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 - On What dates of March 2005 did Friday fall?
A - 5th, 12th, 17th and 24th
B - 4th, 11th, 18th and 25th
C - 6th, 13th, 20th and 27th
D - 7th, 14th, 21th and 28th
To begin with we discover the day on 1.3.2005.
1.3.2005= (2004 year+ period from 1.1.2005 to 1.3.2005

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

4 years = (1 jump year+ 3 normal 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 days.

∴ 1.3.2005 was Tuesday. So, Friday lies on 4.3.2005.
Thus, Friday lies on 4th, 11th, 18th and 25th of March, 2005.
Q 2 - Jan.1, 2007 was Monday. What day of the week lies on Jan 1, 2008?
A - Monday
B - Tuesday
C - Wednesday
D - Sunday
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?
A - Monday
B - Wednesday
C - Thursday
D - Friday 
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 - On sixth March, 2005 Monday falls. What was the day of the week on sixth March, 2004?
A - Sunday
B - Saturday
C - Tuesday
D - Wednesday
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?
A - Monday
B - Tuesday
C - Wednesday
D - Thursday
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 will be the day of the week on fifteenth 2010?
A - Sunday
B - Monday
C - Tuesday
D - Friday
fifteenth August,2010 =(2009 years+ Period from 1.1.2010 to 15.8.2010)
Odd days in 1600 years = 0.
Odd days in 400 years = 0.

9 year = (2 jump year +7 conventional years)
= (2*2+7*1) = 11 odd days= 4 odd days

Jan + feb + March + April + May + June + July + august
31+28+31+30+31+30+31+15 = 227 days =(32 weeks +3 days )
= 3 odd days

Aggregate no. of odd days = (0+0+4+3) =7
= 0 odd day

Given day is Sunday.
Q 7 - The last day of a century can't be: 
A - Monday
B - Wednesday
C - Tuesday
D - Friday
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?
A - 7x2 
B - 8x
C - 14x
D - 7
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.
A - Sunday
B - Saturday
C - Friday
D - Wednesday
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.

    87 Lectures 
     22.5 hours 

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