how many liters of 30% saline solution would it take to turn an 8-liter container of 15% saline solution into 25%?

Question

how many liters of 30% saline solution would it take to turn an 8 - liter vessel of 15% saline solution into a 25% solution? I got this question on a test and did not know the answer, I havent been taught anything like this before. It would be helpful if there was a formula but there were not many solutions online except for a few specific combinations

Answer 1

To do these type of problems, you have to distinguish very clearly between the saline volume versus the total volume (of solutions).

At the start, the total volume = $8$, and the saline volume = $8\times 15\% = 1.2$

Now, suppose that you add a total volume of $x$. Then the saline volume will increase by $0.3x$.

Now, the total volume is $8 + x$, and the saline volume is $1.2 + 0.3x$.

Finally, you want the concentration to turn into $25\%$, which means you want

$$\frac{\text{saline}}{\text{total}} = \frac{1.2 + 0.3x}{8 + x} = 0.25$$

I believe you can take it from here.

Answer 2

An equation has already been given to you in an answer: what I'll do is give you a much shorter method for such problems that will help a lot in competitive exams where time is short

Had you been asked to raise the $15$% saline to the mid-point between $15$% and $30$%, you'd at once have answered that $8L$ of $30$% would be needed

But since you need $15$% to rise by $10$% points,
while $30$% is to fall by $5$% points,

by inverse proportion to the distances from desired point,
you need $16L$ of the $30$% solution.