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.

 

 

 

 

 

 

 

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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:  josh@SingingTurtle.com 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.

Conquering Word Problem Confusion w/ “Friendly 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 “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:

TCB / TBC / CTB / CBT / BTC / BCT

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?