a lab technician needs 60 ml of a 50% saline solution.

Question

How many ml of 30% saline solution should she add to 60% saline solution to obtain the reqiured mixture?

Answer 1

Let x = the amount of 30% solution and y = the amount of 60% solution. Since the mixture will be 60 mL, then

 

x + y = 60

 

The equation for the mixture will be .30x + .60y = .50(60) = 30

 

To make this an equation with only one unknown, solve x + y = 60 for y.

 

y = 60 - x

 

Now, re-write the mixture equation as 

 

.30x + .60(60 - x) = 30 (Multiply both sides of the equation by 100 to eliminate the decimals.) 

30x + 60(60 - x) = 3000

30x + 3600 - 60x = 3000

-30x = -600

x = 20

 

If x = 20, then y = 60 - 20 = 40

 

We need 20 mL of 30% solution and 40 mL of 60% solution. 

 

Answer 2

You will end up with 2 unknowns and 2 equations and this will allow you to solve for the 2 unknowns.

 

x: total amount of 30% saline solution in ml

y: total amount of 60% saline solution in ml

Total amount of final solution: x + y = 60  (Equation 1)

 

Concentration of 30% saline solution: 0.3x

Concentration of 60% saline solution: 0.6y

Total concentration of saline solution: 0.3x + 0.6y = 0.5(60) --->0.3x + 0.6y = 30 --->  x + 2y = 100  (Equation 2)

 

Use elimination method to solve

x+2y=100

x+y= 60

 

Subtract equation 1 from equation 2

y = 40

 

Substitute y into equation 1

x = 20

 

Amount of 30% saline solution is 20 ml

Amount of 60% saline solution is 40 ml