Question

A health clinic uses a solution of bleach to sterilize petri dishes in which cultures are grown. The sterilization tank contains 150 gal of a solution of 4% ordinary household bleach mixed with pure distilled water. New research indicates that the concentration of bleach should be 8% for complete sterilization. How much of the solution should be drained and replaced with bleach to increase the bleach content to the recommended level?

Answer 1

4% of 150 gal has bleach: .04*150 = 6 gallons is bleach
We need it to be 8%.
That means .08 * 150 gals needs to be bleach: .08*150 = 12 gallons

When we drain, what will be left (x) will be 4% bleach, what we add (y) will be 100% bleach. Together we still need it to add up to 150 gallons, so

(.04*x + y)/150 = .08

This represents the fraction of fluid that will be bleach after adding y gals of bleach to the fraction of bleach that remains, which is .04x.

We also have that x+y=150.

So we have 2 equations with 2 unknowns, solvable:
$$
\cfrac{.04*x + y}{150} = .08\\
.04*x + y=12\\
x+y=150\\
\implies y=150-x\\
\text{So,}\\
.04x+150-x=12\\
x(.04-1)=12-150\\
x=\cfrac{12-150}{.04-1}\\
x=147.75\\
y=150-x=150-147.75=6.25
$$
So we must drain 6.25 gallons of the original 150 gallons that contained 4% bleach then add back 6.25 gallons of bleach to make a new batch of 150 gallons that now contains 8% bleach.

Let's check:
$$
\cfrac{6.25+.04*147.75}{150}=0.08
$$
Which is what we wanted.

Does this make sense?