Question

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 of

70!​

Answer 1

The number of fish in the pond in 2035 will be: 3200

We can calculate the number of fish in the pond in 2035 using the following formula:

number_of_fish_in_2035 = initial_fish_count * (1 + growth_rate) ^ number_of_years

Where:

initial_fish_count is the number of fish in the pond in the year 2000 (100)

growth_rate is the annual growth rate of the fish population (10%)

number_of_years is the number of years between 2000 and 2035 (35)

Plugging in these values, we get:

number_of_fish_in_2035 = 100 * (1 + 0.10) ^ 35 = 3200

Therefore, the farmer will have 3200 fish in his pond in the year 2035.

Answer 2

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
`10\%` 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
`10\%`, then you have the original x plus
`10\%` of x.

Changing from percentage to decimal form (which is almost always necessary and is necessary this time)
`10 \%=\left((10)/(100)\right)=0.1` so the growth is
`0.1x` and what was kept was 1x.

When you add up what you keep (the original principal) and the interest that you gain you have
`1x + 0.1x = 1.1x`

If you do this for two years it is
`1.1* 1.1* \text{x or } ((1.1)^2)* \text{x }` and after n many years you will have
`(1.1)^n \text{x}`

In this problem, n = 35 and x = 100

So, the answer is
`(100) *(1.1)^(35)=(100) * 28.102437=2,810.2437` which rounds down to 2,810 fish.