SOLUTION: a dealer wishes to produce a 60 pound mixture of candy to sell at $5 a pound, by mixing candy of $7 a pound and candy of $4 a pound. How many pounds of each should be used?
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-> SOLUTION: a dealer wishes to produce a 60 pound mixture of candy to sell at $5 a pound, by mixing candy of $7 a pound and candy of $4 a pound. How many pounds of each should be used?
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Question 499904: a dealer wishes to produce a 60 pound mixture of candy to sell at $5 a pound, by mixing candy of $7 a pound and candy of $4 a pound. How many pounds of each should be used? Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Mixtures and solutions work like this:
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how many liters of a 70% acid solution must be mixed with a 15% acid solution to get 385L of a 60% acid solution?
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s = amount of 70%
f = amount of 15%
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s + f = 385 (total solution)
0.7s + 0.15f = 0.6*385 (total acid)
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f = 385 - s
Sub for f in 2nd equation
0.7s + 0.15(385-s) = 231
70s + 5775 - 15s = 23100
55s = 17325
s = 315 liters of 70%
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f = 70 liters of 15%
