How many times do the hands of a clock point towards each other in a day?

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

As per the given above question , we can say that
In 1 hour it increases by 6 minutes so in 6 hours it increases by 36 minutes.
Thus , required time will be 11 : 42 am .

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.

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.