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Posts tagged ‘Word Problems’

Algebra Survival Guide — Second Edition — Fresh Off The Presses!

Algebra Survival Guide — Second Edition
by Josh Rappaport
illustrated by Sally Blakemore

The Algebra Survival Guide — Second Edition is here, but not yet released to the general public. Now’s your chance to order it at a 25% discount through April 10th. Just go to

But first, let me tell you about what’s new … a massive, 62-page chapter on advanced story problems.

It’s no secret that algebra gives students the ‘jitters,’ and word problems give them the ‘shakes.’ As a dastardly duo, the word problems of algebra are just about as nerve-wracking as anything in the teenage years.

The Algebra Survival Guide — Second Edition takes a hard look at algebra’s word problems and offers time-tested advice for cracking them. With a new 62-page chapter devoted to these word problems, the new edition tackles the ultimate math nightmares of the puberty years: problems involving rate, time and distance, work performed, and mixture formulas, among others. Added to the pre-existing 20-page introduction-to-story-problems chapter (in the Algebra Survival Guide — 1st Edition), it’s like having a book within a book.

The Algebra Survival Guide — Second Edition also includes:
  • 12 additional content chapters that explain fundamental and advanced areas of algebra
  • a unique question/answer format so students hear their own questions echoed in the text
  • conversational style written in the voice of a friendly tutor
  • step-by-step instructions
  • practice problems after each new concept
  • chapter tests
  • an expanded glossay and index
  • lively illustrations by award-winning artist Sally Blakemore
Finished Spit Fire
The many cartoons not only provide well-deserved comic relief for math learners, they also offer a visual way to grasp algebra’s challenging abstractions. Example: The above cartoon illustrates a real-world mixture problem by showing different percent concentrations of paint.

With all of these features, the Second Edition Algebra Survival Guide is ideal for homeschoolers, tutors and students striving for algebra excellence.

The Second Edition also aligns with the current Common Core State Standards for Math, so it’s ideal for today’s teachers, as well. Its content chapters tackle the trickiest topics of algebra:

Properties, Sets of Numbers, Order of Operations, Absolute Value, Exponents, Radicals, Factoring, Cancelling, Solving Equations, and the Coordinate Plane.

So, have some fun learning algebra!

• Updated version of Josh and snake

Everyday Life Sparks Mathematical Puzzles

So here’s the situation: you’re at the breakfast table, enjoying a bowl of steaming-hot steel-cut oats and maple syrup, and you just poured yourself a mug of black coffee. But then you realize you want to pour some milk in the coffee (sorry, purists). But the milk is in the frig, six feet away. So of course you walk to the frig, grab the milk, bring it to the table, pour some in your coffee, return the milk to the frig and sit back down. Question: could you have done this more efficiently?

Yes, of course. You could have brought your cup of coffee with you as you walked to the frig, poured the milk right there at the frig, returned the milk, and then walked back to the table.

Being Smart?

“Morning Joe”

When I realized this this morning, I thought … hmmm. Had I used a bit of forethought, I would save myself an entire round trip from the table to the frig. And while I have no problem making that extra trip (hey, just burned 1.3 calories, right?), the experience made me wonder if anyone has ever developed a mathematics of efficiency for running errands.

I could imagine someone taking initial steps for this. One would create symbols for the various aspects of errands. There would be a general symbol for an errand, and there would be a special ways of denoting: 1) an errand station (like the frig), 2)  an errand that requires transporting an item (like carrying the mug), 3) an errand that requires doing an activity (pouring milk) with two items (mug and milk) at an errand station, 4) an errand that involves picking something up (picking up the mug), and so on. Then one could schematize the process and use it to code various kinds of errands. Eventually, perhaps, one could use such a system to analyze the most efficient way to, say, carry out 15 errands of which 3 involve transporting items, 7 involve picking things up, and 5 involve doing tasks at errand stations. Don’t get me wrong! I have not even begun to try this, but I’ve studied enough math that I can imagine it being done, and that’s one thing I love about math; it allows us to create general systems for analyzing real-world situations and thereby to do those activities more intelligently.

Of course, one reason I’m bringing this up is to encourage people to think more deeply about things that occur in their everyday lives. Activities that appear mundane can become mathematically intriguing when investigated. A wonderful example is the famous problem of the “Bridges of Konigsberg,” explored by the prolific mathematician Leonhard Euler nearly 300 years ago.

Euler in 1736 was living in the town of Konigsberg, now part of Russia. The Pregel River, which flows through Konigsberg, weaves around two islands that are part of the town, and a set of seven lovely bridges connect the islands to each other and to the town’s two river banks. For centuries Konigsberg’s residents wondered if there was a way to take a walk, starting at Point A, crossing each bridge exactly once, and return to Point A. But no one had found a way to do this.

One of the famous Seven Bridges of Konigsberg

One of the famous Seven Bridges of Konigsberg

Enter Euler. The great mathematician sat down and simplified the problem, turning the bridges into abstract line segments and transforming the bridge entrance and exits into points. Eventually Euler rigorously proved that there is no way to take the walk that people had wondered about. This would be just an interesting little tale, but it has a remarkable offshoot. After Euler published his proof, mathematicians took his way of simplifying the situation and, by exploring it, developed two new branches of math:  topology and graph theory. The graph theory ideas that Euler first explored when thinking about the seven bridges sparked a branch of math that’s used today to determine the most efficient ways of connecting servers that form the backbone of the internet!

