Question 932100
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?

<pre>

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:

{{{(matrix(8,1,
Liters,of,pure,alcohol,in,the,weaker,solution))}}}{{{""+""}}}{{{(matrix(8,1,
Liters,of,pure,alcohol,in,the,stronger,solution))}}}{{{""=""}}}{{{(matrix(7,1,
Liters,of,pure,alcohol,in,medium-strength,solution))}}}

{{{0.20+0.40(6)=0.35(x+6)}}}

Solve that and get x=2 liters.

Edwin</pre>