SOLUTION: One solution contains 20% alcohol and another solution contains 60% alcohol. Some of each of the two solutions is mixed to produce 10 liters of a 50% solution. How many liters of t

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Question 1074549: One solution contains 20% alcohol and another solution contains 60% alcohol. Some of each of the two solutions is mixed to produce 10 liters of a 50% solution. How many liters of the 60% solution should be used?
Found 3 solutions by josgarithmetic, MathTherapy, amalm06:
Answer by josgarithmetic(39617) About Me  (Show Source):
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x of 20%
y of 60%

%2820x%2B60y%29%2F10=50
%282x%2B6y%29%2F10=5
%28x%2B3y%29%2F5=5
x%2B3y=25


The two quantitites must be 10 Liters.
x%2By=10

Solve this simpler system:
system%28x%2B3y=25%2Cx%2By=10%29

highlight_green%28y=7%261%2F2%29
-
highlight_green%28x=2%261%2F2%29

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

One solution contains 20% alcohol and another solution contains 60% alcohol. Some of each of the two solutions is mixed to produce 10 liters of a 50% solution. How many liters of the 60% solution should be used?
Let amount of 60% (needed solution) be S

Then amount of 20% solution = 10 - S

We then get the following MIXTURE equation: .6S + .2(10 - S) = .5(10)

.6S + 2 - .2S = 5

.6S - .2S = 5 - 2

.4S = 3

S, or amount of 60% (needed) solution to mix = highlight_green%28matrix%281%2C4%2C+3%2F.4%2C+or%2C+7.5%2C+L%29%29

That's how SIMPLE this is.....nothing COMPLEX and/or CONFUSING!

Answer by amalm06(224) About Me  (Show Source):
You can put this solution on YOUR website!
Use the method of alligation.

Denote the 20% alcohol with x and the 60% alcohol with y.

Then 50-20=30 and 60-50=10

y/x=3/1

y=(3/4)(10)=7.5 L (Answer)