Question 282711
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Four gallons of a 45% acid solution is obtained by mixing a 90% solution with a 30% solution. 
How many gallons of each solution must be used to obtain the desired mixture?
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        The solution by @mananth is fatally wrong due to wrong setup of the basic equation.

        I came to bring a correct solution.



let the quantity of 90% solution added be x
the total solution obtained is 4 gallons
So the remaining volume will be 30% solution
=4-x gallons
The sum of the acid content of mixed solutions is equal to acid content in the mixture
0.9x + 0.3*(4-x) = 0.45*4
0.9x + 1.2 - 0.3x = 1.80
0.6x = 1.8 - 1.2
0.6x = 0.6
x = 0.6 / 0.6
x = 1 gallons which is 90% acid solution
Balance will be 30% acid solution.
4 - 1 = 3 gallons


<U>ANSWER</U>.  &nbsp;&nbsp;1 gallon of the 90% solution and 3 gallons of the 30% solution.


<U>CHECK for concentration</U>.  &nbsp;&nbsp;{{{(1*0.9+3*0.3)/4}}} = 0.45,  or 45%.  &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;! Precisely correct !
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Solved correctly.