SOLUTION: ou have 10 liters of a solution that is 30% alcohol, how many liters of a 80% alcohol solution must you add to obtain a solution that is 50% alcohol?

Question 664662: ou have 10 liters of a solution that is 30% alcohol, how many liters of a 80% alcohol solution must you add to obtain a solution that is 50% alcohol?
Found 2 solutions by stanbon, ReadingBoosters:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
ou have 10 liters of a solution that is 30% alcohol, how many liters of a 80% alcohol solution must you add to obtain a solution that is 50% alcohol?
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Eqution:
alcohol + alcohol = alcohol
0.30*10 + 0.80x = 0.50(10+x)
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30*10 + 80x = 50*10 + 50x
30x = 20*10
x = 6 2/3 liters (amt. of 80% solution needed)
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Cheers,
Stan H.
Answer by ReadingBoosters(3246) (Show Source): You can put this solution on YOUR website!
Establish variable
x = liters of 80%
...
Given
10 l of 30%
x l of 80%
To yield
10+x of 50%
...
Set the equation and solve for x
.3(10) + .8x = .5(10+x)
3 + .8x = 5 + .5x
Group like terms
.8x - .5x = 5 - 3
.3x = 2
Divide both sides by .3
x = 6.7 liters
.....................
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