Question

The toxicity level of a lake is found by dividing the amount of dissolved toxins the lake water currently has per liter by the maximum safe amount of dissolved toxins that the water can hold per liter and then converting it to a percentage. If the river currently has 0.86 milligrams of dissolved toxins per liter of water and the maximum safe amount of dissolved toxins is 1.04 milligrams per liter, what is the toxicity level of the lake water, to the nearest percentage? A. 86% B. 84% C. 83% D. 80% E. 79%

Answer 1

Correct option: (C) 83%.

Explanation:

It is provided that the toxicity level of a lake is found by dividing the amount of dissolved toxins the lake water currently has per liter by the maximum safe amount of dissolved toxins that the water can hold per liter and then converting it to a percentage.

Denote the variables as follows:

x = amount of dissolved toxins the lake water currently has per liter

y = maximum safe amount of dissolved toxins that the water can hold per liter

The amount of dissolved toxins per liter of water in the river is, x = 0.86 mg/lt.

The maximum safe amount of dissolved toxins that the river water can hold is, y = 1.04 mg/lt.

Compute the toxicity level of the river water as follows:

`\text{Toxicity level}=(x)/(y)* 100`

`=(0.86)/(1.04)* 100\\\\=82.69231\%\\\\\approx 83\%`

Thus, the toxicity level of the river water is 83%.

The correct option is (C).