SOLUTION: What quantity of a 60% acid solution must be mixed with a 20% solution to produce 200 mL of a 55% solution? I don't know what to try, everything the book has to say on this pr

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Question 511529: What quantity of a 60% acid solution must be mixed with a 20% solution to produce 200 mL of a 55% solution?

I don't know what to try, everything the book has to say on this problem provides no help what-so-ever.
Answer by oberobic(2304) (Show Source):
You can put this solution on YOUR website!
Keep track of how much 'pure' stuff you need.
You need 200 mL of a 55% solution, so you need
.55*200 = 110 mL of 'pure' acid and 90 mL of water (or other stuff).
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You need to mix two solutions to arrive a defined total volume, so you need to state the two amounts in terms of one unknown.
x mL of 60% acid
200-x mL of 20% acid
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60%(x) + 20%(200-x) = 110 mL
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.6x + .2(200-x) = 110
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multiply by 10 to eliminate decimal.
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6x + 2(200-x) = 1100
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6x + 400 -2x = 1100
4x = 700
x = 700/4
x = 175
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200-x = 25
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So this suggests you need:
175 mL of 60% solution
25 mL of 20% solution
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Calculate the total 'pure' acid to see if this answer is correct.
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.6*175 = 105 mL
.2*25 = 5 mL
105 + 5 = 110 mL
which is exactly how much pure acid you needed in the 200 mL for it to be 55% solution.
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Answer: Mix 175 mL of 60% acid with 25 mL of 20% acid to make 200 mL of 55% acid.
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Done.