**Introduction:**

A normal clock shows 12 hours.

It has three hands: hour, minute and second hands.

Clocks do not follow decimal system.

A day has 24 hours; 1 hour = 60 minutes and 1 minute = 60 seconds.

Hence, 1 hour = (60 x 60) = 3600 seconds

The plane angle of the clock dial is 360^{o} and is covered in 12 hours by the hour hand, in one hour by the minute hand and in one minute by the second hand.

The hour and minute hands meet at 12 00 hrs. There will be 11 such meetings in 12 hours span. Hence, time interval between two meetings = (12/11) = 1(1/11) hrs (or) 60(60/11) minutes (or) 65 (5/11) minutes.

Take 12 00 hrs as the reference point where both the hour and minute hands coincide.

From there, for every 1^{o} movement of the hour hand, the minute hand will move 12^{o}.

If the minute hand of a clock overtakes the hour hand at intervals of M minutes of correct time, the clock gains or loses in a day by [(720/11)−M][(720×2)/M] minutes (gain if positive and lose if negative).

**Example 1:**

Find the angle between the hour and minute hands when the time is 08 30 hrs.

#### Solution:

From 12 00 hrs,

[imagine the hour and min hand of the clock at the time of 08 30]
then from above table

Angle traced by hour hand: (30 x 8.5) = 255^{o} … (08 30 = 8.5 hrs)

Angle traced by the minute hand: (30 x 6) = 180^{o} ,,, (consider only the minutes part)

Angle between the hands = (255 – 180) = 75^{o}.

**Example 2:**

At what time between 4 and 5^{o} clocks, the hour and minute hands will be at right angle to each other, the first time?

#### Solution:

At 4^{o} clock:

Angle of hour hand = (4 x 30) = 120^{o}

Angle of minute hand = 0^{o} … (coincides with 12, the reference point)

Let the time be T and is ‘m’ minutes past 4^{o} clock.

In ‘m’ minutes, the hour hand moves (m/2)^{o} and the minute hand moves (6m)^{o}.

For the first time to be at right angle (90^{o}), the hour hand will be ahead of the minute hand.

Therefore, 90 has to be subtracted from the hour hand angle to get the angle of the minute hand.

120+(m/2) – 90 = (6m)

(11/2)m = 30 and m = 5(5/11) minutes.

Hence, T = 4 hrs and 5(5/11) minutes.

**Example 3:**

Find the time at which the hour and minute hands are at equal angles between 7 and 8^{o} clocks, with respect to the vertical (line joining the numbers 6 and 12 in a clock)

#### Solution:

Let the angle with respect to the vertical be θ.

At 7^{o} clock:

Angle of hour hand = (7 x 30) = 210^{o} (or) 30^{o} after vertical

Angle of minute hand = 0^{o} … (coincides with 12, the reference point)

Let the time be T and is ‘m’ minutes past 7^{o} clock.

In ‘m’ minutes, the hour hand moves (m/2)^{o} from 7 (30^{o} from vertical) and the minute hand moves (6m)^{o}.

The condition to be satisfied is:

The hour hand will be at an angle of (180+θ)^{o} and the minute hand will be at (180-θ)^{o}.

6m = 180 – θ (or) θ = 180 – 6m … (i)

(m/2)+30 = θ … (ii)

Equating (i) and (ii);

θ = 180 – 6m = (m/2)+30

(13/2)m = 150 and m = (300/13) = 23(1/13) minutes past 7.

T = 7 hrs 23(1/13) minutes.

**Example 4:**

What is the time at which the hour and minute hands coincide between 5 and 6^{o} clocks?

#### Solution:

The hands coincide at 12 00 hrs. Afterwards, they will coincide every (12/11) hrs.

At the fifth meeting, they will be between 5 and 6^{o} clocks.

That is after 5x(12/11) hrs = 60/11 hrs = 5(5/11) hrs = 5 hrs (5×60/11) minutes

= 5 hrs 27(3/11) minutes.

**Example 5:**

At what time between 7 and 8 ^{o}clock will the hands of a clock be in the same straight line but not together?

#### Solution:

At 7^{o} clock:

Angle of hour hand = (7 x 30) = 210^{o}

Angle of minute hand = 0^{o} … (coincides with 12, the reference point)

Let the time be T and is ‘m’ minutes past 7^{o} clock.

The hour hand will be ahead of the minute hand and the angle between them is 180^{o}.

Therefore, 180 has to be subtracted from the hour hand angle to get the angle of minute hand.

Let the time be T and is ‘m’ minutes past 7^{o} clock.

In ‘m’ minutes, the hour hand moves (m/2)^{o} from 7 and the minute hand moves (6m)^{o}.

The condition to be satisfied is:

210+(m/2) – 180 = (6m)

(11/2)m = 30 and m = 5(5/11) minutes.

Hence, T = 7 hrs and 5(5/11) minutes.

**Example 6:**

How many times in a day, are the hands of a clock in straight line but opposite in direction?

#### Solution:

The hands of a clock point in opposite directions (in the same straight line, making an angle 180^{°} between them) 11 times in every 12 hours (because between 5 and 7 they point in opposite directions at 6^{o} clock only).

Hence the hands point in the opposite directions 22 times in a day.

#### Note:

Similarly in 12 hrs, the hands will coincide each other 11 times (because between 11 and 1 they coincide at 12^{0} clock only).

Hence the hands coincide each other 22 times in a day.

**Example 7:**

How much does a watch lose per day, if its hands coincide every 64 minutes?

#### Solution:

If the minute hand of a clock overtakes the hour hand at intervals of M minutes of correct time, the clock gains or loses in a day by [(720/11)−M][(720×2)/M] minutes.

M = 64 minutes which is less than 65(5/11) minutes (the time interval between two meetings).

Hence, the clock will gain: [(720/11)-64][(720×2/64]=[(720-(64×11))/11][720×2/11]
= (8×45/11) 360/11 = 32(8/11) minutes a day.

**Example 8:**

A watch which gains 5 seconds in 3 minutes was set right at 7 a.m. In the afternoon of the same day, when the watch indicated quarter past 4^{o} clock, the correct time is:

#### Solution:

Interval between 7 and 4.15^{o} clock = 9 hrs and 15 minutes = 555 minutes

This 555 minutes includes the actual time and gained minutes.

Let the actual time is T minutes after 7 a.m.

The watch gains 5 seconds (5/60 = 1/12 minutes) every 3 minutes.

T + (T/3)(1/12) = 555

36T+T = 12 x 555

37 T = 36 x 555

T = 540 minutes = 9 hrs after 7 a.m.

Correct time is: (7+9) = 16 hrs (or) 4 p.m.

**Example 9:**

How many times are the hands of a clock at right angle in a day?

#### Solution:

In 12 hours, hands of a clock are at right angles 22 times.

In 24 hours, hands of a clock are at right angles 44 times.

**Example 10:**

A clock strikes 4 taking 9 seconds. In order to strike 12 at the same rate, the time taken is:

#### Solution:

After the first strike, there are 3 intervals to complete the remaining strikes

Time taken for 3 intervals = 9 seconds

Time taken for 1 interval = 3 seconds

In order to strike 12, there are 11 intervals.

Hence, required time = 3×11=33 seconds.

**About Author**

#### Dr. BASKAR .A,

a former Research Scientist Government of India, presently working as a Professor in the Dept of Mechanical Engineering, Authored books on Kinematics and Dynamics of Machinery, Also Published Numerous Research papers moreover a Geek in Mathematics