(i) ঘণ্টার কাঁটা এবং মিনিটের কাঁটা পরস্পরের সাথে সমকোণে অবস্থান করে ঘণ্টায় দুইবার।
12 ঘণ্টায় 22 বার
24 ঘণ্টায় 44 বার

(ii) ঘণ্টার কাঁটা এবং মিনিটের কাঁটা পরস্পরের সাথে মিলিত হয় অথবা বিপরীত দিকে থাকে-
1 ঘণ্টায় 1 বার
12 ঘণ্টায় 11 বার
24 ঘণ্টায় 22 বার
∴ 1 দিন বা 24 ঘণ্টায় সমকোণে থাকে 44 বার।

9:00 AM to 10:30 AM = 90 মিনিট 
9:10 AM to 10:30 AM = 80 মিনিট
প্রথম গাড়িটি 60 মিনিটে যায় 40 মাইল 
প্রথম গাড়িটি 90 মিনিটে যায় (40*90)/60= 60 মাইল
দ্বিতীয় গাড়িটি 80 মিনিটে যায় 60 মাইল 
দ্বিতীয় গাড়িটি 60 মিনিটে যায় (60*60)/80= 45 মাইল 

মধ্যবর্তী কোণ = |(11M - 60H) / 2|°
                 = |(11×40 - 60×2)/2|°    
                 = |(440-120)/2|° 
                 = 160°

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

In one hour, the minute hand gains 330° over the hour hand. i.e., 60 minute, the minute hand gains 330° over the hour hand.

∴ In 16 minutes, the minute hand gain over the hour hand by  \(\frac{330}{60}\times16\;=\;88\)

For ticking 6 times, there are 5 intervals.

Each interval has time duration of 30/5 = 6 secs

At 12 o'clock, there are 11 intervals,
so total time for 11 intervals = 11 x 6 = 66 secs.

As minute hand covers, 60 degrees

Minute hand covers 4800/60 = 80°

Angle between hands of a clock

 

When the minute hand is behind the hour hand, the angle between the two hands at M minutes past H 'o clock

 

= \(30\left(\mathrm H-\frac{\mathrm M}5\right)+\frac{\mathrm M}2\) degrees

 

Here H = 4, M = 15 and the minute hand is behind the hour hand.

 

Hence the angle

 

 = 30[4-(15/5)]+15/2 = 30(1)+7.5 = 37.5 degrees

Angle traced by the hour hand in 6 hours = \(\frac{360^o}{12}\times6\) = 180º

Coincide means 00  angle.

This can be calculated using the formulafor time A to B means  [11m/2 - 30 (A)]

Here m gives minutes after A the both hands coincides.

Here A = 3, B = 4

0 =11m/2 –30 × 3
11m = 90 × 2 = 180
m= 180/11 = 16 4/11

So time = 3 : 16 4/11

On straight line means 180 degree angle.
180 = 11/2 min – 30 hrs
180 = 11/2 m – 30 × 3
180 = 11/2 m – 90
(180 + 90) 2 = 11 m
m = 540/11 = 49 1/11 minutes.

The watch gains 5 seconds in 3 minutes = 100 seconds in 1 hour.

From 8 AM to 10 PM on the same day, time passed is 14 hours.

In 14 hours, the watch would have gained 1400 seconds or 23 minutes 20 seconds.

So, when the correct time is 10 PM, the watch would show 10 : 23 : 20 PM

55 min. spaces are covered in 60 min.

60 min. spaces are covered in \(\frac{60}{50}\times60\)  min. = \(65\;\frac5{11}\) min

Loss in 64 min. = \(65\;\frac5{11}-64\;=\;\frac{16}{11}\) min

Loss in 24 hrs = \(\frac{16}{11}\times\frac1{64}\times24\times60\) min = \(32\frac8{11}\) min

Minute hand covers 5400/60 = 90°

Angle traced by hour hand in \(\frac{13}3\) hrs = \(\left(\frac{360}{12}\times\frac{13}3\right)^0\) = \(130^0\)

Angle traced by min. hand in 20 min = \(\left(\frac{360}{60}\times20\right)^0\) = \(120^0\)

Required angle = (130 - 120)º = 10º

Angle traced by hour hand in 12 hrs = 360º

 

