How do I find the total number of bell tollings and avoid the flagpole error? The problem is as follows:

In a certain town there is a clocktower which happens to sound its
  bell the same number of times as the hour that is registered on the
  clock. If the time between each chime is always the same and to
  announce that it is $\textrm{3 hours}$ spends $\textrm{6 seconds}$.
  How long will it take to announce that it is $\textrm{11 hours}$?

The alternatives given are:


*

*30 sec

*33 sec

*45 sec

*25 sec

*50 sec


Initially I thought to solve this problem by "calling" the convertion factor that it is indicated as the problem mentions:
$$\frac{\textrm{6 sec}}{\textrm{3 hour}}$$
therefore if says $\textrm{11 hour}$ then
$$\textrm{11 hour}\times\frac{\textrm{6 sec}}{\textrm{3 hour}}=22\,\textrm{sec}$$
But this answer $22$ does not appear in any of the alternatives.
Therefore what I did was to separate by sketching the situation in a ruler.
$$1,\,2,\,3,\,4,\,5,\,6,\,7,\,8,\,9,\,10,\,11$$
Between $1$ to $3$ is $\textrm{6 seconds}$
Between $3$ to $5$ is also $\textrm{6 seconds}$
Between $5$ to $7$ is $\textrm{6 seconds}$ again
Between $7$ to $9$ is $\textrm{6 seconds}$
Between $9$ to $11$ is $\textrm{6 seconds}$
Therefore summing all of those would become into:
$6+6+6+6+6=30\,\textrm{sec}$
And this answer does appears in the alternatives in my book. Would this answer be okay?
My question arises why it didn't work in the first place? Did I incurred in a flagpole error?. I do not like the method that I used. Because it might be impractical if the question would involved a longer range, let's suppose maybe a digital 24 hour clock or another kind of problem which would take an even larger interval.
I tried to "force" my method of understanding of the situation by modifying the equation into:
$$\left ( 11-1 \,\textrm{hour}\right ) \times \frac{6\,\textrm{sec}}{\left( \textrm{3-1 hour} \right)}$$ 
this "new equation" does seem to concur with what I have found by "expanding" the whole problem and counting individually each segment.
$$\left ( 11-1 \,\textrm{hour}\right ) \times \frac{6\,\textrm{sec}}{\left( \textrm{3-1 hour} \right)}=\left ( 10\,\textrm{hour} \right ) \times \frac{6\,\textrm{sec}}{2\,\textrm{hour}}=30\,\textrm{sec}$$
So I got to $30$ as mentioned.
But again, is there any way to avoid counting individually?, as it can be impractical and tedious if the problem would had involver a much larger interval. Can somebody help me with a clear and easier way to understand solving this problem?. Finally it would help me a lot if somebody could explain me which kind of situations produces a flagpole error and how we can avoid incurring in one? Does this problem can led someone into this particular error?.
 A: The number of time periods to achieve three rings is two.
The number of time periods to achieve eleven rings is ten.
So $5\times 6=30$
A: Yes, with your first approach:

$$\textrm{11 hour}\times\frac{\textrm{6 sec}}{\textrm{3 hour}}=22\,\textrm{sec}$$

you did succumb to the flagpole error, because you used the total number of flagpoles ($11$ and $3$) rather than the number of flagpoles minus one. 
Indeed, when you later corrected for the flagpole error, you got:
$$\textrm{(11-1) hour}\times\frac{\textrm{6 sec}}{\textrm{(3-1) hour}}=30\,\textrm{sec}$$
and now it works.
You need to do the latter, since you are asked the amount of time it takes to announce the hour, and assuming that each individual chime is instantaneous, you shouldn't focus on the number of chimes, but rather on the intervals between the chimes. But, as the title of your question reveals, you did focus on the "total number of bells tollings", and thus succumbed to the flagpole error.
You ask about what happens when there are larger number involved.  I am not sure why that would change the math, really, other than that you are dealing with larger numbers. But the formulas should all remain exactly the same.
Finally, I think a little more straightforward is to just figure out the time it takes for each ring, which is
$$\frac{\textrm{6 sec}}{(3-1)}=\frac{\textrm{6 sec}}{2}=3 \textrm{ sec}$$
So to signal $11$ hours, it takes:
$$\textrm{(11-1)}\times3 \textrm{ sec}=10\times3 \textrm{ sec}=30 \textrm{ sec}$$
