Clock based problems are one of the frequently asked
questions in most of the competitive

exam. To solve these problems, it is
always better to understand some of the basic principles and the types of
problems that get asked. In this post I hereby explained simple tricks and some
simple formulas for solving clock based problems.
In every competitive exams clock questions are
categorized in to two ways.

- Problems in angles
- Problems on incorrect clocks

**Problems in angles**

*Method :1*

Before we actually start solving problems on angles, we
need to know couple of basic facts clear:

- Speed of the hour hand = 0.5 degrees per minute (dpm)
- Speed of the minute hand = 6 dpm
- At ‘n’ o’ clock, the angle of the hour hand from the vertical is 30n

The questions based upon these could be of the
following types

Example : 1

*What is the angle between the hands of the clock at 7:20*

At 7 o’ clock, the hour
hand is at 210 degrees from the vertical.

In 20 minutes,

Hour hand = 210 + 20*(0.5)
= 210 + 10 = 220 {The hour hand moves at 0.5 dpm}

Minute hand = 20*(6) = 120
{The minute hand moves at 6 dpm}

Difference or angle
between the hands = 220 – 120 = 100 degrees

*Method : 2*

Example :2

*Find the reflex angle between the hands of a clock at 05.30?*

The above problem are solved by the bellow formula

Angle between X and Y =|(X*30)-((Y*11)/2)|

Angle between hands at
5:30

Step 1: X=5 , Y=30

Step 2: 5*30=150

Step 3: (30*11)/2 = 165

Step 4: 165-150=15

Thus, angle between hands at 5:30 is 15 degrees.

*Method : 3*

Example : 3

*At what time 3&4’o clock in the hands of clock together.*

Approximately we know at
03:15 hands of the clock together

So 15*60/55=16.36 min

**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.

Example : 4

*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

**Important Notes**

- Two right angles per hour(Right angle = 90, Straight angle=180)
- Forty four right angles per day
- Between every two hours the hands of the clock coincide with each other for one time except between 11, 12 and 12, 1.In a day they coincide for 22 times.
- Between every two hours they are perpendicular to each other two times except between 2, 3 and 3, 4 and 8, 9 and 9, 10.In a day they will be perpendicular for 44 times.
- Between every two hours they will be opposite to each other one time except between 5, 6 and 6, 7.In a day they will be opposite for 22 times.