Word problem

Question

• A pharmacist needs 8 ounces of a solution that is 50% saline. How many ounces of 60% saline solution and 20% saline solution must be mixed to obtain the mixture needed?

Answer 1

Hi Ashley,

These can be tricky.

 

Let x = ounces of 60% saline. 0.6x = ounces of saline

Let y = ounces of 20% saline. 0.2y = ounces of saline

 

Total solution needed is 8 oz., so x + y = 8

 

To get a 50% saline solution, (total ounces of saline)/(total ounces of solution) = 0.5

 

We have a system of equations:

x + y = 8

(0.6x +0.2y)/(x + y) = 0.5

so, (0.6x + 0.2y)/8 = 0.5

      0.6x + 0.2y = 4

 

Since x + y = 8, y = 8 - x

 

     0.6x + 0.2(8 - x) = 4

     0.6x + (1.6 - 0.2x) = 4

     0.4x + 1.6 = 4

     0.4x = 2.4

         x = 6 ounces

         y = 8 - x = 2 ounces