SOLUTION: How many liters of pure dye must be added to 45 liters of a 6% dye solution to obtain a 10% dye solution?

Question 508791: How many liters of pure dye must be added to 45 liters of a 6% dye solution to obtain a 10% dye solution?
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
Always keep track of how much 'pure' stuff you need.
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You have 45 liters of 6% solution, so it contains .06*45 = 2.7 liters of pure dye in solution
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You need to add 'x' liters of pure dye.
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6%(45) + 100%(x) = 10%(45+x)
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.06(45) + 1.00(x) = .10(45+x)
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multiply through by 100 to get rid of decimals
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6(45) + 100x = 10(45+x)
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270 +100x = 450 +10x
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subtract 270 from both sides
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100x = 180 +10x
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subtract 10x from both sides
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90x = 180
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divide both sides by 90
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x = 2
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Add 2 liters of pure dye to obtain 47 liters of 10% dye.
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Check the amount of pure dye.
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.10*47 = 4.7 liters of dye are supposed to be in 47 liters of 10% solution.
In the original 45 liters we found we had 2.7 liters (.06*45) of pure dye.
So, adding 2 liters of pure dye does it.
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Answer: Add 2 liters of pure dye.
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Done.

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