What quantity of a 50% acid solution must be mixed with a 20% solution to produce 600 mL of a 40% solution?

Suppose the amount of 50% solution is x ml, and the amount of 20% solution = y ml then x + y = 600. We make a second equation using the percentages so that 50% of x + 20%of y = 40% of x+y and that written using decimals is 0.5x + 0.2y = 0.4(x + y) or multiplying by 10, 5x + 2y = 4(x + y) = 4x + 4y and this reduces to x - 2y = 0 or x = 2y so if we use that last line in our first equation x + y = 600, it becomes 2y + y = 600 so y = 200 and x = 400 i.e. we need 400 ml of ther 50% solution and 200 of the 20% solution.

Let x be the amount of the 50% solution. Then the amount of the 20% solution is (600 - x).

.50x + .20(600 - x) = .40(600)

.50x + 120 - .20x = 240

.50x - .20x = 240 - 120

.30x = 120

x = 120/.30 = 400

400 ml of the 50% solution should be mixed with 200 ml of the 20% solution to make 600 ml of a 40% solution.

here is how you set it up (i think)

.5x+.2y=.4(600)

and x and y are the amounts of solution you need of each type

or just ask your teacher for the formula, I am sure he/she will be happy to tell you it