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Brine solution

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by grandh01 » Tue Sep 11, 2012 9:11 pm
How much water must be added to
10 gallons of 10% brine solution
to decrease the concentration to 7%?

1) 0---1.5gal
2) 1.5---3gal
3) 3-----4.5gal
4) 4.5----6gal
5) 6+

OA IS C
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by neelgandham » Tue Sep 11, 2012 10:25 pm
Volume of solution = 10 Units
Volume of water = 9 Units(90/100*10)
Volume of salt(eh?) = 1 Unit

Let the volume of water added be V. Then, the concentration of the new brine solution = 1/(10+V) = 7/100

70 + 7V = 100
7V = 30
V ~ 4.3 Units(Gallons)
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by Anurag@Gurome » Tue Sep 11, 2012 10:29 pm
The amount of salt in 10gallons of 10% brine solution = 1 gallon
we are adding only water so that the concentration becomes 7%.
the concentration after adding x gallons of water = 1 gallon / (10 gallons + x gallons) = 7 /100
=> x =30/7 = 4.28
hence, it is C
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by everything's eventual » Tue Sep 11, 2012 10:45 pm
The initial solution has 1 gallon salt + 9 gallons water.

We are adding water to this solution to get new solution with less concentration. So the quantity of salt will not change in the new solution.

So 1 gallon will be 7% of the new solution.

1 = (7* x)/100 where x = total gallons of new solution.

x = 14.28

So new solution is 14.28 gallons out of which 1 gallon is salt. Water = 13.28 gallons out of which 9 gallons were already present.

So water added = 13.28 - 9 = 4.28 gallons.

Answer = C
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by GMATGuruNY » Wed Sep 12, 2012 12:35 am
grandh01 wrote:How much water must be added to
10 gallons of 10% brine solution
to decrease the concentration to 7%?

1) 0---1.5gal
2) 1.5---3gal
3) 3-----4.5gal
4) 4.5----6gal
5) 6+

OA IS C
Brine is a solution of salt and water.
Since only water is to be ADDED, the amount of SALT in the original solution -- 1 gallon -- will not change.
Thus, for the resulting solution to be 7% brine, 1 gallon of salt must constitute 7% of the new total:
1 = (7/100)t
t = 100/7.

Amount of water added = new total - old total = 100/7 - 10 = 30/7 = 4 + 2/7.

The correct answer is C.
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