Problem related to a clock

Question

I faced the following problem:

At what time after 4 o'clock, the hour and the minute hand will lie opposite to each other?

1. $\quad$ 4-50'-31"
2. $\quad$ 4-52'-51"
3. $\quad$ 4-53'-23"
4. $\quad$ 4-54'-33"

Can someone point me in the right direction?

Answer 1

Figure out how far (how many degrees) the hour hand moves in an hour; in a minute; in a second; then you'll be able to tell where the hour hand is (how far from straight up) at, say, 4:50:31.

Then, figure out how far the minute hand moves in a minute; in a second; and you'll be able to tell where the minute hand is at, say, 4:50:31.

And once you know where both hands are, you can tell whether they are opposite, because you know how many degrees apart they have to be to be opposite, right?

Answer 2

At $4$ o'clock the angle between the hour & the minute hand is $-\frac4{12}\cdot360^\circ=-120^\circ$

To be at $180^\circ,$ the difference of angle to be generated will be $180^\circ-(-120^\circ)=300^\circ$

Now, the hour hand moves in $12$ hours $360^\circ$

So, it moves in $1$ hour $=60$ minutes $\frac{360^\circ}{12}=30^\circ$

Similarly, the minute hand moves in $1$ hour $=60$ minutes $360^\circ$

So, the minutes hand moves $(360-30)^\circ=330^\circ$ faster in $60$ minutes

So, it will move $300^\circ$ faster in $60\cdot\frac{300^\circ}{330^\circ}$ minutes $=\frac{600}{11}$ minutes $=54+\frac6{11}$ minute $=54$ minute $32+\frac8{11}$ second

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Alternatively, let's start with $12$ o'clock when the angle between the hands is $0^\circ$

Now, the hands will lie opposite to each other if the angle between them is $n360^\circ+180^\circ=(2n+1)180^\circ$ where $n$ is any integer

As we have seen the difference will be $330^\circ$ in $60$ minutes

So, the difference will be $(2n+1)180^\circ$ in $\frac{(2n+1)180^\circ}{330^\circ}60$ minutes $=\frac{(2n+1)360}{11}$ minutes after $12$ o'clock

Now, as we need the time to after $4$ o'clock i.e., $4$ hours $=4\cdot60$ minutes after $12$ o'clock,

$\frac{(2n+1)360}{11}$ must be $>240\implies n>\frac{19}6$

As $n$ is an integer, $n_{\text{min}}=4$

So, the time will $\frac{(2\cdot4+1)360}{11}$ minutes after $12$ o'clock

i.e., $\frac{(2\cdot4+1)360}{11}-240$ minutes after next $4$ o'clock

$$\text{Now, }\frac{(2\cdot4+1)360}{11}-240=\frac{120}{11}(27-2\cdot11)=\frac{600}{11}$$

Answer 3

At $4$ o'clock the large hand is $120^\circ$ behind. The next time the two hands are in opposition the large hand has a lead of $180^\circ$. The question therefore is: How long does it take for the large hand to make good $300^\circ$.

Now the large hand moves ${360^\circ\over 60}=6^\circ$ per minute, but during this minute the small hand moves a twelfth of that in the same direction. Therefore the net gain per minute of the large hand is only $5.5^\circ$.

These hints should suffice to finish off the problem.