Chapter 11.3, Problem 41PPS

To determine

Find the percent of the original contaminants will be removed from the water sample.

Answer to Problem 41PPS

  $91.9\%$

Explanation of Solution

Given:

A certain water filtration system can remove 70% of the contaminants each time a sample of water is passed through it. If the same water is passé through the system four times, what percent of the original contaminants will be removed from the water sample?

Concept Used:

Let 100 % contaminated water passes.

It means that $a_1=1$

As the system removes 70% of the contaminant, that is 30% of contamination still in the water.

So, $r=30\%=0.3,\text{and}n=4$

Calculation:

Now, we can calculate $a_4$ by using the formula: $a_n=a_1·r^{n-1}$

  $\begin{array}{l}{a}_1=1;r=30\%=0.3,\text{and}n=4;\text{find}{a}_4\\ a_n=a_1·r^{n-1}\\ $ a_4=a_1\cdot r^{4-1}=a_1\cdot r^3=1\cdot {(0.3)}^3=0.081$a_4==0.081\end{array}$

  $a_4==0.081=8.1\%$

  $100-8.1=91.9\%$ of the original contaminants will be removed.

Thus, $91.9\%$ of the original contaminants will be removed.