Percent mixture problem

Question

A chemical company makes two brands of antifreeze. The first brand is 20% pure antifreeze, and the second brand is 70% pure antifreeze. In order to obtain 50 gallons of a mixture that contains 60% pure antifreeze, how many gallons of each brand of antifreeze must be used?

Answer 1

Hi Morgan,

This is a mixture problem that's related to systems of equations. To solve these, there are two equations required: the "volume" equation, and the "contribution" equation.

You're ultimately solving for the volumes of each anti-freeze stock solution. We should give these well-defined variables. You have a 70% antifreeze and 20% antifreeze as your stock solutions. Let's call these S for seventy and T for twenty.

S = volume of 70% antifreeze

T = volume of 20% antifreeze

Your volume equation comes from the beginning of the third sentence. You want 50 gallons total of new solution. You don't know how much of each antifreeze you'll need yet, but you at least know the volume of each will add to 50.

S + T = 50

Your contribution equation can be put together by using the known information from your stock solution and the concentration of your new solution. Each gallon of 70% antifreeze will contribute 0.7 gallons of antifreeze to your solution. Each gallon of 20% antifreeze will contribute 0.2 gallons of antifreeze to your solution. You want 50 gallons of 60% antifreeze, meaning you want 50 * 0.6 = 30 gallons of pure antifreeze at the end. Your second equation is therefore:

0.7*S + 0.2*T = 30

Now you have two equations and two unknowns. It can be easily solved by elimination.

S + T = 50

0.7*S + 0.2*T = 30

S + T = 50

7S + 2T = 300

-2S - 2T = -100

7S + 2T = 300

5S = 200

S = 40

So we need 40 gallons of 70% antifreeze. That means we need 10 gallons of 20% antifreeze.

We can check this for accuracy:

40 + 10 = 50. So the volume equation checks out.

40*0.7 + 10*0.2 = 28 + 2 = 30. So the contribution equation also checks out.

Every mixture problem is solved with this exact method, regardless of what the mixture is. Hope this helps!

Answer 2