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)   (Show Source): You can put this solution on YOUR website!
x of 20%
y of 60%







The two quantitites must be 10 Liters.


Solve this simpler system:



-
Answer by MathTherapy(10552)   (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 = 

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

Answer by amalm06(224)   (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)  

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