## Kiss those Math Headaches GOODBYE!

### Conquering Mixture Problems, Part 2

If you read my last post, Conquering “Mixture” Problems, Part 1, you’re “sitting pretty” for solving these word problems. If you haven’t, go for it, and you’ll kiss those “mixture-problem blues” away.

In this post we will get to the heart of the matter  —  setting up and solving mixture problems. Here’s the problem we’ll solve:

Tuning up your car in late fall, you realize that you need to adjust the antifreeze level with the cold weather just around the corner. With some fancy equipment, you find that you have 12.0 liters of a 40% antifreeze solution, but you need to wind up with a 60% solution. So you need to figure out how much pure antifreeze to add to the 40% solution to wind up with the 60% solution.

At first glance it appears that you would need to be a genius to solve such a problem. How could anyone figure out anything so tricky?

But as it turns out … you are a genius — because you just learned how to use “master equations”  through my last post  —  so not to worry.

As you learned, master equations give you a super-clear outline for solving word problems. In the last post you learned two key facts about mixture problems:

1) There is a simple formula for calculating the amount of stuff you are interested in, the equation:  Stuff =  (Concentration)  x  (Volume of liquid). Or, more simply:
S  =  C  x  V

2)  For most mixture problems the master equation is this:

(Original Amount of Stuff)  +  (Amount of Stuff Added)  =  (Amount of Stuff at End)

In such word problems, the stuff we’re interested in — what we’re keeping track of — is the antifreeze. That being the case, our master equation will look like this:

(Original Amount of ANTIFREEZE)  +  (Amount of ANTIFREEZE Added)  =  Amount of ANTIFREEZE at End)

Now here is where the first part comes in. The equation  S  =  C   x  V   gives us a nifty way of calculating the stuff for all three parts of the equation. Let’s take this one step at a time.

(Original Amount of Antifreeze):  Couldn’t be easier! The problem tells us that we start with 12 liters of a 40% solution. Using S =  C  x   V, we get:

(Original Amount of Antifreeze)  =  (.4)  x  (12.0)

(Amount of Antifreeze Added):  The problem asks HOW MUCH pure antifreeze to add. Since this is the unknown, we’ll let V  =  the volume of pure antifreeze that we’ll add. And since this is pure antifreeze, the concentration is 100% which, written as a decimal number, is just 1.0. Plugging in to S  =  C  x  V, we get:

(Amount of Antifreeze Added)  =   (1.0) x (V)

(Amount of Antifreeze at End):  We know two facts about the antifreeze at the end.
1)   Its volume will be the sum of what we start with, (12) and what we add (V). So the end volume will be (12 + V).  We also know that our target concentration at the end is 60%, and we convert this to the decimal number, .6. Again, plugging in to S  =  C  x  V, we get:

(Amount of Antifreeze at End)  =  (.6) x (12  +  V)

Now going back to our master equation:

(Original Amount of Antifreeze)  +  (Amount of Antifreeze Added)  =  (Amount of Antifreeze at end),

we plug in the three quantities we just found, and we get this:

(.4)  x  (12.0)  +  (1.0)  x  (V)  =  (.6)  x  (12.0  +  V)

We can solve this algebraically for V, as follows:

4.8  +  V    =    7.2  +  .6V

.4V  =  2.4

V  =  6

Since we let the letter V stand for the amount of pure antifreeze we add, this means that we add exactly 6.0 liters. And that is the solution; it is no harder than this.

So take a little time to let this sink in, and in my next post I’ll give you a few practice problems, to let you sharpen your skills.

#### Comments on: "Conquering Mixture Problems, Part 2" (3)

1. Please I got lost here, how did you get rid of the V on the left hand side?

4.8 + V = 7.2 + .6V

.4V = 2.4

V = 6

Like

• I just divided both sides by 0.4, to get V all by itself.
.4 ÷ .4 = 1, so you get 1V = V, on the left side.
2.4 ÷ .4 = 6, so you get 6, on the right side.
Hope that helps you follow the final steps.

— Josh

Like

2. Thanks Josh, how careless of me though I later figured it out:)
Still, I tried to do this problem below from a textbook and their answer was different from mine eventhough I got all three of your practice questions correct.

A mixture of 12 ounces of vineger and oil is 40% vineger (by weight). How many ounces of oil must be added to the mixture to produce a new mixture that is only 25 percent vineger.

p.s
The previous blog on distance was very enlightening thanks again.

Like