At a certain industrial manufacturer, an open tank contains 6% fluorin

Track on fluorine.

1) Not much algebra
1,000 gallons liquid
6% fluorine = (.06 * 1,000) = 60 gals fluorine

Water evaporates. Water is 0.0% fluorine
NO EFFECT on quantity of fluorine

10 gals of water/day evaporate for 25 days: 250 gallons of liquid are gone

(1,000 - 250) = 750 gals liquid remain

Percent fluorine after =\(\frac{60}{750}=\)
\(\frac{6}{75}=.08*100\)= 8 percent

Answer C

2) Algebra
Let x/100 = percent fluorine

Water is 0.0% fluorine

Water evaporation goes on the RHS in this equation. Water decrease = fluorine concentration increase

\(.06(1,000)=\frac{x}{100}*(1,000 - 250)\)
\(60=\frac{750}{100}x\)
\(60=7.5x\)
\(x=\frac{60}{7.5}=8\) percent

(Percent is already accounted for)

Answer C