Aptitude - Calendar Online Quiz


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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.

Questions and Answers

Answer : B

Explanation

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

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?

A - Sunday

B - Thursday

C - Tuesday

D - Friday

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

A - 2014

B - 2016

C - 2017

D - 2018

Answer : D

Explanation

Count the number of odd days from the year 2007 onwards,
to get the sum equal to 0 odd days.
years20072008200920102011201220132014201520162017
Odd Days12111211121
Sum =14 odd days=0 odd day. ∴ Calendar for the year 2018 will be the same as for the year 2007.

Answer : D

Explanation

We might discover the day on first April, 2001.
First April , 2001=(2000 year + Period structure 1.1.2001 to 1.4.2001)
Odd days in 1600 years =0.
Odd days in 400 year =0.

Jan Feb March April
31 + 28 + 31 + 1 = 91 days =0 odd day.

Aggregate number of odd days= (0+0+0) =0.

On first April, 2001 it was Sunday.
In April, 2001 Wednesday falls on 4th, 11th, 18th and 25th.

Q 6 - What will be the day of the week on fifteenth 2010?

A - Sunday

B - Monday

C - Tuesday

D - Friday

Answer : A

Explanation

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

A - Wednesday

B - Saturday

C - Tuesday

D - Thursday

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 - Which of the accompanying is not a leap year?

A - 700

B - 800

C - 1200

D - 2000

Answer : A

Explanation

The century divisible by 400 is a leap year.
∴ The year 700 is not a jump year.

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

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.


aptitude_calendar.htm

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