Question 220184: how much pure acid should be mixed with 3 gallons of a 20% solution in order to get a 60% acid solution? Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! Let S = amount of new solution
Then .6*S = the amount of acid in the new solution.
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Amount of Acid in existing 3 gallon solution is equal to .2 * 3 = .6 gallons of acid.
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Let X be the amount of acid We need to add to the existing solution.
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The new solution is equal to the old solution plus X because we are adding pure acid which is represented by X.
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This means that S = X + 3
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The amount of acid in the existing solution is .6 gallons
We will be adding X gallons of acid.
The Total amount of acid in the new Solution will be X + .6
That will equal to 60% of the new solution which equals .6*S
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This means that (X+.6) = .6*S
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Since we know that S = (X+3), we can susbtitute in our equation to get:
(X+.6) = .6*(X+3)
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This reduces one equation in two unknown into one equation in one unknown which we can then solve.
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We remove parentheses to get:
X + .6 = .6*X + .6*3
We Subtract .6 from both sides of the equation and we simplify to get:
X = .6*X + .6*3 - .6 which becomes:
X = .6*X + 1.2
We subtract .6*X from both sides of this equation to get:
X - .6*X = 1.2 which simplifies to:
.4*X = 1.2
We divide both sides of this equation by .4 to get:
X = 1.2/.4 = 3
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Answer is you have to add 3 gallons of acid to 3 gallons of 20% acid to get a solution that is 60% acid.
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.6 gallons of acid in the original mixture + 3 gallons of acid = 3.6 gallons in the new mixture.
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3 gallons of the original mixture + 3 gallons of of acid = 6 gallons of the new mixture.
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3.6 / 6 = .6 * 100% = 60% of acid in the new mixture.
