The problem is as follows:
A news director at a local tv station orders to break the regular schedule programming between $\textrm{7 pm}$ and $\textrm{7:30 pm}$ to cover an Olympics event exactly when the hour and minute hands of an analog watch at his desk made an angle of $100 ^{\circ}$. If we know that the station ended its coverage when the hands of the watch made an angle of $110 ^{\circ}$ for the last time during that day. How long has lasted the Olympic event?
To solve this problem I defined that the "speed" the minute hand in an analog watch is:
$$\frac{360^{\circ}}{60\,\textrm{min}}=\frac{6^{\circ}}{\textrm{min}}$$
and for the hour hand it would be:
$$\frac{30^{\circ}}{60\,\textrm{min}}=\frac{0.5^{\circ}}{\textrm{min}}$$
or it could had been
$$\frac{6^{\circ}}{12\,\textrm{min}}=\frac{0.5^{\circ}}{\textrm{min}}$$
as there are $60\times 6^{\circ}=360^{\circ}$ during those $60$ minutes.
Since now I have both speeds all that is left to do is to add up the angles and compare it if they form a full circle ($360^{\circ}$) or half one ($180^{\circ}$).
From the first part of the problem it mentions that the station interrupted its programming at some point between $\textrm{7:00 pm}$ and $\textrm{7:30 pm}$ so I assumed the situation which has been drawn in figure A.
I also defined $\textrm{m= number of minutes elapsed}$ but since I cannot relate minutes and angles directly I'm using the speed so I'm left only with angles.
From the starting point until the "desired" destination there are some $6m$ angles which had elapsed, but this doesn't seem to be useful for me, as I'm not given a number to which relate it.
However I'm given the angle between both hands and that's where I'm aiming at, so I used the $180^{\circ}-6m$ which is something that can be added up to the rest which is the $30^{\circ}$ until the hour hand is at $\textrm{7 pm}$ plus a little bit which is $0.5m$ because the hour hand has also moved and finally I equate this to the $100 ^{\circ}$ as I'm given this angle in the problem. This whole process became into the following equation:
$$180^{\circ}-6m+30^{\circ}+0.5m=100^{\circ}$$
$$5.5m=110$$
$$m=20$$
So the Olympics event started at $\textrm{7:20 pm}$ which makes sense as it is in the range stated in the problem.
Now this is the part where I'm a bit confused.
To get the destination, the last time during that day the hour hand and the minute hand of the watch made an angle of $110^{\circ}$ must be somewhere in the other half of the watch as it is referred to as the last time of the day so it must be of the hour too, therefore I assumed that this time must be somewhere before midnight and it must be between $\textrm{11:30 pm}$ and $\textrm{11:59 pm}$.
The following reasoning seen in Figure B is described below:
The number of minutes elapsed from $\textrm{11:00 pm}$ until that time must be 6m but as mentioned for the previous case, it is insufficient to say when, so we need to add up the $110 ^{\circ}$ which is the angle between both hands of the watch plus the little bit of space between the hour hand and midnight so we can complete a $360 ^{\circ}$. In order to get to know the value of this space I figured out to subtract the $30^{\circ}$ minus the space which has advanced the hour hand which is $0.5m$ in other words $30^{\circ}-0.5m$. By joining altogether what it has been described becomes in the following equation:
$$6m+110^{\circ}+\left( 30^{\circ}-0.5m\right)=360^{\circ}$$
$$5.5m=220$$
$$m=40$$
Therefore the number of minutes until that time must be $40$ and by comparing this to the range I previously defined being $\textrm{11:30 pm}$ to $\textrm{11:59 pm}$. It makes sense.
So finally all that is left to do is to make the difference between them.
$$\left(11+\frac{40}{60}\right)-\left(7+\frac{20}{60}\right)=4+\frac{20}{60}$$
So it becomes into $\textrm{4 hours and 20 minutes}$ the time which has lasted the Olympic event.
However there is an additional detail that it made me confused and it is sketched in Figure C.
In short I assumed that there is an angle of $180^{\circ}$ between $\textrm{11:00 pm}$ and $\textrm{11:30 pm}$ this angle plus $6m$ plus $110^{\circ}$ and $30^{\circ}-0.5m$ should become into $360^{\circ}$, therefore making this equation:
$$180^{\circ}+6m+110^{\circ}+30^{\circ}-0.5m=360^{\circ}$$
$$5.5m=360-30-110-180$$
$$5.5m=360-320$$
$$55m=400$$
$$m=\frac{80}{11}$$
But $\frac{80}{11}$ does not produce a number of minutes in the range I defined so I ignored this "solution" and used the other instead.
However I feel that the last reasoning was due the fact if $6m$ only works when the minute hand begins from zero, that is from the very beginning of the hour or can be used as intended in the previous equation?.
Overall did my way of solving this problem is correct or is there any other way to solve it without incurring into errors of perception?.
There's some good stuff here, but you might be overcomplicating a little. At exact hour times, you know exactly where both hands are and the angle between them.
At 7:00 p.m., the minute hand points at the "12" and the hour hand points at the "7", so there are $150^\circ$ between them (going clockwise from the hour hand to the minute hand).
That angle increases by $\frac{360^\circ}{60 \,\mathrm{min}} = 6^\circ/\mathrm{min}$ due to the motion of the minute hand and decreases by $\frac{360^\circ}{12 \,\mathrm{hr} \cdot 60 \,\mathrm{min}/\mathrm{hr}} = 0.5^\circ/\mathrm{min}$ due to the motion of the hour hand, for a net rate of increase of $5.5^\circ/\mathrm{min}$ until reaching $180^\circ$ at 7:00 p.m.${} + \frac{30^\circ}{5.5^\circ/\mathrm{min}}$, which is $5\frac{5}{11}$ minutes after 7:00 p.m.
From then on, the angle decreases from $180^\circ$ at a rate of $5.5^\circ/\mathrm{min}$ (the minute hand moves to decrease the angle and the hour hand moves to increase the angle). To decrease by $80^\circ$ to $100^\circ$ requires $\frac{80^\circ}{5.5^\circ/\mathrm{min}} = 14\frac{6}{11}$. Together, $5\frac{5}{11} + 14\frac{6}{11} = 20$ minutes have passed since 7:00 p.m. That is, the coverage of an Olympic event started at 7:20 p.m.
At midnight, the hour and minute hands point at the "12", so the angle between them is $0^\circ$.
Going backward in time, the minute hand increases the angle between the hands by $6^\circ/\mathrm{min}$ and the hour hand decreases the angle between the hands by $0.5^\circ/\mathrm{min}$, so the angle between the hands increases by $5.5^\circ/\mathrm{min}$. We want to know the last time prior to midnight that the angle between the hands was $110^\circ$, so we calculate $\frac{110^{\circ}}{5.5^\circ/\mathrm{min}} = 20 \,\mathrm{min}$. That is, the coverage ended at 11:40 p.m.
Between 7:20 p.m. and 11:40 p.m., 4 hours and 20 minutes elapse. So the Olympic event has lasted 4 hours and 20 minutes.