The Following Section consists Maths Aptitude Test on Calendar. Take the Quiz and improve your overall Maths on Calendar.
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Maths Aptitude Test on Calendar 1
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Question 1 of 15
1. Question
1 pointsIt was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
Correct
Incorrect
On 31st December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
∴ On 31st December 2009, it was Thursday.
Thus, on 1st Jan, 2010 it is Friday.

Question 2 of 15
2. Question
1 pointsWhat was the day of the week on 28^{th} May, 2006?
Correct
Incorrect
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

Question 3 of 15
3. Question
1 pointsWhat was the day of the week on 17^{th} June, 1998?
Correct
Incorrect
17^{th} June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)
Odd days in 1600 years = 0
Odd days in 300 years = (5 x 3) 1
97 years has 24 leap years + 73 ordinary years.
Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.
Jan. Feb. March April May June (31 + 28 + 31 + 30 + 31 + 17) = 168 days
∴ 168 days = 24 weeks = 0 odd day.
Total number of odd days = (0 + 1 + 2 + 0) = 3.
Given day is Wednesday.

Question 4 of 15
4. Question
1 pointsWhat will be the day of the week 15^{th} August, 2010?
Correct
Incorrect
15^{th} August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)
Odd days in 1600 years = 0
Odd days in 400 years = 0
9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days 4 odd days.
Jan. Feb. 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 = (0 + 0 + 4 + 3) = 7 0 odd days.
Given day is Sunday.

Question 5 of 15
5. Question
1 pointsToday is Monday. After 61 days, it will be:
Correct
Incorrect
Each day of the week is repeated after 7 days.
So, after 63 days, it will be Monday.
∴ After 61 days, it will be Saturday.

Question 6 of 15
6. Question
1 pointsIf 6^{th} March, 2005 is Monday, what was the day of the week on 6^{th} March, 2004?
Correct
Incorrect
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 6^{th} March, 2005 will be 1 day beyond the day on 6th March, 2004.
Given that, 6^{th} March, 2005 is Monday.
∴ 6^{th} March, 2004 is Sunday (1 day before to 6^{th} March, 2005).

Question 7 of 15
7. Question
1 pointsOn what dates of April, 2001 did Wednesday fall?
Correct
Incorrect
We shall find the day on 1^{st} April, 2001.
1^{st} April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)
Odd days in 1600 years = 0
Odd days in 400 years = 0
Jan. Feb. March April (31 + 28 + 31 + 1) = 91 days 0 odd days.
Total number of odd days = (0 + 0 + 0) = 0
On 1,sup>st April, 2001 it was Sunday.
In April, 2001 Wednesday falls on 4^{th}, 11^{th}, 18^{th} and 25^{th}.

Question 8 of 15
8. Question
1 pointsHow many days are there in x weeks x days?
Correct
Incorrect
x weeks x days = (7x + x) days = 8x days.

Question 9 of 15
9. Question
1 pointsThe last day of a century cannot be
Correct
Incorrect
100 years contain 5 odd days.
∴ Last day of 1^{st} century is Friday.
200 years contain (5 x 2) 3 odd days.
∴ Last day of 2^{nd} century is Wednesday.
300 years contain (5 x 3) = 15 1 odd day.
∴ Last day of 3^{rd} century is Monday.
400 years contain 0 odd day.
∴ Last day of 4^{th} century is Sunday.
This cycle is repeated.
∴ Last day of a century cannot be Tuesday or Thursday or Saturday.

Question 10 of 15
10. Question
1 pointsOn 8^{th} Feb, 2005 it was Tuesday. What was the day of the week on 8^{th} Feb, 2004?
Correct
Incorrect
The year 2004 is a leap year. It has 2 odd days.
The day on 8^{th} Feb, 2004 is 2 days before the day on 8^{th} Feb, 2005.
Hence, this day is Sunday.

Question 11 of 15
11. Question
1 pointsThe calendar for the year 2007 will be the same for the year:
Correct
Incorrect
Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.
Year : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 Odd day : 1 2 1 1 1 2 1 1 1 2 1
Sum = 14 odd days 0 odd days.
Calendar for the year 2018 will be the same as for the year 2007.

Question 12 of 15
12. Question
1 pointsWhich of the following is not a leap year?
Correct
Incorrect
The century divisible by 400 is a leap year.
The year 700 is not a leap year.

Question 13 of 15
13. Question
1 pointsOn 8^{th} Dec, 2007 Saturday falls. What day of the week was it on 8^{th} Dec, 2006?
Correct
Incorrect
The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 8^{th} Dec, 2007 will be 1 day beyond the day on 8^{th} Dec, 2006.
But, 8^{th} Dec, 2007 is Saturday.
∴ 8^{th} Dec, 2006 is Friday.

Question 14 of 15
14. Question
1 pointsJanuary 1, 2008 is Tuesday. What day of the week lies on Jan 1, 2009?
Correct
Incorrect
The year 2008 is a leap year. So, it has 2 odd days.
1^{st} day of the year 2008 is Tuesday (Given)
So, 1^{st} day of the year 2009 is 2 days beyond Tuesday.
Hence, it will be Thursday.

Question 15 of 15
15. Question
1 pointsJanuary 1, 2007 was Monday. What day of the week lies on Jan. 1, 2008?
Correct
Incorrect
The year 2007 is an ordinary year. So, it has 1 odd day.
1^{st} day of the year 2007 was Monday.
1^{st} day of the year 2008 will be 1 day beyond Monday.
Hence, it will be Tuesday.
Calendar aptitude questions are common in CAT and NET, apart from many other entrance examinations. Using the right logic, it is easy to correctly answer these questions and secure higher grades in the process. ExamTime Quiz has an impressive range of questions with multiple choice answers on the said topic. Thus, preparing for mathematical aptitude test on calendar  to crack any exam  is easier than before.