Question

A lake initially contains 3000 fish. Suppose that in the absence of predators or other causes of removal, the fish population increases by 6% each month. However, factoring in all causes, 400 fish are lost each month. How many fish will be in the pond after 9 months? (Don't round until the very end.)

Answer 1

Let x = number of months.

Let y = total number of fish in pond.

6% of 3,000 = 180 fish

We can use an equation to help us with this problem.

y = 3,000 + 180x - 400x. Combine like terms.

y = 3,000 - 220x. Plug 9 in for x.

y = 3,000 - 220(9). Solve for y

y = 1,020.

There are 1,020 fish in the pond after 9 months.