SOLUTION: A chemist mixes two different solutions with concentrations of 20% and 40%HCL to obtain 21 L of 33 1/3 HCL. How much of each solution was used in the inverse matrix solution.

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Question 888426: A chemist mixes two different solutions with concentrations of 20% and 40%HCL to obtain 21 L of 33 1/3 HCL. How much of each solution was used in the inverse matrix solution.
Answer by lwsshak3(11628) About Me  (Show Source):
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A chemist mixes two different solutions with concentrations of 20% and 40%HCL to obtain 21 L of 33 1/3% HCL. How much of each solution was used in the inverse matrix solution.
let x=amt of 20% solution to mix
21-x=amt of 40% solution to mix
20%x+40%(21-x)=(100/3)%*21
.20x+8.4-.40x=7
.20x=1.4
x=7
21-x=14
amt of 20% solution to mix=7 L
amt of 40% solution to mix=14 L