Problems on Calendar

Problems on Calendar - Solved Examples

1. What day of the week does May 28 2006 fall on
A. Saturday	B. Monday
C. Sunday	D. Thursday

answer with explanation

Answer: Option C
Explanation:
28th May 2006 = (2005 years + period from 1-Jan-2006 to 28-May-2006)

We know that number of odd days in 400 years = 0
Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)

Number of odd days in the period 2001-2005
= 4 normal years + 1 leap year
= 4 x 1 + 1 x 2 = 6

Days from 1-Jan-2006 to 28-May-2006 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 28(may)
= 148

148 days = 21 weeks + 1 day = 1 odd day

Total number of odd days = (0 + 6 + 1) = 7 odd days = 0 odd day
0 odd day = Sunday

Hence May 28 2006 is Sunday.

2. What will be the day of the week 15th August, 2010?
A. Thursday	B. Sunday
C. Monday	D. Saturday

answer with explanation

Answer: Option B
Explanation:
15th Aug 2010 = (2009 years + period from 1-Jan-2010 to 15-Aug-2010)

We know that number of odd days in 400 years = 0
Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)

Number of odd days in the period 2001-2009
= 7 normal years + 2 leap year
= 7 x 1 + 2 x 2 = 11 = (11 - 7x1) odd day = 4 odd day

Days from 1-Jan-2010 to 15-Aug-2010
= 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 31(may) + 30(Jun) + 31(Jul) + 15(Aug)
= 227

227 days = 32 weeks + 3 day = 3 odd day

Total number of odd days = (0 + 4 + 3) = 7 odd days = 0 odd day
0 odd day = Sunday
Hence 15th August, 2010 is Sunday.

3. Today is Monday. After 61 days, it will be
A. Thursday	B. Sunday
C. Monday	D. Saturday

answer with explanation

Answer: Option D
Explanation:
61 days = 8 weeks 5 days = 5 odd days

Hence if today is Monday, After 61 days, it will be = (Monday + 5 odd days)
= Saturday

4. On what dates of April, 2001 did Wednesday fall?
A. 2nd, 9th, 16th, 23rd	B. 4th, 11th, 18th, 25th
C. 3rd, 10th, 17th, 24th	D. 1st, 8th, 15th, 22nd, 29th

answer with explanation

Answer: Option B
Explanation:
We need to find out the day of 01-Apr-2001

01-Apr-2001 = (2000 years + period from 1-Jan-2001 to 01-Apr-2001)

We know that number of odd days in 400 years = 0
Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)

Days from 1-Jan-2001 to 01-Apr-2001 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 1(Apr) = 91
91 days = 13 weeks = 0 odd day

Total number of odd days = (0 + 0) = 0 odd days
0 odd day = Sunday. Hence 01-Apr-2001 is Sunday.

Hence first Wednesday of Apr 2011 comes in 04^th and successive Wednesdays come in 11^th, 18^th and 25^th

5. How many days are there in x weeks x days
A. 14x	B. 8x
C. 7x2	D. 7

answer with explanation

Answer: Option B
Explanation:
$x$ weeks $x$ days $=(7\times x)+x=7x+x=8x$ days

6. The calendar for the year 2007 will be the same for the year
A. 2017	B. 2018
C. 2014	D. 2016

answer with explanation

Answer: Option B
Explanation:
For a year to have the same calendar with 2007 ,the total odd days from 2007 should be 0.

Year:	2007	2008	2009	2010	2011	2012	2013	2014	2015	2016	2017
Odd Days:	1	2	1	1	1	2	1	1	1	2	1

Take the year 2014 given in the choice.

Total odd days in the period 2007-2013 = 5 normal years + 2 leap year
= 5 x 1 + 2 x 2 = 9 odd days
= 2 odd day (As we can reduce multiples of 7 from odd days which will not change anything)

Take the year 2016 given in the choice.

Number of odd days in the period 2007-2015 = 7 normal years + 2 leap year
= 7 x 1 + 2 x 2 = 11 odd days
= 4 odd days
(Even if the odd days were 0, calendar of 2007 will not be same as the calendar of 2016 because 2007 is not a leap year whereas 2016 is a leap year. In fact, you can straight away ignore this choice due to this fact without even bothering to check the odd days)

Take the year 2017 given in the choice.

Number of odd days in the period 2007-2016 = 7 normal years + 3 leap year
= 7 x 1 + 3 x 2 = 13 odd days
= 6 odd days

Take the year 2018 given in the choice.

Number of odd days in the period 2007-2017 = 8 normal years + 3 leap year
= 8 x 1 + 3 x 2 = 14 odd days
= 0 odd day (As we can reduce multiples of 7 from odd days which will not change anything)

Also, both 2007 and 2018 are not leap years.
Since total odd days in the period 2007-2017 = 0 and both 2007 and 2018 are of same type, 2018 will have the same calendar as that of 2007

7. Which of the following is not a leap year?
A. 1200	B. 800
C. 700	D. 2000

answer with explanation

Answer: Option C
Explanation:
Remember the leap year rule (Given in the formulas)

1. Every year divisible by 4 is a leap year, if it is not a century.
2. Every 4th century is a leap year, but no other century is a leap year.

800,1200 and 2000 comes in the category of 4th century (such as 400,800,1200,1600,2000 etc).
Hence 800,1200 and 2000 are leap years

700 is not a 4th century, but it is a century. Hence it is not a leap year

8. 01-Jan-2007 was Monday. What day of the week lies on 01-Jan-2008?
A. Wednesday	B. Sunday
C. Friday	D. Tuesday

answer with explanation

Answer: Option D
Explanation:
Given that January 1, 2007 was Monday.

Odd days in 2007 = 1 (we have taken the complete year 2007 because we need to find out the odd days from 01-Jan-2007 to 31-Dec-2007, that is the whole year 2007)

Hence January 1, 2008 = (Monday + 1 Odd day) = Tuesday

9. 8th Dec 2007 was Saturday, what day of the week was it on 8th Dec, 2006?
A. Sunday	B. Tuesday
C. Friday	D. Tuesday

answer with explanation

Answer: Option C
Explanation:
Given that 8th Dec 2007 was Saturday

Number of days from 8th Dec, 2006 to 7th Dec 2007 = 365 days
365 days = 1 odd day

Hence 8th Dec 2006 was = (Saturday - 1 odd day) = Friday

10. On 8th Feb, 2005 it was Tuesday. What was the day of the week on 8th Feb, 2004?
A. Sunday	B. Friday
C. Saturday	D. Monday

answer with explanation

Answer: Option A
Explanation:
Given that 8th Feb, 2005 was Tuesday

Number of days from 8th Feb, 2004 to 7th Feb, 2005 = 366 (Since Feb 2004 has 29 days as it is a leap year)

366 days = 2 odd days
Hence 8th Feb, 2004 = (Tuesday - 2 odd days) = Sunda
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