A pharmacist wants to mix a 16% saline solution with a 26% saline solution to get 15 L of a 20% saline solution. How much of each solution should she use?

Question

A pharmacist wants to mix a 16% saline solution with a 26% saline solution to get 15 L of a 20%
saline solution. How much of each solution should she use?

Answer 1

Andy did good until he put 'mL', rather than 'L', for the amounts of the 16% and 26% solutions.

 

The correct answers should be:

 

9 L of the 16% saline solution; and

6 L of the 26% saline soluation

 

 

Answer 2

X is the amount of 16% saline and 16% = 0.16

Y is the amount of 26% saline and 26% = 0.26

Likewise 20% = 0.20

 

The amount of saline is:

 

 0.16X + 0.26Y = 0.20(15)

 

0.16X + 0.26Y = 3 <--- first equation

 

X+Y = 15  is the amount of total liquid <--- second equation

 

Solving the second equation for Y and substituting into the first equation:

 

0.16X + 0.26 ( 15 - x) = 3

 

16X + 26( 15 - x) = 300  <--- multiplies both sides by 100 to clear away the decimals

 

16X + 390 - 26X = 300

 

0 = 300 - 16X - 390 + 26X <--- everybody to the right side

 

0 = 10X - 90

 

90 = 10X

 

X = 9

 

9 mL of the 16% saline solution

6 mL of the 26% saline solution

 

9(0.16) + 6(0.26) = 3 = 15(0.2)