Calendar Questions and Answers

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: B

Multiples of 4 in a century:

\[ \frac{100}{4} = 25 \]

But the century year at the end is a leap year only if divisible by 400, and in three centuries out of four it is not:

\[ 25 - 1 = 24 \]

So an ordinary century has 24 leap years - the standard figure behind the 5 odd days in 100 years.

Answer: B

We know that,

1 year = 365 days

1 day = 24 hours

1 hour = 60 minutes

1 minute = 60 seconds.

Then, 1 year = 365 x 24 x 60 x 60 seconds.

= 8760 x 3600

1 year = 31536000 seconds.

Hence, 10 years = 31536000 x 10 = 315360000 seconds.

Answer: C

15th August, 1947 = (1946 years + Period from 1st Jan., 1947 to 15th )

Counting of odd days:

1600 years have 0 odd day. 300 years have 1 odd day.

47 years = (11 leap years + 36 ordinary years)= [(11 x 2) + (36 x 1) ]odd days = 58 odd days = 2 odd days.

Jan  Feb  Mar  Apr  May  Jun  Jul  Aug.

31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 days = (32 weeks + 3 days) = 3,

Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.

Hence, the required day was 'Saturday'.

Answer: C

First,we count the number of odd days for the left over days in the given period.

Here,given period is 12.2.1986 to 1.1.1987

Feb Mar Apr May June July  Aug Sept Oct Nov Dec Jan

16   31    30  31   30    31   31   30   31  30   31   1 (left days)

2 + 3 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 1(odd days) = 1 odd day

So,given day Wednesday + 1 = Thursday is the required result

Answer: C

25 Jan, 1981 = (1980 years + Period from 1.1.1981 to 25.1.1981)

Odd days in 1600 years = 0

Odd days in 300 years = 1

80 years = (60 ordinary years + 20 leap year) = (60 x 1 + 20 x 2)= 2 odd days

Jan = 25 days = (3 weeks + 4 days) = 4 odd days.

Total number of odd days = (0 + 1 + 2 + 4) = 7= 0 odd day.

Given day is Sunday.

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: 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: A

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

\[ 350 = 50 \times 7 \quad\Longrightarrow\quad 0 \text{ odd days} \]

350 days is exactly fifty weeks, so the day of the week is unchanged: Saturday.

Answer: B

The day of the week repeats every 7 days.

Given today is Friday. Again Friday is repeated on the 7th day, 14th,... on 7 multiple days.

Hence, Friday is on the 63rd day, as 63 is multiple of 7.

Now, the required day of the week on the 65th day is Sunday.