Question 1184101
Suppose that one solution contains 30% alcohol and another solution contains
80% alcohol. How many liters of each solution should be mixed to make 12.5
liters of a 60% alcohol solution?
<pre>
This is mixing a weaker solution of alcohol with a stronger solution of alcohol and ending up with a medium strength solution of alcohol.

It can be done with one unknown, as the above shows, but very often, word
problems that have two answers are much easier to think through and set up
if you use two unknowns.


                          | percent as |       liters of
                   liters |  decimal   | pure alcohol contained
-------------------------------------------------------------------
weaker solution  |   x    |     0.3    |      0.3x
stronger solution|   y    |     0.8    |      0.8y
-------------------------------------------------------------------
medium-strength  |  12.5  |     0.6    | 12.5(0.6) which is 7.5
       solution  |        |            |
So the system comes from the 1st and 3rd columns:

{{{system(x+y=12.5,0.3x+0.8y=7.5)}}}

Solve that and get x = 5 liters of 30% and y = 7.5 liters of 80%.

As a very rough check, make sure the answer makes sense.  That is, the
medium solution, 60%, is closer to the stronger, 80%, than it is to the
weaker 30%. So common sense tells you that it will take more of the stronger
than the weaker.  So make sure to observe that 7.5 liters is more than 5 liters.

Edwin</pre>