SOLUTION: A chemist has one solution that is 80% acid and another solution that is 30% acid. How much of the first (80%) solution is needed to make a 400L solution that is 62% acid?
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Polynomials-and-rational-expressions
-> SOLUTION: A chemist has one solution that is 80% acid and another solution that is 30% acid. How much of the first (80%) solution is needed to make a 400L solution that is 62% acid?
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Question 40867: A chemist has one solution that is 80% acid and another solution that is 30% acid. How much of the first (80%) solution is needed to make a 400L solution that is 62% acid?
__________liters of the 80% solution is required. (I found out this answer is not 128)
How much of the second (30%) solution is needed to make a 400L solution that is 62% acid?
_____ liters of the 80% solution is required. (I found out this answer is not 272)
Thanks soooo much!!!!! Answer by fractalier(6550) (Show Source):
You can put this solution on YOUR website! Let x be the amount of 80% acid. Thus 400-x is the amount of 30% acid. Then the setup is
x(.80) + (400-x)(.30) = 400(.62)
.8x + 120 - .3x = 248
.5x = 128
x = 256 liters of 80%
400-x = 144 liters of 30%
