Problem related to a clock I faced the following problem:

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

*

*$\quad$ 4-50'-31"

*$\quad$ 4-52'-51"

*$\quad$ 4-53'-23"

*$\quad$ 4-54'-33"


Can someone point me in the right direction?
 A: 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?
A: 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

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