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Aptitude - Calendar Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to Calendar. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Q 1 - What was the week's day on fourth June, 2002?

Answer : A

Explanation

4Th.june, 2002 = (2001 years+ period from 1.1.2002 to. 4.6.2002.
Odd days in 1600 years = 0
Odd days in 400 years = 0
Odd days in 1 customary years = 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 days

Aggregate no. of odd days = (1+1) = 2
∴ required day is Tuesday.

Q 2 - Jan.1, 2007 was Monday. What day of the week lies on Jan 1, 2008?

Answer : B

Explanation

The year 2007 is a common year.
So, it has 1 odd day.

First day of the year 2007 was Monday.
First day of the year 2008 will be 1 day past Monday.
Subsequently, it will be Tuesday.

Q 3 - Jan.1 2008 is Tuesday. What date of the week lies on Jan 1, 2009?

Answer : C

Explanation

The year 2008 is a jump year.
So, it has 2 odd days.
First day of the year 2008 is Tuesday (Given).

In this way, first day of the year 2009 is 
2 days past Tuesday.
Subsequently, it will be Thursday.

Q 4 - The calendar for the year 2007 will be the same for the year:

Answer : D

Explanation

Count the number of odd days from the year 2007 onwards,
to get the sum equal to 0 odd days.

years 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017
Odd Days 1 2 1 1 1 2 1 1 1 2 1

Sum =14 odd days=0 odd day.
∴ Calendar for the year 2018 will be the same as for the year 2007.

years	2007	2008	2009	2010	2011	2012	2013	2014	2015	2016	2017
Odd Days	1	2	1	1	1	2	1	1	1	2	1

Q 5 - What was the week's day on seventeenth June, 1988?

Answer : D

Explanation

17th June, 1998= (1997 year Period structure 1.1.1998 to 17.6.1998)
Odd days in 1600 year=0.
Odd days in 300 year = (5*3) =1.
97 year has 24 jump year +73 standard years.
= 24*2 + 73*1
= 48 + 73
= 121 odd days
= 17 weeks + 2 odd days = 2 odd days
= 2 odd days.
∴ 1998 years have (0+1+2) = 3 odd days

Jan to June = (31+29+31+30+31)
= 152 days

Add 17 days of June.
= 152 + 17
= 169 days
= 24 weeks + 1 days
= 1 odd day.

∴ Total number of odd day = 3 + 1 = 4 odd days.
Hence 17.06.1998 was Thursday.

Q 6 - What was the week's day on 28th may, 2006?

Answer : D

Explanation

28.May,2006 =(2005 years +Period structure 1.1.2006 to 28.5.2006)
Odd days in 1600 years=0.
Odd days in 400 years = 0.

5 years = (4 common years+ 1 jump years)=(4*1+1*2)odd days
= 6 odd days.

Jan + Feb. + March + April + May
31 + 28 + 31 + 30 + 28 =148 days.

148 days = (21 weeks+1 day) = 1 odd day
Aggregate no. of odd days = (0+0+6+1) = 7= 0 odd day.
 
Given day is Sunday.

Q 7 - Today is Monday. Following 61 days, it will be?

Answer : B

Explanation

Each day of the week is rehashed following 7 days.
In this way, after 63 days it will be Monday.
∴ Following 61 days , it will be Saturday.

Q 8 - How number of days are present in x weeks x days?

Answer : B

Explanation

x weeks x days=(7x+x) days=8x days.

Q 9 - It was Sunday on Jan 1, 2006. Discover the week's day on Jan 1, 2010.

Answer : C

Explanation

On 31st December,2005 it was Saturday.
Number of odd days frame the year 2006 to the year 2009=(1+1+2+1)=5 days.

∴ On 31st December 2009 , it was Thursday.
In this way, on first jan, 2010 it is Friday.

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