Question 1200676
.
Ulysses made two alcohol solutions of different concentrations. 
When he mixed 4 liters of the first solution with 6 liters if the second, 
he obtained 35% solution of alcohol. 
But when he mixed 6 liters of the first solution with 4 liters of the second, 
he obtained a 50% solution of alcohol. 
What would be the percentage of alcohol if he mixed 2 liters of first solution 
with 8 liters of the second?
~~~~~~~~~~~~~~


<pre>
Let x be the concentration of the alcohol in the first solution,
      expressed as a decimal number.
Let x be the concentration of the alcohol in the second solution.


Write equations for the concentrations as you read the problem

    4x + 6y = 0.35(4+6),  or  4x + 6y = 3.5,   (1)

    6x + 4y = 0.5*(6+4),  or  6x + 4y = 5.     (2)


To solve equations (1), (2), multiply eq(1) by 3; multiply eq(2) by 2.  You will get

    12x + 18y = 10.5    (1')

    12x +  8y = 10      (2')


Subtract eq(2') from eq(1')

          10y = 0.5  --->  y = 0.05.


Then from equation (2)

     6x = 5 - 4*0.05 = 4.8  --->  x = 4.8/6 = 0.8.


Thus the concentrations of the original mixtures are  x= 0.8,  y= 0.05.


Then the concentration of the desired mixture is  

    C = {{{(2x + 8y)/(2+8)}}} = {{{(2*0.8 + 8*0.05)/10}}} = {{{2/10}}} = 0.2 = 20%.    <U>ANSWER</U>
</pre>

Solved.