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Question 166423: If you have supplies of 2% alcohol and 6% alcohol. How much of each would you mix to obtain 60 liters of 3.2$ alcohol?
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! let x = amount of 2% alcohol solution.
let y = amount of 6% alcohol solution.
let z = amount of 3.2% alcohol solution.
z = 60 liters.
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x amount of 2% alcohol solution + y amount of 6% alcohol solution = 60 liters of 3.2% alcohol solution.
x + y = z
z = 60
equation for that is:
x + y = 60 (equation 1)**********************************
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.02 * x = amount of alcohol in 2% solution.
.06 * y = amount of alcohol in 6% solution.
.032 * z = amount of alcohol in 3.2% solution.
z = 60
.032 * 60 = 1.92 liters of alcohol in 60 liters of 3.2% solution.
the amount of alcohol in the 2% solution plus the amount of alcohol in the 6% solution must total up to be 1.96 liters of alcohol.
equation for that is:
.02*x + .06*y = 1.96 (equation 2)***********************************
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two equations to solve are:
x + y = 60 (equation 1)
.02*x + .06*y = 1.92 (equation 2)
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multiply equation 2 by 50:
x + y = 60 (equation 1)
x + 3*y = 96 (equation 2)
subtract equation 1 from equation 2:
2*y = 36
y = 18
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if y = 18, then x must = 42 since x + y = 60 (equation 1)
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answer is:
42 liters of 2% solution plus 18 liters of 6% solution equals 60 liters of 3.2% solution.
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x = 42
y = 18
x + y = 60 (equation 1)
42 + 18 = 60
60 = 60
equation 1 is good
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.02*x + .06*y = .032*z (equation 2)
.02*42 + .06*18 = .032*60
.84 + 1.08 = 1.92
1.92 = 1.92
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both equations are satisfied so answer is good.
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