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30 Jun 2018, 06:50
Kudos
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C
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Difficulty:
35%
(medium)
Question Stats:
72% (01:49) correct
28%
(02:07)
wrong
based on 651
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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%
A. 4%
B. 6%
C. 8%
D. 10%
E. 15%
30 Jun 2018, 09:13
Kudos
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Quantity of fluorine - 6% of 1000= 60 gallons
10 gallons of water evaporate everyday
So in 25 days - 250 gallons of water evaporates
remaining quantity -1000-250=750
concentration of fluorine after 25 days - (60/750 )*100=8% Ans C
Posted from my mobile device
10 gallons of water evaporate everyday
So in 25 days - 250 gallons of water evaporates
remaining quantity -1000-250=750
concentration of fluorine after 25 days - (60/750 )*100=8% Ans C
Posted from my mobile device
30 Jun 2018, 18:46
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wishalsp333
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
