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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.

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 - What was the week's day on sixteenth July, 1776?

16 July, 1776 = (1775 years + period from 1.1.1776 to 16.7.1776. Checking of odd days. No. of odd days in 1600 years =0 No. of odd days in 100 years = 5 75 years= 18 jump years+57 conventional years = (18*2+57*1) odd days = 93 odd days = (13 weeks +2 days) = 2 odd days. ∴ 1775 years have= (0+5+2) odd days= 7 days= 0 days Jan + Feb. + March + April + May + June + July 31 +29 + 31 + 30 + 31 + 30 +16 = 198 days 198 days = (28 weeks+2 days) = 2 odd days =Total no. of odd days = (0+2) = 2 Thus, the required day is Tuesday.

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

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:

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 |

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

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.

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.

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.

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.

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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