Of course, there’s also the classic example of Archimedes shouting “Eureka!” and running through the streets naked after seeing water rise in his bathtub. In that moment, Archimedes, who had been trying to help his king figure out if the crown that was just made for him had been created with pure gold, or with an alloy, saw that the water displacement would help him solve the problem. In the end, Archimedes determined that the crown was not pure gold, and the king rewarded the great thinker for his efforts.

As I write this, I find myself wondering if any of you readers can think of other situations in which everyday life experiences led mathematicians or scientists to major discoveries. It would be enlightening to hear more of these stories.

And, if no such stories spring to mind, check out this site, which lists several such stories.

In any case, the way that such discoveries occur shows that you never know where a seemingly trivial idea might lead … so it’s good to keep your eyes and mind open.

How to Solve Word Problem Using Easy Numbers

Recently I’ve been using a new technique to help students solve word problems, and nearly every day I am amazed at how helpful it is.

The technique helps students overcome their confusion with word problems.

The approach involves giving students permission to replace the numbers in a word problem with what I call easy, or “friendly numbers.” Essentially “friendly numbers” are just numbers that are easy to think about because they are simple, round numbers.

Here’s an example of the replacement process.

Word Problem as written:  Of the people who voted, 90 percent of them voted for Sammy. If 1930 people voted, how many of them voted for Sammy.

I was tutoring a student. Her response after reading this:  Huh?

Then I told her that it’s ok to temporarily replace the numbers in the problem  with “friendly numbers,” just to make the problem easier to grasp. I helped her see that in this problem she could temporarily replace the 90% with 50% and replace the 1,930 figure with a nice round number, like 600.

The student then picked up her pencil and wrote the problem like this:

Of the people who voted, 50 percent of them voted for Sammy. If 600 people voted, how many of them voted for Sammy.

Then I asked the student if she could figure out this problem. She said it now made sense. She went on to say that if 50 percent of the people voted for Sammy, that meant that half of the 600 people voted for Sammy. So that means that 300 people voted for Sammy.

Then I asked the student if she could come up with an equation to solve this problem. With a bit of help, she came up with:

.5 x 600 = # voting for Sammy

She solved this using decimal multiplication and got the right answer: 300 voted for Sammy.

Then I asked her if she could make a similar equation for the original problem, using the following questions as prompts:

What number in the original problem corresponds to your 50%? Answer:  90%

What number in the original problem corresponds to 600? Answer:   1,930

Once she saw these correspondences, I had the student write her equation for the “friendly numbers” problem. Then, just below that I had her write the corresponding equation for the original problem. Her work looked like this:

.5 x 600  =  # voting for Sammy

.9 x 1,930   =  # voting for Sammy

I asked her to now solve this using decimal multiplication, and she got the correct answer, 1,737

After going through this process I often ask students what made the original problem seem so hard . Usually they will say they don’t know, or they will sometimes say that they just couldn’t understand it.

From my work with “friendly numbers” I’ve come up with a theory. I believe that for many students, merely looking at “unfriendly numbers” has a “psych-out” factor. When kids get “psyched out” by those numbers, they go into a mental panic. And in that panic they lose their intuitive sense of what they need to do.

While this is a problem, it is not insurmountable. All we educators need to do is help the student re-cast the problem with “friendly numbers.” When they do, the “psych-out” factor vanishes, and students see what needs to be done. And generally students can transfer their sense of what needs to be done from the easier problem to the original problem. At that point they are on their way to solving it.

So I encourage you to teach students how to use “friendly numbers” when solving word problems. Perhaps you will also find that students can succeed once they first make the original problem easy to grasp.

Conquering Mixture Problems — Answers

Answers to Mixture Problems

In my last post I provided three mixture problems for all of you to do.

Here again are the problems, with the answers to them italicized.

1.  Kendra starts with 10 liters of a 40% antifreeze solution. How many liters of pure antifreeze would she need to add to end up with a solution that is 60% antifreeze?

Kendra would need to add 5 liters of pure antifreeze.

2.  Keith the chemist has a solution that is 25 quarts of 20% Boric Acid. How many quarts of 70% Boric Acid would Keith need to add to end up with a solution that is 50% Boric Acid?

Keith would need to add 37.5 quarts of 70% Boric Acid.

3.  Erin has a 2-liter solution that is 15% alcohol. How much pure alcohol would she need to add to it to end up with a solution that is 40% alcohol?

Erin would need to add 5/6 of a quart.

Conquering Mixture Problems — Practice

In my last two blogs I showed how to solve mixture problems. So now I want to give you some practice, so you can become an expert at solving these kinds of problems.

The answers will be stated in the next blog.

1.  Kendra starts with 10 liters of a 40% antifreeze solution. How many liters of pure antifreeze would she need to add to end up with a mixture that is 60% antifreeze?

2.  Keith the chemist has a mixture that is 25 quarts of 20% Boric Acid. How many quarts of 70% Boric Acid would Keith need to add to end up with a mixture that is 50% Boric Acid?

3.  Erin has a 2-liter mixture that is 15% alcohol. How much pure alcohol would she need to add to it to end up with a solution that is 40% alcohol?

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.