so the clock is running fast - and the conventional measure of the discrepancy is

\[ \frac{720}{11} - 65 = \frac{5}{11} \text{ minutes per interval} \]

There are \( \tfrac{1440}{65} \) such intervals in a day:

\[ \frac{5}{11} \times \frac{1440}{65} = \frac{7200}{715} = \frac{1440}{143} \text{ minutes} \]

That is about 10.07 minutes - a much smaller discrepancy than the 64-minute case, because 65 is far closer to \( 65\tfrac{5}{11} \).

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: A

9th May, 2001 = (2000 years + Period from 1.1.2001 to 9.5.2001)

Odd days in 1600 years = 0

Odd days in 400 years = 0

Jan. Feb. March April May 

(31 + 28 + 31 + 30 + 9) = 129 days

129 days = (18 weeks + 3 day)= 3 odd day.

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

Given day is Wednesday.

Answer: A

1856 is a leap year, and leap-year calendars repeat on a 28-year cycle:

\[ 1856 - 28 = 1828 \]

So the calendar of 1828 was used again in 1856. The cycle is unbroken here because no skipped century year falls between them.

Answer: B

1856 is a leap year, so its calendar returns after 28 years:

\[ 1856 + 28 = 1884 \]

Both must be leap years beginning on the same weekday, which the 28-year cycle ensures - and 1884 is indeed divisible by 4.