Question 504987: How much pure acid should be mixed with four gallons of a 50% acid solution in order to get a 60% acid solution?
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! With mixture problems, always keep track of how much 'pure' stuff you need.
In this case you have 50% solution, so it is 1/2 pure stuff and 1/2 other stuff.
You have 4 gallons, so the include 2 gallons of pure stuff and 2 gallons of other stuff (perhaps water).
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You need 60% mixture.
And you have pure stuff to add.
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So,
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x = amount of pure stuff to add
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50%(4) + 100%(x) = 60%(4+x)
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Notice the final volume is 4+x.
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.5(4) + 1(x) = .6(4) + .6(x)
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multiply by 10 to eliminate decimals
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5(4) + 10x = 6(4) + 6(x)
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20 + 10x = 24 + 6x
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4x = 4
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x = 1
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This suggests you add 1 gallon of pure stuff to 4 gallons of 50% mixture to end up with 5 gallons of 60% mixture.
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Always check your solution.
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5 * 60% = 3 gallons of pure stuff and 2 gallons of water
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4 * 50% = 2 gallons of pure stuff
1 * 100% = 1 gallon of pure stuff
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Correct!
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Answer: Add 1 gallon of pure stuff.
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Done.
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