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Solution problem

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by oks » Sat May 16, 2009 11:40 am
A 48 gallon solution of salt and water is 10% salt. How many gallons of water must be added to the solution in order to decrease the salt to 8% of the volume?
(a) 8
(b) 12
(c) 13
(d) 14
(e) 16

OA: B

Thanks!
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by truplayer256 » Sat May 16, 2009 12:31 pm
Of the 48 gallons of solution, 48(1/10) or 4.8 gallons of it is made up of salt. The question is asking us to find out how much water we must add to the 48 gallons so that salt only makes up 8 percent of the solution instead of 10. This is one way to go about solving it:

4.8/48+x=2/25
120=96+2x
24=2x
x=12 B.
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by dumb.doofus » Sat May 16, 2009 11:06 pm
8% of 48+x = 4.8

solve it to get x = 12
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Re: Solution problem

by Vemuri » Sun May 17, 2009 9:12 am
The equation you form should be either for only salt or only for water. The above solutions are based on salt quantity. Lets check how to arrive at the answer comparing only water.

Let x --> be the gallons of water being added.

90%(solution) + x gallons of water = 92%(solution)
0.9(48)+x = 0.92(48+x)
==> 8x=96
==> x=12 gallons

Hope this helps
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