How many ML of a 35% acid mixture and a 95% acid mixture should be mixed to get 120 ML of a 40% acid mixture?

ANSWER: 110ML of 35% acid solution must be mixed with 10ML of 95% solution to obtain 120ML of 40% solution. SOLUTION:   First, set up table. fill in the unknowns with variables x and y.  The table is attached below.From the table shown below, we can easily set up the two equations.  Sum of values of two acids = Value of mixture   0.35x + 0.95y = 48  For convenience, we will multiply the entire equation by 100, 35 x + 95y = 4800     ------ (1)  Now, Sum of amounts of each acid = Amount of mixture  x + y = 120    --------- (2) Multiply eqn 2  with 35 for easy calculation and derive the equation into one variable. 35 = 35x + 35y = 4200  Subtracting equation (2) from (1),  we get 0 + 60y = 600  Thus, 60y = 600  [tex]y = \frac{600}{60} = 10[/tex]Substituting y = 10 in (2), 35x + 35(10) = 4200  35x + 350 = 4200  35x = 4200 - 350  35x = 3850  [tex]x = \frac{3850}{35} = 110[/tex]So, we have x = 110 and y = 10 We can conclude that 110ML of 35% acid solution must be mixed with 10ML of 95% solution to obtain 120 ML of 40% solution.