Clock problems can be broadly classified in two categories:
a) Problems on angles
b) Problems on incorrect clocks
Problems on angles
Before we actually start solving problems on angles, we need to get couple of basic facts clear:
Speed of the hour hand = 0.5 degrees per minute (dpm) {The hour hand completes a full circle or 360 degrees in 12 hours or 720 minutes}
Speed of the minute hand = 6 dpm {The minute hand completes a full circle in 60 minutes}
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
Example 2: At what time do the hands of the clock meet between 7:00 and 8:00
Ans: At 7 o’ clock, the hour hand is at 210 degrees from the vertical.
In ‘t’ minutes
Hour hand = 210 + 0.5t
Minute hand = 6t
They should be meeting each other, so
210 + 0.5t = 6t
=> t = 210/5.5 = 420/11= 38 minutes 2/11th minute
Hands of the clock meet at 7 : 38 : 2/11th
Example 3: At what time do the hands of a clock between 7:00 and 8:00 form 90 degrees?
Ans: At 7 o’ clock, the hour hand is at 210 degrees from the vertical.
In ‘t’ minutes
Hour hand = 210 + 0.5t
Minute hand = 6t
The difference between them should be 90 degrees. Please note that it can be both before the meeting or after the meeting. You will get two answers in this case, one when hour hand is ahead and the other one when the minute hand is ahead.
Case 1: 210 + 0.5t – 6t = 90
=> 5.5t = 120
=> t = 240/11 = 21 minutes 9/11th of a minute
Case 2: 6t – (210 + 0.5t) = 90
=> 5.5t = 300
=> t = 600/11 = 54 minutes 6/11th of a minute
So, the hands of the clock are at 90 degrees at the following timings:
7 : 21 : 9/11th and 7 : 54 : 6/11th
Some other results which might be useful:

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?
Ans: 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
Example 5: A watch gains 5 seconds in 3 minutes and was set right at 8 AM. If it shows 5:15 in the afternoon on the same day, what is the correct time?
Ans: The watch gains 5 seconds in 3 minutes => 1 minute in 36 minutes
From 8 AM to 5:15, the incorrect watch has moved 9 hours and 15 minutes = 555 minutes.
When the incorrect watch moves for 37 minutes, correct watch moves for 36 minutes.
=> When the incorrect watch moves for 1 minute, correct watch moves for 36/37 minutes
=> When the incorrect watch moves for 555 minutes, correct watch moves for (36/37)*555 = 36*15 minutes = 9 hours
=> 9 hours from 8 AM is 5 PM.
=> The correct time is 5 PM.
I am sure you would have heard the proverb that even a broken clock is right twice a day. However, a clock which gains or loses a few minutes might not be right twice a day or even once a day. It would be right when it had gained / lost exactly 12 hours.
Example 6: A watch loses 5 minutes every hour and was set right at 8 AM on a Monday. When will it show the correct time again?
Ans: For the watch to show the correct time again, it should lose 12 hours.
It loses 5 minutes in 1 hour
=> It loses 1 minute in 12 minutes
=> It will lose 12 hours (or 720 minutes) in 720*12 minutes = 144 hours = 6 days
=> It will show the correct time again at 8 AM on Sunday.
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