Problems on Clock Hand

Hours, Minutes and Seconds

a) Minute Spaces:

The face or dial of watch is a circle whose circumference is divided into 60 equal parts, called minute spaces.

A clock has two hands :

Hour Hand: The hour hand (or short hand ) indicates time in hours .

Minute Hand: The minute hand or long hand indicates time in minutes.

**Points to Note : **

- When the hands are coincident , the angle between them is 0^{0}

- When the hands point in opposite direction , the angle between them is 180^{0}

- The hands are in the same straight line , when they are coincident or opposite to each other .
So the angle between the two hands is 0^{0} or 180^{0}

The dial of the clock is circular in shape and was divided into 60 equal minute spaces

- Minute Hand covers 360^{0} in 1 hour i.e. in 60 minutes. Hence , MINUTE HAND COVERS 6^{0} PER MINUTE

- Hour hand covers 360^{0} in 12 hours. Hence hour hand covers 30^{0} per hour. Hence , HOUR HAND COVERS ( 1/2 )^{0} PER MINUTE.

- Thus in 1 minute , the minute hand gains 5 (1/2)^{0} than the hour hand.

- The minute hand moves 12 times as fast as the hour hand.

In a period of 12 hours , the hands make an angle of :

- 0^{0} with each other (i.e. they coincide with each other ) 11 times.

- 180^{0} with each other (i.e. they lie on the same straight line) 11 times

- 90^{0} or any other angle with each other 22 times

In every hour:

- Both the hands coincide once

- The hands are straight (point in opposite directions) once . In this position , hands are 30 mins spaces apart .

- The hands are twice at right angles . In this this position , hands are 15 mins spaces apart .

The time gap between any 2 coincidences is 12/11 hours or 65 (5/11) minutes.

In every exams clock questions are categorized in to two ways :

a) Problems in Angles

b) Problems on Incorrect Clocks

a) Problems in Angles

Angle between Hours (H) and Minutes (M) = ** 1/2| 60*H - 11*M | **

Example 1: Find the angle between hour hand and minute hand of a clock at 05:30.

Given: Hour (H) = 05 Minute (M) = 30

Angle between Hours (H) and Minutes (M) = 1/2|60 H - 11 M|

Angle between Hours (H) and Minutes (M) = 1/2|60*05 - 11*30|

= 1/2|60*05 - 11*30|

= 1/2|300 - 330|

= |-15|

= 15^{0}

Example 2: At what time between 5 and 6 O' clock are the hands of a clock together ?

Given : H = 5

Both the Hands of clock together means 0^{0}

We know that angle between Hours and Minutes = 1/2 |60H - 11M|

So 0^{0} = 1/2 |60*5 - 11 M|

0 = 300 - 11 M

So M = 300/11

Therefore Hands of a clock are together at 300/11 = 27 (3/11)

Example 3: At what time between 5 and 6 O' clock will the hands of a clock be at right angle?

Given H = 5

Right Angle means 90^{0}

We know that angle between Hours and Minutes = 1/2 |60H - 11M|

So 90^{0} = 1/2 |60*5 - 11 M|

180 = | 300 -11 M |

Here 300 -11 M can be positive or negative depending on the value of M.

If 300 -11 M is positive

180 = 300 -11 M

11 M = 120

So M = 120/11

If 300 -11 M is negative

180 = - ( 300 -11 M )

180 = - 300 + 11 M

11 M = 480

So M = 480/11 = 43 (7/11)

Therefore Hands of a clock are at right angle at 10 (10/11) mins past 5 and 43 (7/11) mins past 5

Example 4: Find the time between 4 and 5 O' clock when the two hands of a clock are 4 mins apart ?

We know that Minute Hand covers 6^{0} in 1 minute .

Given that two hands of a clock are 4 mins apart , means 4 * 6 = 24^{0} apart.

Given : H = 4

We know that angle between Hours and Minutes = 1/2 |60H - 11M|

So 24 = 1/2 |60 * 4 - 11 * M|

48 = | 240 - 11 M|

Here 240 - 11 M can be positive or negative depending on the value of M.

If 240 - 11 M is positive

48 = 240 - 11 M

So M = 192 / 11 = 17(5/11)

If 240 - 11 M is negative

48 = - (240 - 11 M)

48 = -240 + 11M

11 M = 288

M = 288 / 11 = 26 (2/11)

So hands will be 4 mins apart at 26(2/11) mins past 4 and 17 (5/11) mins past 4 O' clock.

Example 5: Find the difference between two time instants , one before and one after 12 pm when the angle between the hour hand and the minute hand is 115^{0} ?

Here we have to find the 2 time instant, one before the 12 pm and one after 12 pm.

1) Let's take first instant before 12 pm.

The time for this instant will be between 11 and 12.

So H = 11 Angle = 115^{0}

We know that angle between Hours and Minutes = 1/2 |60H – 11M|

So 115^{0} = 1/2 |60 * 11 - 11 * M|

230 = | 660 - 11 M|

Here 660 - 11 M can be positive or negative depending on the value of M.

a) If 660 - 11 M is positive

230 = | 660 - 11 M|

230 = 660 -11 M

11 M = 430

M = 430 / 11 = 39.09 Min

b) If 660 - 11 M is negative

230 = | 660 - 11 M|

230 = - (660 -11 M)

230 = -660 + 11 M

11 M = 890

M = 80.90 Min

Here 80.90 Minutes is not possible , since the minute (M) cannot have value more than 60.

So time before 12 pm when the angle between the hour hand and the minute hand is 1150 is 11:39 ---------------(1)

2) Let's take second instant after 12 pm.

The time for this instant will be between 12 and 01.

So H = 0 Angle = 1150

Note : ** If Hour is between 12 and 01 pm then value of H is 0. **

We know that angle between Hours and Minutes = 1/2 |60H - 11M|

115^{0} = 1/2 |60 * 0 - 11 M|

230 = 11 M

M = 230 /11 = 20.9 Minutes

So time after 12 pm when the angle between the hour hand and the minute hand is 1150 is 12:20 --------------------(2)

We need to find difference between 1 and 2 .

Therefore the time gap between two instants = 12:20 - 11:39 = 41 Minutes

b) Problems on Incorrect Clocks

Such sort of problems arise when a clock runs faster or slower than expected pace.

When solving these problems it is best to keep track of the correct clock.

If the hands of a clock (which do not show the correct time) coincide every p minutes , then :

- If p > 65 ( 5/11 ) , then the watch is going slow or losing time

- If p < 65 (5/11) , then the watch is going fast or gaining time.

Example :

If a clock indicates 6:10 , when the correct time is 6 , it is said to be 10 mins too fast .

And if it indicates 5:50 , when the correct time is 6 , it is said to be 10 mins too slow.

Example 6: A watch gains 5 seconds in 3 minutes and was set right at 8 AM. What time will it show at 10 PM on the same day?

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

Note 1 :

The minute hand of a clock overtakes the hour hand at intervals of M minutes of correct time. The clock gains or losses in a day by:

Example 7: The minute hand of a clock overtakes the hour hand at intervals of 65 mins. How much a day does the clock gain or loss ?

Here M = 65

Therefore the clock gains or loss in a day by : = (720/11 - M) (60*24/M) mins

= (720/11 - 65) (60*24/65) mins

= 1440/143 mins = 10 (10/143) mins