At what time between 4 : 30 and 5 will the hands of a clock be in a straight line?

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 .

As per the given above question , we have
Birth date of Sashikant = September 28
Difference in number of days from August 15 to September 28 = 16 + 28 = 44
Number of odd days in 44 days = 2 odd days
Hence , Birthday of Shashikant = Sunday + 2 odd days = Tuesday.

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.

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 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.