A watch which gains uniformly, is 5 min slow at 7 o'clock in the morning on Sunday and it is 5 min 48 s fast at 7 pm on following Sunday. Which day was it correct?

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.

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

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.

Try yourself. It was Wednesday.

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.

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.

Number of odd days between 6th November, 1987 and 4th April, 1988 .
= (30 - 6) days of November 1987+ 31 days of December 1987 + 31 days of January 1988 + 29 days of February
1988 + 31 days March 1988 + 4 days of April 1988
= {(30 - 6 ) + 31 } + (31 + 29 + 31 + 4)
= 24 + 31 + 31 + 29 + 31 + 4 = 150
= 21 weeks + 3 days = 3 odd days
? 6th November, 1987 will be three days backward from Monday.
? Required day is Friday.

Period upto 17th July, 1776 = 1775 yr + Period from 1st January 1776 to 17th July 1776

Counting of odd days
In 1600 yr = 0
In 100 yr = 5
75 yr = 18 leap years + 57 ordinary years = (18 x 2 + 57 x 1) odd days
= 93 odd days
= 13 weeks + 2 days
= 2 odd days

? 1775 yr have(0 + 5 + 2) odd days
= 7 odd days = 1 week + 0 odd day
= 0 odd day

Number of days between 1.1.1776 to 17.7.1776
January + February + March + April + May + June + July
31 + 29 + 31 + 30 + 31 + 30 + 17 = 199 days = 28 weeks + 3 odd days

? Total number of odd days = (3 + 0) = 3
Hence, the required day is Wednesday.

Period up to 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 year + 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.

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.