SOLUTION: how many liters of a 20% alcohol should be added to 6 liters of a 40% alcohol solution so that the mixture is a 35% alcohol solution?

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Question 932100: how many liters of a 20% alcohol should be added to 6 liters of a 40% alcohol solution so that the mixture is a 35% alcohol solution?
Found 2 solutions by ewatrrr, Edwin McCravy:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
x the amount of the 20% solution
.20x + .40(6L) = .35(6L + x)
.05(6L)/.15 = x = 2L, the amount of the 20% solution

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
ow many liters of a 20% alcohol should be added to 6 liters of a 40% alcohol solution so that the mixture is a 35% alcohol solution? 

Make this chart:

                         |  liters   | percent as |  liters of   |
                         | of liquid | a decimal  | pure alcohol |
-------------------------|-----------|------------|--------------|
      weaker solution    |           |            |              |
     stronger solution   |           |            |              | 
-------------------------|-----------|------------|--------------|
medium strength solution |           |            |              | 


Let x be the answer, x = no. of liters of weaker solution.
And fill in what is given:

                         |  liters   | percent as |  liters of   |
                         | of liquid | a decimal  | pure alcohol |
-------------------------|-----------|------------|--------------|
      weaker solution    |     x     |   0.20     |              |
     stronger solution   |     6     |   0.40     |              | 
-------------------------|-----------|------------|--------------|
medium strength solution |           |   0.35     |              | 


Now we get the number of liters of liquid by adding x and 6, getting x+6,


                         |  liters   | percent as |  liters of   |
                         | of liquid | a decimal  | pure alcohol |
-------------------------|-----------|------------|--------------|
      weaker solution    |     x     |   0.20     |              |
     stronger solution   |     6     |   0.40     |              | 
-------------------------|-----------|------------|--------------|
medium strength solution |    x+6    |   0.35     |              | 


Next we get the number of liters of pure alcohol in each by multiplying
the number of liters of liquid by the percentage of each:

                         |  liters   | percent as |  liters of   |
                         | of liquid | a decimal  | pure alcohol |
-------------------------|-----------|------------|--------------|
      weaker solution    |     x     |   0.20     |   0.20x      |
     stronger solution   |     6     |   0.40     |   0.40(6)    | 
-------------------------|-----------|------------|--------------|
medium strength solution |    x+6    |   0.35     | 0.35(x+6)    |

The equation comes from:

%28matrix%288%2C1%2C%0D%0ALiters%2Cof%2Cpure%2Calcohol%2Cin%2Cthe%2Cweaker%2Csolution%29%29%22%22%2B%22%22%22%22=%22%22

0.20%2B0.40%286%29=0.35%28x%2B6%29

Solve that and get x=2 liters.

Edwin