P.V. Narsihma Rao was elected party leader on 29th may, 1991. What was the day of the week?

First we look for the leap years during this period 1997, 1998, 1999 are not leap years.
1998 and 1999 together have net 2 odd days No. of days remaining in 1997 = 365 - 16 = 349 days = 49 week 6 odd days
January 4, 2000 gives 4 odd days.
? Total no. of days = 2 + 6 + 4 = 12 days = 7 Days (1 week) + 5 odd days

Hence, January 4, 2000 will be 5 days beyond Thursday i.e., it will on Tuesday.

1600 years contains 0 odd day; 300 years conains 1 odd day. Also, 83 years contains 20 leap years and 63 ordinary years and therefore, (40 + 0) odd days i.e., 5 odd days.

? 1983 years contain (0 + 1 + 5) i.e., 6 odd days
Number of days from Jan. 1984 to 31st Oct 1984 = (31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31) = 305 days = 4 odd days
? Total number of odd days = 6 + 4 = 3 odd days
So, 31st Oct 1984 was wednesday

We go on counting the odd day from 1991 onwards till the sum is divisible by 7 . The number of such days are 14 up to the year 2001. So, the calendar for 1991 will be repeated in the year 2002.

We can say that ,
August 1, 1988 means = 1987 years + 7 months
Number of odd days in 1987 years = 1600 years have 0 odd days + 300 years have 1 odd day + 87 years have 21 leap years and 66 ordinary years.
So, there are [ ( 21 × 2 ) + ( 66 × 1 ) ] = 108 days, = 15 weeks and 3 odd days.
Number of days between January 1, 1988 to August 1, 1989 = January + February + March + April + May + June + July + August = 31 + 29 + 31 + 30 + 31 + 30 + 31 + 1 = 241 days = 30 weeks and 4 odd days.
Total number of odd days = 0 + 1 + 3 + 4 = 8 odd days = ( 1 x 7 + 1 ) = 1 odd day.
Thus, Friday falls on 5th, 12th, 19th and 26th in August 1988.

Time from 10 am to 3 pm of the next day = 29 h
Now, 24 h 10 min of the clock in question = 24 h of the correct clock
i.e., 145 h of the clock in question = 24 h of the correct clock
29 h of the clock in question = 24 x (6/145) x 29) h of correct clock = 28 h 48 min correct clock
So, the correct time is 28 h 48 min after 10 am i.,e 48 min past 2 pm.

Count the number of odd days from the year 2008 onward to get the sum equal to 0 odd day.

Let us see
2008 = 2
2009 = 1
2010 = 1
2011 = 1
2012 = 2
2013 = 1
2014 = 1
2015 = 1
2016 = 2
2017 = 1
2018 = 1

Sum = 2 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 + 1 = 14 days = 2 weeks = 0 odd days
? Calendar for the year 2019 will be the same as for the year 2008

Given that n = 5 and n + 1 = 6
The hands will make right angle at
(5n ± 15) x 12/11 min past 5 .
? (5 x 5 ± 15) x 12/11 min past 5.
? (25 + 15) x 12/11 min past 5 and (25 - 15) x 12/11 min past 5.
? 40 x 12/11 min past 5 and 10 x 12/11 min past 5.
? 4801/11 min past 5 and 120/11 min past 5.
? 43⁷/₁₁ min past 5 and 10¹⁰/₁₁ min past 5.

Time from 7 am on Sunday to 7 pm on following Sunday
= 7 days 12 h = 180 h
? Watch gains (5 + 5⁴/₅) min or 54/5 in 180 h.
Now, 54/5 min are gained in 180 h.
? 5 min gained in (180 x 5/54 x 5) h = 83 h 20 min.
= 3 days 11 h and 20 min

? Watch is correct after 3 days 11 h and 20 min after 7 am of Sunday.
? It will be correct at 20 min past 6 pm on Wednesday.

6th March 2005 = Monday.
Then, 6th March 2004 = Monday - 1 day = Sunday
[ ? 2004 is a leap year but it does not cross 29th February of 2004, so only 1 is taken as odd day. ]

6th March 2004 = Sunday
7th march 2004 = Monday.