Angle traced by hour hand in 5 hrs 10 min. i.e  \(\frac{31}6\) hrs = \(\left(\frac{360}{12}\times\frac{31}6\right)^0\) =  \(155^0\)

The minute hand angle is the easiest since an hour (i.e. 60 minutes) corresponds to the entire 360 degrees, each minute must correspond to 6 degrees. So just multiply the number of minutes in the time by 6 to get the number of degrees for the minute hand. 
Here 15 minutes corresponds to 15 x 6 = 90 degrees

Next, you have to figure out the angle of the hour hand. Since there are 12 hours in the entire 360 degrees, each hour corresponds to 30 degrees. But unless the time is EXACTLY something o'clock, you have to write the time as a fractional number of hours rather than as hours and minutes.
Here the time is 9:15 which is (9 + 15/60) = 37/4 hours. Since each hour corresponds to 30 degrees, we multiply 30 to get (37/4)(30) = 277.5 degrees.

Since the hour hand is at 277.5 degrees and the minute hand is at 90 degrees, we can subtract to get the angle of separation. 277.5 - 90 = 187.5 =~ 360 - 187.5 = 172.5 degrees.

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

 

Angle traced by it in \(\frac{11}3\) hrs = \(\left(\frac{360}{12}\times\frac{11}3\right)^0\) = \(110^0\)

Angle traced by minute hand in 60 min. = \(360^0\)

Angle traced by it in 40 min = \(\left(\frac{360}{60}\times40\right)^0\) = \(240^0\)

Required angle (240 - 110)º = \(130^0\)

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

 

Therefore, The watch gains \(\left(2+4\frac45\right)\) min. or  34/5 min. in 170 hrs.

Now, 34/5 min. are gained in 170 hrs.

Therefore, 2 min are gained in  \(\left(170\times\frac5{34}\times2\right)\) hrs = 50 hrs

 

Therefore, 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.

point to be remembered

To be together between 9 and 10 o'clock, the minute hand has to gain 45 min. spaces.
55 min. spaces gained in 60 min.

 

45 min. spaces are gained in \(\frac{60}{55}\times45\) min  or \(49\frac1{11}\) min

To be together between 9 and 10 o'clock, the minute hand has to gain 45 min. spaces.
55 min. spaces gained in 60 min.

 

45 min. spaces are gained in \(\frac{60}{55}\times45\) min  or \(49\frac1{11}\) min

The hands of a clock coincide 11 times in every 12 hours (Since between 11 and 1, they
coincide only once, i.e., at 12 o'clock).

The hands overlap about every 65 minutes, not every 60 minutes.
The hands coincide 22 times in a day.

In 12 hours, they are at right angles 22 times.

In 24 hours, they are at right angles 44 times.

At 3 o'clock, the minute hand is 15 min. spaces apart from the hour hand.
To be coincident, it must gain 15 min. spaces.55 min. are gained in 60 min.
15 min. are gained in
= (60/55 x 15)min
=16+4/11
The hands are coincident at 16 + 4/11 min past 3

this type of problems the formulae is
(5*x + or -15)*(12/11)
Here x is replaced by the first interval of given time
here i.e 4
Case 1 : (5*x + 15)*(12/11)
(5*4 +15)*(12/11)
(20+15)*(12/11)
35*12/11=420/11=38 2/11 min.
Therefore they are right angles at 38 2/11 min .past4
Case 2 : (5*x-15)*(12/11)
(5*4-15)*(12/11)
(20-15)*(12/11)
5*12/11=60/11 min=5 5/11min
Therefore they are right angles at 5 5/11 min.past4.

At 5 o'clock, the hands are 25 min. spaces apart.
To be at right angles and that too between 5.30 and 6, the minute hand has to gain (25 + 15) = 40 min. spaces.

 

55 min. spaces are gained in 60 min

 

40 min. spaces are gained in \(\frac{60}{55}\times40\) min

In this type of problems the formulae is \(\left(5x-30\right)\times\frac{12}{11}\) 

x is replaced by the first interval of given time Here i.e 8

\(\left(5\times8-30\right)\times\frac{12}{11}\)

= \(\frac{120}{11}\)min

Therefore the hands will be in the same straight line but not
together at \(\frac{120}{11}\) min.past 8.