Conquering “Mixture” Problems, Part 1

In the last blog you learned how to use a cool tool, “the master equation,” to slay (rate) x (time) = (distance) problems, R x T = D.

Now that you are initiated into the wonders of master equations, you might like to know that you can also use them for problems that many find even trickier:  those dreaded “mixture” problems.

Think for a sec, if you dare, and you’ll recall these little beasts, problems like this:

You start out with 5 liters of a 40% antifreeze solution. How many liters of pure antifreeze would you need to add to wind up with a mixture that is 73% anti-freeze.

The nightmares coming back to you now?

But as I mentioned, you can now use a “master equation” to solve these problems, just as we  did with R x T  =  D problems.

First, though, you need to understand something fundamental about mixture problems. And it helps if you can relate it to what we just learned about R x T = D problems.

With R x T  = D problems, a key was seeing that any distance can be represented by a rate multiplied by a time. For example, if a car travels 60 mph for 4 hours, we can express the distance it travels as the (rate)  x the (time):  (60 mph)  x  (4 hours) = 240 miles. The distance IS the product.

With “mixture problems,” there is a similar situation. For any mixture, we can express the amount of stuff that we care about through this basic but all-important equation:  Stuff =  (Concentration) x  (Volume of liquid). Or, still more shorthand: Stuff  =  (Concentration)  x  (Volume), which I like to abbreviate as
S  =  C  x  V.

What does this mean?  Well, here’s an example. Suppose in a word problem you’re told that you have 4 liters of a 50% antifreeze solution. You need to know how much actual antifreeze is in that solution. The antifreeze is the “stuff” we care about here. Use your new equation:  Stuff  =  (Concentration)  x  (Volume). So just multiply the (concentration) by the (volume) of liquid. That means you multiply  (50% concentration)  x  (4 liters), which is the same as (.5)  x  (4.0)  =  2.0. This means that in those four liters of solution there are exactly 2 liters of antifreeze. Wondering why this is true?  Just remember that 50% means HALF. So a  50% antifreeze solution means that half the liquid is antifreeze. Since you have 4 liters, half of that, 2 liters, is antifreeze.

What’s great is that you use this same principle and equation no matter how complicated the numbers might become (and you know that they don’t always stay easy, right?). So suppose you’re dealing with 12 liters of a 35% antifreeze solution. No problem. To see how much antifreeze is in those 12 liters, just use your new equation:  S  =  C  x  V. Antifreeze =  (.35) x (12)  =  4.2. This means that in those 12 liters of solution there are exactly 4.2 liters of antifreeze.

Taking this one step further, suppose that you need an algebraic expression to stand for a certain volume of liquid, an expression like (12 – x). And suppose you know that this liquid is 65% antifreeze. To express the amount of antifreeze in this solution, you still multiply the concentration by the volume, but now it looks like this:
Antifreeze  =  (.65) (12 – x).

That is all there is to it …  S  =  C  x  V. Burn that idea into your mind, right next to  R x T  =  D, and the rest will be “cake.”

One other thing to know about “mixture” problems. All you really care about in these problems is the amount of the solution whose % concentration you are given. So, for example, in a problem about antifreeze, the “master equation” you would use is this:

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

In my next blog I will show how you put these ideas together to actually interpret and solve a mixture problem. Trust me, now that you know S  =  C  x  V, it won’t be difficult.

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