a pharmacist mixed some 10% saline solution with some 15% saline solution to obtain 100ml of a 12%saline solution.How much of 10% saline solution did the pharmist use in the mixture.dO NOT UNDERSTAND TO MUCH INFO.

.024 divided by .235

how did you get that

let the amount of 10% solution be x ml

then the amount of 15% solution must be 100-x ml

.10x + .15(100-x) = .12(100)
10x + 15(100-x) = 12(100)
10x + 1500 - 15x = 1200
-5x = -300
x = 60

31.25 liters

To solve this problem, we can use the concept of the concentration of a solution.

Let's assume that the pharmacist used x milliliters of the 10% saline solution.

Since the final volume of the mixture is 100 ml, the remaining volume would be (100 - x) ml.

Now, let's calculate the amount of salt (saline) in the 10% saline solution.

In a 10% solution, 10% of the volume is salt. Therefore, the amount of saline in x ml of the 10% solution is 0.10x ml.

Similarly, the amount of salt in the (100 - x) ml of the 15% saline solution is 0.15(100 - x) ml.

According to the problem, the mixture should have a 12% concentration, which means the amount of salt in the mixture is 0.12 * 100 ml.

Since the amount of salt in the total mixture should be equal to the sum of the amounts of salt in the individual solutions,

0.10x + 0.15(100 - x) = 0.12 * 100

Now you can solve this equation to find the value of x, which represents the amount of the 10% saline solution used by the pharmacist.