How many times do a couple of bells ring close enough that they are heard as one? This is a problem from the book: Recreations in the theory of numbers by Albert Beiler.
The problem is this: You have 2 bells $B_1,B_2$. 
$B_1$ rings every $\frac{4}{3}$ seconds while $B_2$ rings every $\frac{7}{4}$seconds. 
How many strokes are heard during 15 minutes if 2 strokes following each other within an interval of $\frac{1}{2}$ second or less are perceived as one sound?
I'm not sure how to formalize this problem into a mathematical statement.
Since $\frac{4}{3} = \frac{16}{12}$ and $\frac{7}{4}=\frac{21}{12}$, calculating the least common multiple of $(16,21)=336,$ I arrived at the conclusion that $B_1$ and $B_2$ will ring at exactly the same time after $28$ seconds. In this interval, $B_1$ will have rang $21$ times while $B_2$ $16$ times, so if I find how many times in this interval, both $B_1$ and $B_2$ stroke within $\frac{1}{2}$ second, I could get the total number of strokes heard during the whole period, but I'm not sure how to do this without explicitly finding it out manually. 
I did the calculation and found out that the total strokes heard in one 28 second period is $24$, and they meet $12$ times. I get the correct answer but I would like to know if there is a less tedious way of finding the answer 
Any help would be appreciated!

In $28''$ bell $B_1$ rings $21$ times, and its sound covers $21''$ of the $28''$. The probability that a random sound of bell $B_2$ (a Dirac $\delta$) is not heard then is ${3\over4}$. During these  $28''$ bell $B_2$ emits $16$ Dirac-sounds, only $4$ of which will then actually be heard (elementary number theory takes care of that). It follows that during $28''$ we will actually hear $21+4=25$ strokes; makes $803.57$ in $15'$.
Note that we cannot assume that the bells are "in phase". In fact we have to assume that at the beginning of the $28''$ the phases of $B_1$ and $B_2$ are uniformly distributed in $\bigl[0,{4\over3}\bigr]$, resp. $\bigl[0,{7\over4}\bigr]$. This is taken care of in the above approach.
