Let x be the pure acid volume (in liters), and

let y be the volume of the 10% acid solution to be mixed.


Then you have these two equations:


    for the total liquid volume

        x + y = 90             (1)


    for the pure acid volume

        x + 0.1y = 0.32*90     (2)


Subtract equation (2) from equation (1). You will get


            0.9y = 90 - 0.32*90.

Next divide both sides by 0.9. You will get


               y =  = 68.


Thus you need 68 liters of the 10% acid solution.


To find x, use equation (1)


    x + 68 = 90


and get from it


    x = 90 - 68 = 22.


The problem is just solved.  You need  22 liters of the pure acid and  68 liters of the 10% acid solution.    ANSWER