Question 994900
A chemist needs 10L of 40% solution of alcohol. He plans to 
mix a 60% solution with a 20% solution. How many liters of 
each solution must he mix to produce the required solution?
<pre>
Actually in this problem ONLY, since 40% just happens to be 
exactly half-way between 60% and 20%, we know immediately that 
it would take exactly the same amount of each.  So we know that
to get 10 liters, we must use 5 liters of each.  However, below 
is the way we would solve such a problem in ALL cases, whether 
the desired percentage is half-way between the given percentages 
or not.
 
Let x = the number of liters of the 40% solution, and
Let y = the number of liters of the 60% solution.

Then we make this chart:


                    number        percent       number  
                      of          of pure       of liters 
                    liters        alcohol       of pure
                      of          in each       alcohol  
                     each         solution      in each  
                   solution     as a decimal    solution
--------------------------------------------------------
First solution        x             0.60        0.60x       
Second solution       y             0.20        0.20y
-------------------------------------------------------
Final solution       10             0.40        0.40(10)

The first equation comes from the number of liters of 
solution in each solution:

                        x + y = 10

The second equation comes from the the number of liters
of pure alcohol in each solution:

                0.60x + 0.20y = .40(10) 

We remove the decimals by multiplying each term by 100,
which involves moving the decimals two place right.

                    60x + 20y = 40(10)
                    60x + 20y = 400

We can simplify that further by dividing through by 20

                       3x + y = 20

Now we have the system of equations

                     {{{system(x+y=10,3x+y=20)}}}

We solve that either by substitution or elimination.  I'll
choose substitution.  Solve the first equation for y:

                            y = 10-x

Substitute in the second equation:
                      
                    3x+(10-x) = 20
                      3x+10-x = 20
                        2x+10 = 20
                           2x = 10
                            x = 5 liters of 60%

          Substitute x=5 in y = 10-x
                            y = 10-5
                            y = 5 liters of 20%

 
Edwin</pre>