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Calender - Aptitude Questions and Answers - RejinpaulPlacement

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1.
What was the day of the week on 16th July, 1776 ?
Answer & Solution
Answer: (B) Tuesday
Solution:   16th July 1776 = (1775 years + Period from 1.1.1776 to 16.7.1776)

Counting of odd days :
Number of odd days in 1600 years = 0
Number of odd days in 100 years = 5

75 years = 18 leap years + 57 ordinary years
= (18 * 2 + 57 * 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 days

Jan. + Feb. + March + April + May + June + July
(31 + 29 + 31 + 30 +31 + 30 +16) = 198 days

198 days = (28 weeks + 2 days) = 2 odd days
Total number of odd days = (0 + 2) = 2

Hence, the required day is Tuesday
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2.
What was the day of the week on 15th August 1947?
Answer & Solution
Answer: (C) Friday
Solution:   15th August 1947 = (1946 years + Period from 1.1.1947 to 15.8.1947)

Counting of odd days:
Odd days in 1600 years = 0
Odd days in 300 years = (5 * 3) = 15 = 1

46 years = (11 leap years + 35 ordinary years)
= (11 * 2 + 35 * 1) odd days = 57 odd days
= (8 weeks + 1 day) = 1 odd day
Odd days in 1946 years = (0 + 1 + 1) = 2 days

Jan. + Fab. + 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 = (2 + 3) = 5

Hence, the required day is Friday.
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3.
What was the day of the week on 4th June, 2002?
Answer & Solution
Answer: (C) Tuesday
Solution:   4th June, 2002 = (2001 years + Period from 1.1.2002 to 4.6.2002)

Counting of odd days:
Odd days in 1600 years = 0
Odd days in 1 ordinary year = 1
Odd days in 2001 years = (0 + 0 + 1) = 1

Jan. + Feb. + March + April + May + June
(31 + 28 + 31 + 30 + 31 + 4) = 155 days
= 22 weeks + 1 day = 1 odd day

Total number of odd days = (1 + 1) = 2
Required day is Tuesday.
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4.
On what dates of March 2005 did Friday fall?
Answer & Solution
Answer: (D) 4th, 11th, 18th, 25th
Solution:   First we find the day on 1.3.2005

1.3.2005 = (2004 years + Period from 1.1.2005 to 1.3.2005

Odd days in 1600 years = 0
Odd days in 400 years = 0

4 years = ( 1 leap year + 3 ordinary years)
= (1 * 2 + 3 * 1) odd days = 5 odd days

Jan. + Feb. + March
(31 + 28 + 1) = 60 days = (8 weeks + 4 days) = 4 odd days.

Total number of odd days = (0 + 0 + 5 + 4) = 9 = 2 odd day
1.3.2005 was Tuesday.
So, Friday lies on 4.3.2005

Hence, Friday lies on 4th, 11th, 18th, 25th of March, 2005.
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5.
Prove that the Calendar for the year 2003 will serve for the year 2014.
Answer & Solution
Answer: (B) True
Solution:   We must have same day on 1.1.2003 and 1.1.2014
So, number of odd days between 31.12.2002 and 31.12.2013 must be 0.
This period has 3 leap years and 8 ordinary years.
Number of odd days = (3 * 2 + 8 * 1) = 14 ≡ 0 odd day.

Calendar for the year2003 will serve for the year 2014
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6.
January 1, 2007 was Monday. What day of the week lies on January, 2008?
Answer & Solution
Answer: (B) Tuesday
Solution:   The year 2007 is an ordinary year. So, it has 1 odd day.

1st day of the year 2007 was Monday.
1st day of the year 2008 will be 1 day beyond Monday.

Hence, it will be Tuesday.
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7.
January1, 2008 is Tuesday. What day of the week lies on January 1, 2009 ?
Answer & Solution
Answer: (C) Thursday
Solution:   The year 2008 is a leap year. So, it has odd days.

1st day of the year 2008 is Tuesday. (Given)
So, 1st day of the year 2009 is 2 day beyond Tuesday.

Hence, it will be Thursday.
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