Percent mixture problem using linear equation

Question

In the lab, Lisa has two solutions that contain alcohol and is mixing them with each other. Solution A is 2% alcohol and Solution B is 6% alcohol. She uses 200

milliliters of Solution A. How many milliliters of Solution B does she use, if the resulting mixture is a 5% alcohol solution?

Answer 1

Assume that the amount of solution B required is x ml.

Now we have to find the amount of alcohol present in ml present in solution A and B using the percent concentrations and volume of solutions.

The amount of alcohol in solution A = 2*200/100= 4 ml

(amount = %concentration *volume of solution/100)

The amount of alcohol in solution B = 6*x/100= 0.06x ml

The total alcohol present in the mixture=4+0.06x ml

Total volume of the mixture=200+x ml

The total concentration of the mixture needs to be 5%. i.e.

(4+0.06 x)/(200+x)=5/100

=> 4+0.06x=0.05(200+x)

=> 4+0.06x=10+0.05x

=> 0.01x=6

=> x=600 ml

So 600 ml of solution B is required.