The following question was part of my mock test and I was unable to solve it.

Assuming that:
A: The hour hand and minute hand of the clock move without jerking
B: The clock shows a time between 8 o'clock and 9 o'clock
C: The two hands of the clock are one above the other
After how many minutes (nearest integer) will the two hands will be lying again on one another?

I can think of that it would happen after 60 minutes as hour hand will also move but I am unable to formulate it mathematically in terms of angles.

Can you please help?


3 Answers 3


We don't need the information that current time is between 8 and 9 o'clock.

Here is a solution using the equation of angles:

We know that the hour and minute hands overlap right now. If they're to overlap again, the minute hand must be 360 degrees (one cycle) ahead of the hour hand. Since we know that the hour hand moves at 0.5 degrees per minute and the minute hand moves at 6 degrees per minute, let's say $m$ minutes have passed and set up an equation.$$ 360 + \text{movement of hour hand} = \text{movement of minute hand} \\360 + 0.5m = 6m \\m = \frac{720}{11} = 65\frac{5}{11} \approx 65 \text{ minutes}$$


They lie on one another at 12, between 1 and 2 (just after 1:00), between 2 and 3, … and between 10 and 11 (just before 11:00), so eleven times in every twelve-hour period. Since the clock is symmetric around its center (meaning the same amount of time passes between any two pairs of times when the hands lie on one another), there's 1/11 of twelve hours between any two such passes, or 65 minutes to the nearest integer.


A minute hand makes one turn per hour.
An hour hand makes 1/12 of turn per hour.

Their relative speed is (1 - 1/12) turn/hour = 11/12 tph.
So one will make a full turn relative to the other after 12/11 hours
which is 720/11 ≈ 65,45 minutes.


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