<PRE><B>Question 821488</B>

two vessels contain mixtures of sulfuric acid and distilled water,
we can do this in percent acid
:
 one of which contains twice as much as water as acid,
this solution is 33% acid (.33)
:
 and the other three times as much acid as water.
this one is 75% acid (.75)
:
 how much must there be taken from each to fill a one litre cup, in which the acid and water shall be equally mixed? 
:
let x = amt of the 1st solution (in liters)
the resulting amt is to 1 liter, therefore
(1-x) = amt of the 2nd solution
:
A typical mixture equation
.33x + .75(1-x) = .50(1)
.33x + .75 - .75x = .5
.33x - .75x = .5 - .75
-.42x = -.25
x = -.25/-.42
x ~ .6 liters of the 1st solution
then
1 - .6 = .4 liters of the 2nd solution

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