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QUESTION: 1

It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?

Solution:

On 31^{st} December, 2005 it was Saturday.

Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.

∴ On 31^{st} December 2009, it was Thursday.

Thus, on 1^{st} Jan, 2010 it is Friday.

QUESTION: 2

What was the day of the week on 28th May, 2006?

Solution:

28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)

Odd days in 1600 years = 0

Odd days in 400 years = 0

5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) ≡ 6 odd days

Jan. Feb. March April May

(31 + 28 + 31 + 30 + 28 ) = 148 days

∴ 148 days = (21 weeks + 1 day) ≡ 1 odd day.

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

Given day is Sunday.

QUESTION: 3

What was the day of the week on 17th June, 1998?

Solution:

17^{th} June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)

Odd days in 1600 years = 0

Odd days in 300 years = (5 x 3) ≡ 1

97 years has 24 leap years + 73 ordinary years.

Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.

Jan. Feb. March April May June

(31 + 28 + 31 + 30 + 31 + 17) = 168 days

Therefore 168 days = 24 weeks = 0 odd day.

Total number of odd days = (0 + 1 + 2 + 0) = 3.

Given day is Wednesday.

QUESTION: 4

What will be the day of the week 15th August, 2010?

Solution:

15^{th} August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)

Odd days in 1600 years = 0

Odd days in 400 years = 0

9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days ≡ 4 odd days.

Jan. Feb. March April Mayb June July Aug.

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days

∴ 227 days = (32 weeks + 3 days) ≡ 3 odd days.

Total number of odd days = (0 + 0 + 4 + 3) = 7 ≡ 0 odd days.

Given day is Sunday.

QUESTION: 5

Today is Monday. After 61 days, it will be:

Solution:

Each day of the week is repeated after 7 days.

So, after 63 days, it will be Monday.

∴ After 61 days, it will be Saturday.

QUESTION: 6

If 6^{th} March, 2005 is Monday, what was the day of the week on 6^{th} March, 2004?

Solution:

The year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.

∴ The day on 6^{th} March, 2005 will be 1 day beyond the day on 6^{th} March, 2004.

Given that, 6^{th} March, 2005 is Monday.

∴ 6^{th} March, 2004 is Sunday (1 day before to 6^{th} March, 2005).

QUESTION: 7

On what dates of April, 2001 did Wednesday fall?

Solution:

We shall find the day on 1^{st} April, 2001.

1^{st} April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)

Odd days in 1600 years = 0

Odd days in 400 years = 0

Jan. Feb. March April

(31 + 28 + 31 + 1) = 91 days ≡ 0 odd days.

Total number of odd days = (0 + 0 + 0) = 0

On 1^{st} April, 2001 it was Sunday.

In April, 2001 Wednesday falls on 4^{th}, 11^{th}, 18^{th} and 25^{th}.

QUESTION: 8

How many days are there in x weeks x days?

Solution:

x weeks x days = (7x + x) days = 8x days.

QUESTION: 9

The last day of a century cannot be

Solution:

100 years contain 5 odd days.

∴ Last day of 1^{st} century is Friday.

200 years contain (5 x 2) ≡ 3 odd days.

∴ Last day of 2^{nd} century is Wednesday.

300 years contain (5 x 3) = 15 ≡ 1 odd day.

∴ Last day of 3^{rd} century is Monday.

400 years contain 0 odd day.

∴ Last day of 4^{th} century is Sunday.

This cycle is repeated.

∴ Last day of a century cannot be Tuesday or Thursday or Saturday.

QUESTION: 10

On 8th Feb, 2005 it was Tuesday. What was the day of the week on 8th Feb, 2004?

Solution:

The year 2004 is a leap year. It has 2 odd days.

∴ The day on 8^{th} Feb, 2004 is 2 days before the day on 8^{th} Feb, 2005.

Hence, this day is Sunday.

QUESTION: 11

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

Solution:

Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.

Sum = 14 odd days ≡ 0 odd days.

∴ Calendar for the year 2018 will be the same as for the year 2007.

QUESTION: 12

Which of the following is not a leap year?

Solution:

The century divisible by 400 is a leap year.

∴ The year 700 is not a leap year.

QUESTION: 13

On 8th Dec, 2007 Saturday falls. What day of the week was it on 8th Dec, 2006?

Solution:

The year 2006 is an ordinary year. So, it has 1 odd day.

So, the day on 8th Dec, 2007 will be 1 day beyond the day on 8th Dec, 2006.

But, 8th Dec, 2007 is Saturday.

∴ 8th Dec, 2006 is Friday.

QUESTION: 14

January 1, 2008 is Tuesday. What day of the week lies on Jan 1, 2009?

Solution:

The year 2008 is a leap year. So, it has 2 odd days.

1^{st} day of the year 2008 is Tuesday (Given)

So, 1^{st} day of the year 2009 is 2 days beyond Tuesday.

Hence, it will be Thursday.

QUESTION: 15

January 1, 2007 was Monday. What day of the week lies on Jan. 1, 2008?

Solution:

The year 2007 is an ordinary year. So, it has 1 odd day.

1^{st} day of the year 2007 was Monday.

1^{st} day of the year 2008 will be 1 day beyond Monday.

Hence, it will be Tuesday.

QUESTION: 16

It was Thursday on 12^{th} January 2006. What day of the week it will be on January 12^{th} 2007 ?

Solution:

There is exactly 1 year, (365 days) between two dates.

2006 is an ordinary year. It has one odd day.

The day of the week on January 12^{th} 2007 is one day beyond Thursday ⇒ Friday

QUESTION: 17

What day of the week will 22 Apr 2222 be?

Solution:

22 Apr 2222 = (2221 years + period from 1-Jan-2222 to 22-Apr-2222)

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 2001-2200

= Number of odd days in 200 years

= 5 x 2 = 10 = 3

(As we can reduce perfect multiples of 7 from odd days without affecting anything)

Number of odd days in the period 2201-2221

= 16 normal years + 5 leap years

= 16 x 1 + 5 x 2 = 16 + 10 = 26 = 5 odd days

Number of days from 1-Jan-2222 to 22 Apr 2222

= 31 (Jan) + 28 (Feb) + 31 (Mar) + 22(Apr) = 112

112 days = 0 odd day

Total number of odd days = (0 + 3 + 5 + 0) = 8 = 1 odd day

1 odd days = Monday

Hence 22 Apr 2222 is Monday

QUESTION: 18

If today is Monday, what will be the day 350 days from now?

Solution:

350 days, 350/7 = 50, no odd days, so it will be a Monday.

QUESTION: 19

What was the day of the week on 24th July 2011?

Solution:

Formula : (Date + Month code + No.of years + No.of leap year + Century code)/7

=43/7

=1

=Sunday

QUESTION: 20

If the first day of a year (other than leap year) was Friday, then which was the last day of that year?

Solution:

Given that first day of a normal year was Friday

Odd days of the mentioned year = 1 (Since it is an ordinary year)

Hence First day of the next year = (Friday + 1 Odd day) = Saturday

Therefore, last day of the mentioned year = Friday

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