Question 795155
Set up two equations, much like you do for coin problems. One for the value and one for the number of coins. On;y now you use liquids with a different percentage concentration and you have the total volume of the combined concentrations.
Let a = the number of milliliters of the 25% sulfuric acid solution
Let b = the number of milliliters of the 50% sulfuric acid solution
Let c = the number of milliliters of the total 40% sulfuric acid solution
Using the volume we have
(1) a + b = c or
(2) a + b = 160
Now for the mixture we have
(3) .25*a + .5*b = .4*c or
(4) a/4 + b/2 = .4*160 or
(5) a/4 + b/2 = 64 or 
(6) a + 2*b = 4*64 or
(7) a + 2*b = 256 
Now solve (7) and (2) simultaneously to get a and b.
The first step is to subtract (2) from (7) to get
(8) (a + 2*b) - (a + b) = 256 - 160 or
(9) a - a + 2*b - b = 96 or
(10) b = 96
Then from (2) we get
(11) a + 96 = 160 or
(12) a = 160 - 96 or
(13) a = 64
Let's (always) check our answer with (3).
Is (.25*64 + .5*96 = .4*160)?
Is (16 + 48 = 64)?
Is (64 = 64)? Yes
Answer: Hector should mix 64ml of the 25% solution with 96ml of the 50% solution in order to get 160ml at a 40% concentration of sulfuric acid.