SOLUTION: How many liters of a 40%-alcohol solution must be mixed with 10 liters of a solution that is 80% alcohol to get a solution that is 60% alcohol?
Question 263609: How many liters of a 40%-alcohol solution must be mixed with 10 liters of a solution that is 80% alcohol to get a solution that is 60% alcohol? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! How many liters of a 40%-alcohol solution must be mixed with 10 liters of a solution that is 80% alcohol to get a solution that is 60% alcohol?
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let x = amt of 40% solution required
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A typical mixture equation
:
.4x + .8(10) = .60(x+10)
.4x + 8 = .6x + 6
8 - 6 = .6x - .4x
2 = .2x
x =
x = 10 liters of 40% solution required
:
:
Check
.4(10) + .8(10) = .6(10+10)
4 + 8 = .6(20)
