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# Calendar - Solved Examples

Q 1 - What was the day of the week on 15th June, 1776?

**Answer - B**

**Explanation**

15^{th}June 1776 = (1775 years + Period from 01.01.1776 to 15.06.1776) Counting of odd days: No of odd days in 1600 years = 0 No of odd days in 100 years = 5 75 years = 18 leap years + 57 ordinary years = 18*2 + 57*1 = 36 + 57 = 93 odd days = 13 weeks + 2 odd days = 2 odd days ∴ 1775 years have (0+5+2) = 7 odd days = 0 odd days. Jan to May = (31+29+31+30+31) = 152 days Add 15 days of June. = 152 + 15 = 167 days = 23 weeks + 6 days = 6 odd days. ∴ Total number of odd days = 0 + 6 = 6 odd days. Hence 15.06.1776 was Saturday.

Q 2 - January 15, 1997 was a Wednesday. What day of the week was on Jan 5, 2000?

**Answer - A**

**Explanation**

1997, 1998 and 1999 are not leap years. 1998 and 1999 has 2 odd days. No of days remaining in 1997 = 365 - 15 = 350 = 50 weeks of 0 odd days. 05.01.2000 = 5 odd days. Total no of odd days = 2 + 0 + 5 = 7 7 days from Wednesday is Wednesday. ∴ Jan 5, 2000 was also Wednesday.

Q 3 - The calendar for the year 2007 will be the same for the year:

**Answer - A**

**Explanation**

We will count the no of odd days from the year 2007 onwards to get the sum equal to 0 odd days.

Year | 2007 | 2008 | 2009 | 2010 | 2011 | 2012 | 2013 | 2014 | 2015 | 2016 | 2017 |
---|---|---|---|---|---|---|---|---|---|---|---|

Odd day | 1 | 2 | 1 | 1 | 1 | 2 | 1 | 1 | 1 | 2 | 1 |

**Answer - B**

**Explanation**

We must have same day on 1.1.2003 and 1.1.2014. Along these lines, number of odd days somewhere around 31.12.2002 and 31.12.2013 must be 0. This period has 3 jump years and 8 common years. Number of odd days = (3*2+8*1) =14=0 odd days. ∴ Calendar for the year 2003 will serve for the year 2014.

Q 5 - What was the week's day on fifteenth august, 1947?

**Answer - C**

**Explanation**

fifteenth Aug.1947 =(1946 years +period from 1.1.1947 to 15.8.1947) Odd days in 1600 years =0 Odd days in 300 years = (5*3) =15 =1946 years = (11 jump years+35 customary years) = (11*2 +35*1) odd days= 57 days = (8 weeks +1 day) = 1 odd day ∴ odd days in 1946 years= (0+1+1) =2 Jan + Feb. + March + April + May + June + July + Aug (31 + 28 +31+ 30 + 31 +30+31+15) = 227 days 227 days = (32 weeks +3 days) = 3 odd days. Aggregate no. of odd days = (2+3) = 5 Consequently the required day is Friday.

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