If a farmer stocked his pond with 100 fish in the year 2000 and their population grows at a fixed rate of 10% how many fish will he have in the year 2035? Hint: Use the Rule of70!​

By 2035 he will have 2810 fishes.Solution:Given, if a farmer stocked his pond with 100 fish in the year 2000 and their population grows at a fixed rate of [tex]10\%[/tex] a year We have to find number of fish he will have in the year 2035. Hint: Use the Rule of 70!  Let the amount of fish in the tank be x. initially x = 100 If x grows [tex]10\%[/tex], then you have the original x plus [tex]10\%[/tex] of x.  Changing from percentage to decimal form (which is almost always necessary and is necessary this time) [tex]10 \%=\left(\frac{10}{100}\right)=0.1[/tex] so the growth is [tex]0.1x[/tex] and what was kept was 1x.  When you add up what you keep (the original principal) and the interest that you gain you have [tex]1x + 0.1x = 1.1x[/tex]  If you do this for two years it is [tex]1.1\times 1.1\times \text{x or } ((1.1)^2)\times \text{x }[/tex] and after n many years you will have [tex](1.1)^n \text{x}[/tex]In this problem, n = 35 and x = 100  So, the answer is [tex](100) \times(1.1)^{35}=(100) \times 28.102437=2,810.2437[/tex] which rounds down to 2,810 fish.