How do I solve this ?

Question

A pharmacist needs to obtain a 60% alcohol solution. How may ounces of a 30% alcohol solution must be mixed with 50 ounces of an 80% alcohol solution to obtain a 60% alcohol solution?

Answer 1

Let x = number of ounces of 30% solution

 

Then, x + 50 = number of ounces of 60% solution

 

Alcohol content of 30% solution = 0.30x

Alcohol content of 80% solution = 0.80(50) = 40

Alcohol content of mixture = 0.60(x + 50) = 0.60x + 30

 

So, 0.30x + 40 = 0.60x + 30

 

                 10 = 0.30x

 

                  x = 10/0.30 = 33 1/3 ounces