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Archive for December, 2009

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.

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.

Using “Master Equations”

In my last blog I described what  master equations are and how you can use them to solve word problems. I then promised to show you how to use master equations to actually solve word problems.

Here is the blog that shows how you use them to solve equations.

The two master equations I described were:

1)  Distance 1  =  Distance 2


2)  Distance 1 +  Distance 2  =  Distance Total

Let’s see how you solve a word problem with one of these master equations.

Here’s the word problem that we will be solving:

Tino and Gino get into an argument and drive away from one another. Tino leaves first, heading north at 65 kilometers per hour. Two hours later Gino heads south, traveling 45 kilometers per hour. The question:  at what time will Tino and Gino be 460 kilometers apart?

Step 1: Decide which master equation to use. Since Tino and Gino are traveling in opposite directions, they are covering different distances. Since their distances are different, we would not use the master equation Distance 1 = Distance 2 . The only other option at this time is Distance 1 + Distance 2 = Distance Total.

We can use this master equation, calling the distance that Tino travels  Distance 1, and  calling the distance that Gino travels Distance 2. Then the Distance Total would be the 460 kilometers.

At this point we can specify the master equation for this problem like this:

(Distance of Tino)  +  (Distance of Gino)  =  460

The next step is extremely useful, and it makes everything start coming together. To use this step we rely on the fact that distance = (rate) x (time). That being the case, we can express (Distance of Tino) as (rate Tino) x (time Tino), and we can similarly express (Distance of Gino) as (rate Gino) x (time Gino).  Using this step, the original master equation morphs into:

(rate Tino) x (time Tino)  +  (rate Gino) x (time Gino)  = 460

Once we reach this level of specificity, we can start filling in the blanks, as follows:

(rate Tino)  =  65

(rate Gino)  =  45

Of course we also need to come up with expressions for (time Tino) and (time Gino), and this is a bit more tricky, but not too bad. Notice that the problem says that Tino leaves first, and that Gino leaves two hours later. That means that Gino drives two hours LESS than Tino. In algebra-ese, we can express this idea by letting t = the time Gino drives, and then (t + 2)  for the time that Tino drives.

So now we have:

(time Tino)  =  t + 2

(time Gino)  = t

Putting it all together we make this grand substitution:

(rate Tino) x (time Tino)  +  (rate Gino) x (time Gino)  =     460
(65)         x          (t + 2)        +       (45)        x         (t)           =      460

Do you see what is great about this equation? We have one equation and just one variable. In the world of algebra that means “Hallelujah” because it tells us that we can solve for the variable — which we do as follows:

65t  +  130                       +                   45t                          =    460

110t  +  130   =   460

110t                 =  330

t  =  3

Since we let the variable t stand for Gino’s time driving, this means that Gino has been on the road for three hours when he is 460 kilometers from Gino.

Since Tino drove two hours more than Gino, Tino must have been on the road for five hours when he and Gino were 460 kilometers apart.

Problem solved.

Again, the main point is simply that understanding master equations gives you a guideline that makes it simple to understand problems that otherwise would have left us scratching our heads.

I’ll probably write a bit more about master equations, as they are so useful that everyone should really know what they are and how to use them.

Conquer Word Problems

How to Conquer Word Problems —

Easy, you just need to use “Master Equations.”

A “what?” you ask. A “Master Equation,” I reply.

That  is a template that helps you solve word problems in two key ways:

1)  It lets you know what kind of word problem you are dealing with in the first place — you’ll be able to categorize it, AND …

2)  It gives you a “blueprint” for solving the problem, providing a quick and direct path between the problem and the answer.

Let me give you examples of two Master Equations that can help you solve those oft-vexing rate-time-distance problems.

The first master equation, Master Equation “A” is:
Distance 1  =  Distance 2

This is the one to use for word-problem situations like these:

—  Mike bikes from home to the Slippery Rock Climbing Gym, then he  bikes back, from the gym to home.

—  Candace climbs up Pike’s Peak, then she runs down Pike’s Peak.

—  Little Alvin skateboards from his house to Tiny Ted’s house on Monday. On Tuesday he makes the same trip but on bicycle.

Notice that in all three of these situations the distance covered is the same. The distance from home to a gym is the same as the distance from the gym to home; the distance up a hill is the same as the distance down a hill. That’s what makes these all cases of Distance 1 = Distance 2. The first distance equals the second distance.

The second master equation, Master Equation “B” is:
Distance 1 + Distance 2  = Distance Total

This is the one for situations like these:

—  Mike bikes from home to the Slippery Rock Climbing Gym, then he  bikes a bit more, from the gym to Cheesman Park.

—  Candace and James are 400 miles apart. They drive toward each other furiously fast. When will their cars pass each other (assuming, hopefully, that they are in different lanes)?

—  Little Alvin and Tiny Ted stand back to back. Little Alvin walks east, and Tiny Ted walks west. When will they be seven miles apart?

Notice that in all of these situations there are two different distances, and the distances can be added together to give you a total distance. Mike first goes from his home to the gym, then from the gym to the park; Candace and James each drive separate distances, but together they go 400 miles. That is why you need to use the formula:  Distance 1 + Distance 2 =  Distance Total for these word problems.

So o.k., in this blog entry I am not going to explain how to use the “Master Equations” to finish solve the problems. If I did that, the entry would be too LONG.

But I will do this in my next blog.

Stay tuned!