How much pure acid should be mixed with 6 gallons of a 50 ac

How much pure acid should be mixed with 6 gallons of a 50% acid solution in order to get an 80% acid solution? A. 9 gal B. 24 gal C. 15 gal D. 3 gal

Solution

We know we have 6 gallons of 50% acid solution. Multiplying 6 by .5 = 3 gallons of pure acid in the solution.
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We need to know how much more pure acid we need to add to get an 80% acid solution.
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Set the unknown to x.
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Since we are adding x gallons of pure acid, the final solution will be 6+x gallons at 80% acid. We can state this algebraically as: .8(6+x)
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It will consist of 6 gallows of .5 acid solution + x gallons of 100% acid. We can show this as
.5(6) + 1(x)
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Of course, we don\'t normally show 1 times anything. But I just wanted to be clear.
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Combining we have:
.8(6+x) = .5(6) + x

Multiplying what we have in parentheses
4.8 + .8x = 3 + x
1.8+ .8x = x
1.8= .2x
Dividing both sides by .2
9= x
x = 9
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So this suggests we need to add 9 gallons of pure acid to get to an 80% solution.
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If we add 9 gallons to 6 gallons, we have 15 gallons of solution in total.
If it is 80% acid, then we would have .8(15) = 12 gallons of pure acid in the solution.
We started with 3 gallons of pure acid and added 12 gallons of pure acid = 15.