Kiss those Math Headaches GOODBYE!

Archive for the ‘Word Problems’ Category

“Algebra Survival” Program, v. 2.0, has just arrived!

The Second Edition of both the Algebra Survival Guide and its companion Workbook are officially here!

Check out this video for a full run-down on the new books, and see how — for a limited time — you can get them for a great discount at the Singing Turtle website.


Here’s the PDF with sample pages from the books: SAMPLER ASG2, ASW2.

And here’s the website where you can check out the books more fully and purchase the books.









Problem of the Week, 9/17/10: “Flying Flora”

Flying Flora, late as usual for her business meeting, speeds from Santa Fe to Las Cruces at 105 mph.

Off the Plaza in Santa Fe, New Mexico

Santa Fe Plaza

After arriving in Las Cruces, she gets an email alerting her that she was caught by a radar gun and received a speeding ticket (she knows the local DA; otherwise she would have been thrown in jail!). Much chastened, Flora drives back from Las Cruces to Santa Fe at just 60 mph.

Las Cruces, New Mexico

Organ Mountains, Las Cruces

Your task:  Without using a specific value for the distance between the towns, find Flying Flora’s average rate for the round trip. Please show your work and round off your answer to the nearest tenth of a mile per hour. Post your answers on Facebook, or send to: All entries must be received by 10 p.m. this Wednesday, Sept. 22, to qualify for getting your name posted among the five first to get it right, on Friday.

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.

Challenge Problem – the ANSWER

Hi everyone,

Here is the answer to yesterday’s challenge problem, the probability problem about catching fish.

First, there are 6 ways to catch the three fish in three casts, getting exactly one trout, one carp and one bass.

You could get the fish in any of these six orders:


Next you find the probability of getting one of these possibilities. Let’s take the first one, TCB.

Keep in mind that after you catch the first fish, the number of fish left goes down by one, to 11; after catching the second fish, the number of fish goes down to 10.

The probability for the TCB possibility is calculated by multiplying:  6/12 x 4/11 x 2/10 = 2/55

When you think about the five other ways to catch the fish, you’ll see that the order of the numerators changes, but the denominators remain 12, 11, 10. So the probability for catching the three fish in any of the six ways is always the same:  2/55.

To get the probability for all six catches, just multiply the probability of one catch by 6:

6/1 x 2/55  =  12/55

And that is the answer: the probability of catching exactly one trout, one carp and one bass is 12/55, which works out to about 21.8%, meaning that this should happen a little more than 1/5 of the time.

I didn’t see anyone submit any answers, but feel free to send them in. Remember that I will post only the correct answers, so no one has to worry about seeing an incorrect answer posted.

Have a great day!

—  Josh

Interactive Challenge Problem — Send in responses

O.K., time to wake up the ol’ brain cells.

Here’s a little challenge problem. I’ll post the problem today, and then I’ll post the answer the next day. My thought is that this would be relevant for many kinds of people.

Teachers can use this as a fun class-opener. Homeschoolers can use it to start their math studies. And anyone who enjoys math can use it to sharpen math skills. So enjoy.

It would also be fun to see how many people get it right, so please send in your answers as comments. I will post only the correct answers.

The problem:

A lake contains exactly 12 fish: 6 trout, 4 carp, and 2 bass.
On any given cast, you catch exactly one fish, and no kind of fish is biting any more than any other kind. (i.e.:  Your odds of catching fish are governed by mathematical probability alone.)
What is the probability that in three casts you will catch exactly one trout, one carp and one bass?
To get credit, you must provide the correct answer and show how you solved the problem.

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.