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

Answer: Option C

Explanation:

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.

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

Answer: Option D

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.

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

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

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

168 days = 24 weeks = 0 odd day.

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

Given day is Wednesday.

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

Answer: Option A

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

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

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.

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

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

Answer: Option B

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.