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- 1. What was the day of the week on 24
^{th}July 2011?

Answer: Option B

Explanation:**Formula:-(Date + Month code + No.of years + No.of leap year + Century code)/7**

= (24 + 0 + 11 + 2 + 6)/7 = 43/7 = 1

= Sunday

= (24 + 0 + 11 + 2 + 6)/7 = 43/7 = 1

= Sunday

- 2. What was the day of the week on, 16
^{th}July, 1776?

Answer: Option B

Explanation:**16**

Counting of odd days :

1600 years have 0 odd day. 100 years have 5 odd days.

75 years = (18 leap years + 57 ordinary years)

= [(18 x 2) + (57 x 1)] odd days = 93 odd days

= (13 weeks + 2 days) = 2 odd days.

∴ 1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day.

Jan. Feb. March April May June July

31 + 29 + 31 + 30 + 31 + 30 +16 = 198days

= (28 weeks + 2 days) =2days

∴ Total number of odd days = (0 + 2) = 2. Required day was 'Tuesday'.

^{th}July, 1776 = (1775 years + Period from 1st Jan., 1776 to 16^{th}July, 1776)Counting of odd days :

1600 years have 0 odd day. 100 years have 5 odd days.

75 years = (18 leap years + 57 ordinary years)

= [(18 x 2) + (57 x 1)] odd days = 93 odd days

= (13 weeks + 2 days) = 2 odd days.

∴ 1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day.

Jan. Feb. March April May June July

31 + 29 + 31 + 30 + 31 + 30 +16 = 198days

= (28 weeks + 2 days) =2days

∴ Total number of odd days = (0 + 2) = 2. Required day was 'Tuesday'.

- 3.
**What was the day of the week on 16**^{th}August, 1947?

**Saturday**

Answer: Option D

Explanation:**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. March April May June July 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'.

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. March April May June July 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'.

- 4. If Aug 10th,2012 falls on Friday then June 10th,2013 falls on which day ?

Answer: Option A

Explanation:**First,we count the number of odd days for the left over days in the given period.
Here,given period is 10.8.2012 to 10.6.2013
Aug Sept Oct Nov Dec Jan Feb Mar Apr May Jun
21 30 31 30 31 31 28 31 30 31 10(left days)
0 + 2 + 3 + 2 + 3 + 3 + 0 + 3 + 2 + 3 + 3 (odd days) = 24 => 24/7 = 3 odd days
So,given day Friday + 3 = 5 + 3 = 8 => 8/7 = 1 odd day. Monday is the required result.**

- 5. If September 21st,1987 falls on Wednesday then November 21st,1987 falls on which day?

Answer: Option

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

Here,given period is 21.9.1987 to 02.11.1987

Sept Oct Nov

9 31 2 (left days)

2 + 3 + 2 (odd days) = 7 => 7/7 = 0 odd days

So,given day Monday + 0 = 1 + 0 = 1. Monday is the required result.

Here,given period is 21.9.1987 to 02.11.1987

Sept Oct Nov

9 31 2 (left days)

2 + 3 + 2 (odd days) = 7 => 7/7 = 0 odd days

So,given day Monday + 0 = 1 + 0 = 1. Monday is the required result.

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