algebra question

Question

In the lab, Hans has two solutions that contain alcohol and is mixing them with each other. He uses 4

 times as much Solution A as Solution B. Solution A is  10%

alcohol and Solution B is  18%

 alcohol. How many milliliters of Solution B does he use, if the resulting mixture has 348

 milliliters of pure alcohol? 

Answer 1

3000 mL of the mixture is needed to have 348 mL of pure alcohol

if 4 times as much A as B is mixed with B

.10(4x) + .18x = ,58x = 5y

y = .58/5 = .116 = 11.6% alcohol for the mixture of A and B

.116 x 3000= 348 mL pure alcohol plus 2652 mL of 0% alcohol

one fifth of 3000 = 600 ml of B mixed with 2400 ml of A