Clock puzzles Questions

The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o'clcok only).

So, in a day, the hands point in the opposite directions 22 times.

Angle traced by hour hand in 12 hrs = 360º.

Angle traced by hour hand in 5 hrs 10 min.  i.e., 31/6 hrs = 36012*316°= 155º

 Time from 12 p.m. on Monday to 2 p.m. on the following Monday = 7 days 2 hours = 170 hours. 

 The Watch gains 2+445 min.  or 345  min.in 170 hrs. 

Now,  345min.are gained in 170 hrs. 

  2 min.are gained in170*534*2 hrs=50 hrs 

 Watch is correct 2 days 2 hrs. after 12 p.m. on Monday i.e., it will be correct at 2 p.m. on Wednesday

Angle traced by hour hand in 12 hrs. = 360º.

Angle traced by it in  11/3 hrs=  36012*113°=110°

Angle traced by minute hand in 60 min. = 360º.

Angle traced by it in 40 min. =(360/60*40)°=240°

  Required angle (240 - 110)º = 130º.

Tomorrow

The day before yesterday was two days ago; the day after the day before yesterday was yesterday; two days after that (yesterday) is tomorrow.

Since, in one hour, two hands of a clock coincide only once, so, there will be value.
Required time T = 211H×30 + Ao minutes past H.
Here H - initial position of hour hand = 2 (since 2 O'clock)
A° = Required angle = 0° (Since it coincides)

T = 2112×30 + 0o minutes past 2 

=> 11011 minutes past 2

At 9 o’clock, the hour hand is at 9 and the minutes hand is at 12, i.e., the two hands are 15 min. spaces apart.
So, the minute hand should gain = (30 - 15) minutes = 15 minutes
55 minutes will be gained in 60 min.
15 minutes spaces will be gained in ((60/55) x 15) min. = 180/11 min.
The hands will be in the same straight line but not together i.e.,in 180 degrees at 180/11 min. past 9.

Since at 4 : 20 the minute hand will be at 4 and the angle between them will be same as the distance covered in degree by the hour hand in 20 minutes.

Required angle = distance of hour hand = speed × time = 12× 20 =10 degrees.