On what date of August, 1988 did Friday fall?

According to question , we know that
The hands of a clock point towards each other 11 times in every 12 hours. (because between 5 and 7, at 6 O'clock only they point towards each other)
So, in a day the hands point towards each other 22 times.
Hence , correct answer will be 22 times.

We know that , In every hour there are two positions in which hands are at right angle.
Each of these positions is repeated 11 times in every 12 hours.
? In every 12 hours, hands are at right angles 11 + 11 = 22 times and in a day hands are at right angles 22 + 22 = 44 times.
Hence , required answer will be 44 times.

As per the given above question , we can say that
1. In 1 min ,minute hand moves 6 Degree
2. In 1 min,hour hand moves 0.5 Degree
Let suppose man went out at y past three i.e. 3 hours y minutes
then, minute hand will be at ( 6y ) Degree from origin.
and hour hand will be at ( 3 × 30 + y × 0.5) Degree = ( 90 + 0.5y ) Degree
Similarly ,man returned at z past eight i.e. 8 hours z minutes
then, minute hand will be at ( 6z ) Degree from origin.
and hour hand will be at ( 8 × 30 + z × 0.5 ) Degree = ( 240 + 0.5z ) Degree
And as per the given condition :-
6y = 90 + 0.5y ...... ( ? )
and 6z = 240 + 0.5z ...... ( ? )
Solve these equations , we get y = 16.36 and z =43.63

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.

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

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.

Try yourself. It was Wednesday.

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.