100 yr have 5 odd days.
∴ Last day of 1st century of is Friday.

200 yr have (5 x 2) = 1 week + 3 odd day = 3 odd day
∴ Last day of 2nd century of Wednesday,

300 yr have (5 x 3) = 2 week + 1 odd day = 1 odd day
∴ Last day of 3rd century is Monday.

400 yr have 0 odd day.
∴ Last day of 4th century is Sunday.

Correct Option: A

100 yr have 5 odd days.
∴ Last day of 1st century of is Friday.

200 yr have (5 x 2) = 1 week + 3 odd day = 3 odd day
∴ Last day of 2nd century of Wednesday,

300 yr have (5 x 3) = 2 week + 1 odd day = 1 odd day
∴ Last day of 3rd century is Monday.

400 yr have 0 odd day.
∴ Last day of 4th century is Sunday.

This cycle is repeated.
∴ Last day of a century cannot be Tuesday of Thursday of Saturday.

First, we find out the day on 1.03.2005. Period upto 1.3.2005.
= (2004 yr + Period from 1.1.2005 to 1.3.2005)
∴ Odd day in 1600 yr = 0
Odd day in 400 yr = 0
4 yr = (1 leap year + 3 ordinary years)
= (1 x 2 + 3 x 1) = 5 odd days

Correct Option: B

First, we find out the day on 1.03.2005. Period upto 1.3.2005.
= (2004 yr + Period from 1.1.2005 to 1.3.2005)
∴ Odd day in 1600 yr = 0
Odd day in 400 yr = 0
4 yr = (1 leap year + 3 ordinary years)
= (1 x 2 + 3 x 1) = 5 odd days

Number of days between 1.1.2005 to 1.3.2005,
= January + February + March
= 31 + 28 + 1
= 60 days = (8 weeks + 4 days)
= 4 odd days

Total number of odd days
= (0 + 0 + 5 + 4) = 9
= 1 weak + 2 odd days
∴ 1.3. 2005 was Tuesday.

So, Saturday will fall on 5.3.2005.
Hence, Saturday lies on 5th, 12th, 19th, and 26th of March 2005.

Period upto 17th August, 2010 = (2009 yr + Period from 1.1.2010 to 17.8. 2010)

Counting of odd days
Odd day in 1600 yr = 0
Odd days in 400 yr = 0
9 yr = (2 leap years + 7 ordinary years)
= ( 2 x 2 + 7 x 1) = 1 week + 4 days = 4 odd days

Number of days between 1.1.2010 to 17.8.2010
January + February + March + April + May + June + July + August.
= (31 + 28 + 31 + 30 + 31 + 30 + 31 + 17) days
= 229 days = 32 weeks + 5 odd days

Correct Option: C

Period upto 17th August, 2010 = (2009 yr + Period from 1.1.2010 to 17.8. 2010)

Counting of odd days
Odd day in 1600 yr = 0
Odd days in 400 yr = 0
9 yr = (2 leap years + 7 ordinary years)
= ( 2 x 2 + 7 x 1) = 1 week + 4 days = 4 odd days

Number of days between 1.1.2010 to 17.8.2010
January + February + March + April + May + June + July + August.
= (31 + 28 + 31 + 30 + 31 + 30 + 31 + 17) days
= 229 days = 32 weeks + 5 odd days

Total number of odd days = (0 + 0 + 4 + 5 ) days
= 9 days = 1 week + 2 odd days
Hence, the required day is Tuesday.

Period upto 5th June, 2002 = (2001 yr + Period from 1.1.2002 to 5.6.2002)
∴ Odd days in 1600 yr = 0
Odd days in 400 yr = 0
Odd days in 1 ordinary year = 1
Odd days in 2001 year = (0 + 0 + 1) = 1

Number of days from 1.1.2002 to 5.6.2002
January + February + March + April + May + June
= 31 + 28 + 31 + 30 + 31 + 5
= 156 days = 22 weeks + 2 days
= 2 odd days

Correct Option: A

Period upto 5th June, 2002 = (2001 yr + Period from 1.1.2002 to 5.6.2002)
∴ Odd days in 1600 yr = 0
Odd days in 400 yr = 0
Odd days in 1 ordinary year = 1
Odd days in 2001 year = (0 + 0 + 1) = 1

Number of days from 1.1.2002 to 5.6.2002
January + February + March + April + May + June
= 31 + 28 + 31 + 30 + 31 + 5
= 156 days = 22 weeks + 2 days
= 2 odd days

∴ Total number of odd days = (1 + 2) = 3
∴ The required day is Wednesday.

Here, H = 8 > 6.
We know that ,

Correct Option: C

Here, H = 8 > 6.
We know that ,