SOLUTION: A chemist has one solution that is 80% acid and another that is 30% acid. He needs 200 liters of a solution that is 62% acid. How much of each should be used?
Let x=number
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Question 353720: A chemist has one solution that is 80% acid and another that is 30% acid. He needs 200 liters of a solution that is 62% acid. How much of each should be used?
Let x=number of liters of 80% solution
Y= number of liters of 30% solution
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A chemist has one solution that is 80% acid and another that is 30% acid. He needs 200 liters of a solution that is 62% acid. How much of each should be used?
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Equation:
acid + acid = acid
0.80x + 0.30(200-x) = 0.62*200
Multiply thru by 100 to get:
80x + 30*200 - 30x = 62*200
50x = 32*200
x = 4*32 = 128 liters (amt. of 80% solution needed in the mixture)
200-x = 72 liters (amt. of 30% solution needed in the mixture)
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Cheers,
Stan H.
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