Question 398200
What quantity x of a 65% acid solution must be mixed with a 20% solution to produce 300 mL of a 45% solution.
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Equations: 
Quantity Eq:  x+y=300 mL
Acid Eq::::: 0.65x+.20y=.45*300
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Multiply thru the 1st by 65
Multiply thru the 2nd by 100
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65x + 65y = 65*300
65x + 20y = 45*300
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Subtract and solve for "y":
45y = 20*300
y = (4/9)300
y = 133 1/3 mL (amt of 20% solution needed in the mixture)
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Solve for "x":
x+y = 300
x = 300 - 133 1/3
x = 166 3/4 mL (amt of 65% solution needed in the mixture)
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Cheers,
Stan H.