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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:

  1. 30 sec
  2. 33 sec
  3. 45 sec
  4. 25 sec
  5. 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?.

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The number of time periods to achieve three rings is two. The number of time periods to achieve eleven rings is ten.

So $5\times 6=30$

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  • $\begingroup$ Okay, but. Mind if you explain it a bit slowly for someone who is not familiar with any of the terms which you had just used. In other words in layman terms. What is the thing that I should do first?, then what I shouldn't do? and more importantly, what are the limitations that should be considering when working with time periods, although I believe that this would had been about other things, maybe lengths. $\endgroup$ – Chris Steinbeck Bell Apr 2 '18 at 9:44
  • $\begingroup$ as I'm sure you are aware by now, the total time for a set of rings is about counting the three-second gaps between the rings, which is one less than the number of rings. $\endgroup$ – David Quinn Apr 2 '18 at 10:58
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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:

$$\textrm{(11-1)}\times3 \textrm{ sec}=10\times3 \textrm{ sec}=30 \textrm{ sec}$$

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  • $\begingroup$ That's a nice explanation. So each time when I want to compute the time interval i do have to subtract one from whatever hour is it?. I mentioned about longer time intervals because I only reached the answer by drawing individually each time and separating them by groups of three. This can be okay if let's say there are few numbers. But this can be problematic if there are many of them, so the most logical way would be to use a multiplication as you had shown. $\endgroup$ – Chris Steinbeck Bell Apr 3 '18 at 22:05
  • $\begingroup$ @ChrisSteinbeckBell exactly ... and your method would only work well if your groups would fit into the larger whole an exact number of times. So, yes, I recommend just finding the time between two consecutive bell tollings, for which you do indeed need to first subtract one. $\endgroup$ – Bram28 Apr 3 '18 at 22:41

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