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:


and for the hour hand it would be:


or it could had been


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.

Sketch of the problem

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:




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:

Sketch of the problem

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:

$$6m+110^{\circ}+\left( 30^{\circ}-0.5m\right)=360^{\circ}$$



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.


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.

Sketch of the second dubious solution

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:






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?.


1 Answer 1


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).

Seven o'clock

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.

Now 180 degrees apart

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.

Now 100 degrees 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.

Backing up to 11:40 p.m to get 110 degrees.

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.

  • $\begingroup$ Probably one of my major concerns with solving these problems is that to "understand" what's going on is by making a graphic or a drawing as if you show it by plain equations is difficult to understand what is what you meant. Maybe for some not, but for me it is. Regardless if making a drawing is tedious, is better doing it than relying on Latex. Now, I suggest that is better to separate the information because as its shown continuous it is too compact and it is not easy to follow the meaning of the sentences. The thing about increasing and decreasing I don't get it. $\endgroup$ Commented Mar 30, 2018 at 8:13
  • $\begingroup$ You've already drawn these diagrams. Repeating them back to you would seem unkind. Consider the $150^\circ$ angle formed by the hands at 7:00 p.m. As time progresses, the minute hand moves away from this region, attempting to increase the angle. Also, the hour hand moves into this region, attempting to decrease the angle. The net effect is that the angle increases by $5.5^\circ/\mathrm{min}$. $\endgroup$ Commented Mar 30, 2018 at 8:19
  • $\begingroup$ Sorry, but I disagree I don't think is unkind, anyways. Thank you for adding some diagrams into your answer. Remember I'm not in your brain so it takes me time to figure out exactly how you pictured out. My method uses angles to obtain the minutes from them not the other way around. It doesn't make much sense on which is the angle that increases? I would say that $150^{\circ}$ decreases due the motion of the minute hand and "resists to change" by increasing a bit the advance of the hour hand so it advances at a rate of $0.5 ^{\circ}$ per minute. $\endgroup$ Commented Mar 30, 2018 at 8:56
  • $\begingroup$ If you use $30^{\circ}$ that means you consider this net change within the 10 minute time frame in the watch. The second part to decrease from $180$ to $100$ was not too obvious if you don't know where to put the starting point but it makes sense in the end to sum both times elapsed. $\endgroup$ Commented Mar 30, 2018 at 8:57
  • $\begingroup$ If you go backwards in time both hands move in the same direction (counterclockwise), hence the minute hand will increase from midnight to its final destination the same for the hour hand. This case seems more logical to subtract one from the other and divide by the rate of $\frac{5.5 ^{\circ}}{\textrm{min}}$ The last part seems to check with what I've found. I'm just curious if is it okay to make a generalization that will always the net change be $\frac{5.5^{\circ}}{\textrm{min}}$?. $\endgroup$ Commented Mar 30, 2018 at 8:58

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