Now, the question is as follows[I'm not very sure which tags is apt for this question so please feel free to edit!]
There are $1000$ trouts in a lake. If none are caught, the population will increase at $10\%$ per year. If more than $10\%$ are caught, the population will fall. As an approximation, assume that if $25\%$ of fish are caught per year, the population falls by $15\%$ per year. Estimate the total catch before the lake is "fished out."
If the catch rate is reduced to $15\%$, what is the total catch in this case?
I am a mathematics graduate and my high-school sister came along to ask me this from class and honestly, I cannot convince myself on what makes sense as a solution. The answer however, is apparently, $6667$.
My questions are as follows
$1.$ How do we know how many percentages of fish were caught each year? All we know is that $a.$ none caught, increase by $1.1$, $b.$ more than $10\%$ caught, they "fall" but no specifications except $c.$ at exactly $25\%$ caught, we have $0.85$. What if they caught $31\%$ of the population? And how frequently would that occur? How frequently will they catch none, and catch $25\%$ etc?
$2.$ Like, say if I assume that they catch $25\%$ of the population every year (which makes the information of no-catch-is-$10\%$-increase), I get something like $250+212.5+...$ a geometric series sum. $250$ is found by $1000 \times 0.25$ and the next term is $1000\times 0.85 \times 0.25$ since the remaining population falls by $15\%$ and we catch another $25\%$ out of it in year $2$(I do this from $1000$ because,it makes no sense for the population to descrease on its own AFTER being caught by humans. So I don't do $(1000-250) \times 0.85$). This keeps going until we "fish out" the lake so I assumed that we go up to infinite number of terms i.e. the lake population tends to zero. But this gives me $1667$ by $\lim_{n \to \infty}S_n=\lim_{n \to \infty} \frac{250(0.85^n-1)}{0.85-1}$
$3.$ The answer, which is $1000+850+...$ (it's $850$ not $1000-250=750$ because, "when $25\%$ is caught, the population becomes $1000 \times 0.85=850$ which makes sense, because the remaining $750$ in the lake reproduces and becomes $850$) also makes sense if I think about it like this; we keep fishing until the lake is empty. So, everything, those that remain those that are reproduced etc will eventually get caught. So this sum "also" makes sense.
Say I accept (although not very comfortable with how we can just assume they catch exactly $25\%$ per annum. More than half of the information provided in the question is then completely unnecessary) that they catch $25\%$ every single year. Then both $2,3$ above makes sense to me as a means to find "the total catch until the lake is empty of trouts."
I just cannot reason to see why $3$ is correct. $2$ is essentially "finding the sum of the catch of each year" and $3$ is "finding the total number of fish that have every existed in the lake."
Actually, $3$ feels even wrong because, for instance $1000+850$ already is double counting the fish in year $1$ and year $2$; the $750$ fish that remained in the second year has also existed in the lake in the first year. So I am doubling this $750$ fish that I claim to be caught in the end.
Can more experienced people and fellow mathematicians come up with a possible explanation and an answer that makes sense?
The wording of the problem isn't perfect, but it was clear to me that the first part of the question was asking how many fish will be caught if they catch $25\%$ per year. Then there are initially $1000$ fish; at each step, there are $0.85^i\cdot 1000$ fish in the population, $25\%$ of which are caught. Therefore, the answer is $$\sum_{i=0}^\infty 0.25\cdot 0.85^i \cdot 1000 = \frac{250}{1-0.85} \approx 1,667$$ I might not be understanding your first question; $25\%$ are caught every year, that's just what the problem is stating.
And your reasoning as to why $2$ is correct and $3$ is incorrect seems mostly sound. Now, "the total number of fish that have ever existed in the lake" is the same amount as fish caught (since the lake is eventually fished out); this is not the error in $3$. You were correct to identify that the logic in $3$ double-counts the fish many times, which is why it leads to the larger and incorrect answer.