What quantity of 60% acid solution must be mixed with a 30% solution to produce 300ml of a 50% solution?

Start by defining variables to represent the amounts of each of the solutions that are going to be mixed.

So let x = the amount of the 60% solution, and y = the amount of the 30% solution.

First of all, you know that x + y = 300, because after they are mixed you want to have 300 ml of the 50% solution.

Second, look at how much acid is in each of the starting solutions and how much is in the ending solution.

The 60% solution is 60% acid, so the amount of acid in it is 60% of x, or .6x.  Similarly, the amount of acid in the 30% solution is 0.3y

At the end, the amount of acid you want to end up with is 50% of 300ml, so that's 150 ml.

When you mix the two solutions, you're not going to change the total amount of acid you started with.  This means that 0.6x + 0.3y = 150

So you have two solutions with two unknowns, and that's something you can solve:

Let's use elimination.  Multiply the first equation through by 0.6:

0.6x + 0.6y = 0.6*300 = 180

Since x + y = 300, that means that x = 200.

So you will need 200 ml of 60% solution and 100 ml of 30% solution.