wishalsp333 wrote:

At a certain industrial manufacturer, an open tank contains 6% fluorine dissolved in water. If the tank holds 1000 gallons, and 10 gallons of the total volume evaporate as water each day, what is the concentration of fluorine after 25 days

A. 4%

B. 6%

C. 8%

D. 10%

E. 15%

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

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