Question

In the tank that is not fully filled is 88% acid. The rest of the tank is filled with 24% acid. After mixing, the same amount that has been filled is taken from the tank. The rest of the tank is fill

Answer 1

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:

x + y = 300

0.6x + 0.3y = 150

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

0.6x + 0.6y = 0.6*300 = 180

Now you have:

0.6x + 0.6y = 180

0.6x + 0.3y = 150

Subtract:

0.3y = 30

Divide by 0.3

y = 30 / 0.3 = 100

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

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