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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 - On eighth Dec, 2007 Saturday falls. What was the day of the week would it say it was on eighth dec, 2006?

Answer : D

Explanation

The year 2006 is a common year. 
Along these lines, it has 1 odd day.
In this way, the day on eighth Dec, 
2007 will be 1 day past the day.

On eighth Dec, 2006.
Yet, eighth Dec,2007 is Saturday.
∴ eighth Dec, 2007 is Friday.

Q 4 - On sixth March, 2005 Monday falls. What was the day of the week on sixth March, 2004?

Answer : B

Explanation

The year 2004 is a jump year. 
In this way, it has 2 odd days.
∴ The day on sixth March, 2005 will be 2 days.

past the day on sixth Walk, 2004.
Yet, sixth March, 2005 is Monday.
∴ sixth March, 2004 is Saturday.

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 - The last day of a century can't be:

Answer : C

Explanation

100 years contain 5 odd days.
∴ Last day of first century is Friday. 

200 years contain (5*2) = 3 odd days.
∴ Last day of second century is Wednesday.

300 year contain (5*3) =15=1 odd day.
∴ Last day of third century is Monday.

400 year contain 0 odd days.
∴ Last day of fourth century is Sunday.

The cycle is repeated.
∴ Last day of a century can't be Tuesday or Thursday or 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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