Calendar Questions and Answers

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.

Answer: B

31st Dec, 2011 = (2010 years + Period from 1.1.2011 to 31.12.2011)

Odd days in 2000 years = 0

10 years has 2 leap years + 8 ordinary years.

Number of odd days in 10 years ( 2 x 2 + 8) = 5 odd days.

31st,Dec => complete year of 2011 (non-leap year) = 1 odd day.

Total number of odd days = (0 + 5 + 1) = 6.

Given day is saturday.

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: A

We know that the day repeats every 7 days, 14 days, 21 days,...

So if today is Monday, after 7 days it is again Monday, after 14 days again it is Monday.

Hence, after 2 weeks i.e, 14 days the day repeats and is the same day.