A question about percentages of acid solution being added to more acid solutions.

Question

how many liters of an 80% acid solution must be mixed with a 15% acid solution to get 390L of a 70% acid solution?

I'm taking a math prep class in college and I never learned this stuff in high school. The professor wont help me

Answer 1

Firstly, let x = ml of 80% solution added, y = ml of 15% solution added. From there we can write down two equations.

$$\begin{eqnarray}x &+ y &= 390 \\ 0.8x &+ 0.15y &= 0.70 \times 390.\end{eqnarray}$$

Are you okay to solve it from there?

Answer 2

$$\begin{array}{|c|c|c|} \hline \text{Percent}& \text{Total Solution} & \text{Acid} \\ \hline 80\% & x & (0.8)x \\ \hline 15\% & 390-x & (0.15)(390-x) \\ \hline 70\% & 390 & (0.70)390 \\ \hline \end{array}$$

You can see we are adding the top two lines to get the bottom line. The acid must add up so you can say $$(0.8)x+(0.15)(390-x)=(0.7)390$$