SOLUTION: How much of the solutions do I need to use when one container has 40% and another has 100% and 50% is needed for one gallon?I came up with a huge number.

Question 234293: How much of the solutions do I need to use when one container has 40% and another has 100% and 50% is needed for one gallon?I came up with a huge number.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How much of the solutions do I need to use when one container has 40% and another has 100% and 50% is needed for one gallon?
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Equation:
0.40x + 1(1-x) = 0.50(1)
Multiply thru by 100 to get:
40x + 100(1-x) = 50
40x + 100 -100x = 50
-60x = -50
x = 5/6 gal. (amount of 40% solution needed in the mixture)
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1-x = 1-(5/6) = 1/6 gal. (amount of 100% solution needed in the mixture)
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Cheers,
Stan H.
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