Traditional Word Problems

Table of Contents

1. Barry Garelick on Traditional Word Problems
2. Wayne Bishop on Number Bases
3. Anna on Homeschooling
4. SteveH on Governing Equations
5. Conclusion

Today is a traditionalists' debate post -- and it's just in time, too. Our favorite traditionalist, Barry Garelick, posted on Friday the 13th. And this post has drawn eight comments, including one from our favorite commenter, SteveH.

https://traditionalmath.wordpress.com/2018/07/13/a-traditional-approach-to-a-traditional-word-problem-in-algebra/

Unlike most Garelick posts, this one doesn't begin with a link to an article, but rather a traditional word problem that's on his mind:

I saw this problem a few days ago and it got me thinking how I would present it in my algebra class: 

“An airplane can fly 550 miles with the wind in the same amount of time it takes to fly 425 miles against the wind. Find the speed of the wind, if the airplane is flying at a constant speed of 195 miles per hour.”

After his students set up an equation to solve, Garelick writes about a common problem that many traditionalist teachers have -- reducing the students' dependence on calculators:

Big numbers, so before they can ask whether they can use calculators, the first thing I ask the students is whether we can whittle down 550 and 425 by dividing both sides by a common factor? Dividing by 5 two times in a row (or 25 once) yields:  22/(195+w) = 17(195-w).

They still want to use their calculators.  I forbid their use for this problem. 

“Let’s just write it out without doing any multiplying,” I tell them.

And after his students have solved the problem, Garelick concludes:

This type of approach helps develop number sense, which some claim traditionally taught math doesn’t do a good job developing. I present it here as an example of what I consider a rich problem.  It’s not the problem that is rich so much as the way it can be used to address different strategies that may be employed within the same problem.

OK, so there are several issues addressed in this post. First of all, Garelick tells us that this is a traditional word problem. Traditional word problems include not just rate-time-distance problems as in this example, but age problems and mixture problems as well.

The relationship between word problems and the Common Core is a bit complex. The Common Core emphasizes mathematical modeling -- converting real-world language to math. Yet traditional word problems such as distance, age, and mixture problems are rare in Common Core math.

For example, back in my May 23rd post, I mentioned the following word problem -- Question 15 from the SBAC Practice Exam:

A store sells new and used video games. New video games cost more than old video games. All used video games cost the same. All new video games also cost the same.

Omar spent a total of $84 on 4 used video games and 2 new video games. Sally spent a total of $78 on 6 used video games and 1 new video game. Janet has $120 to spend.

Enter the number of used video games Janet can purchase after she purchases 3 new video games.

Why is Garelick's distance problem considered "rich," but not the SBAC price problem? Conversely, why does the Common Core approve of this price problem, but not Garelick's distance problem? We will try to discover the answers to these questions throughout this post.

Garelick's method of solving this without a calculator is interesting. Notice that in one step, to divide by 25, he divides by 5 twice. I'd argue that division by 25 is interesting in itself, because we can think of 25 as a quarter. So we can get the students thinking in terms of money -- how many quarters does it take to make $5.50 or $4.25? Otherwise I agree with his steps -- especially the way he avoids multiplying anything by 195.

According to Garelick, this question helps develop number sense -- and the idea that division by 5 twice is equivalent to division by 25 once contributes to this development.

But here's something similar to which traditionalists might object. Instead of dividing by 25, just divide by 5 twice -- therefore instead of dividing by 4, just divide by 2 twice. I found an example of this on YouTube -- the "Half Half Strategy":

As we can plainly see here, Carrie Susong -- the lady in the video -- divides 36 by 4 using this "Half Half Strategy." I can't read Garelick's mind, my feeling is that he'd criticize this division method as a waste of time. So why would Garelick object to dividing 36 by 2 twice yet have no problem with dividing 550 by 5 twice?

That's an easy once -- 36/4 = 9 (and the related 9 * 4 = 36) are basic one-digit math facts that students should have memorized, whereas 550/25 is always a multi-step division problem. Garelick and other traditionalists would especially complain if Susong were a teacher who forces her students to show the work for "Half Half," and awards zero credit to the kid who divides 36/4 = 9 in a single mental step because he can retrieve the multiplication 9 * 4 = 36 in one second. (By the way, nothing in this video suggests that Susong is such a teacher, but you know what I'm talking about here.)

The "Half Half Strategy" of division by 4 might have been common in the old days when everyone used Roman Numerals. But now that we use a simpler numeration system, traditionalists expect students to multiply and divide single digits in one step.

Wayne Bishop on Number Bases

Even though SteveH is the commenter we follow the most closely, let's look at some of the points made by others in this comment thread, in the order they posted. We begin with Wayne Bishop.

Wayne Bishop:

My only criticism/suggestions would be on the integer simplifications. Simple tests for common factors (the divisors of 10) and the sum of the digits divisible by 3 should be known and is usually enough (it makes the 39 much more obvious in this example) and reduction by multiples of 1001 for 7, 11, and 13 simultaneously for big numbers is kinda cute.

Hmm, I'm not quite sure how knowing that 39 is a multiple of three makes 195/39 easier -- except, of course, for the additional knowledge that 195 is also a multiple of three.

Wayne Bishop:

Beyond knowing and using these “tricks”, the “whys” of these simple tests are interesting and worth exploring in an algebra class. Even more, examining their equivalents in other base notation adds another level of understanding.

In my last post, I wrote a little about number bases (in discussing the Pappas quaternary problem for Friday the 13th). It's interesting that Bishop, presumably a traditionalist (since I assume that anyone who comments on the Garelick blog, especially to express agreement with him, is a traditionalist), would advocate spending time in a math class on number bases.

And of course, Bishop is writing about the divisibility rules. For decimal, he describes both the divisor rule (for 2 and 5) and the omega rule for 3 (which works for 9 as well).

He also mentions an obscure rule for 7, 11, and 13. This is technically a cube alpha rule (based on 1001, one more than the cube of the base). Sometimes I wonder whether such a rule is worth it, since for 7, we must memorize three-digit 143 multiples of 7. Well, since Bishop uses the rule, we might as well use it too.

To those of us familiar with divisibility rules, it may be strange to use cube alpha for 11, since 11 can be found using alpha itself. Well, Bishop's cube alpha allows us to resolve three primes, including a prime smaller than 11 (namely 7). Since 11 isn't an important (small) prime, we might as well use one rule to resolve all three primes.

Bishop suggests that we should examine another base in our classes. Suppose we take up Bishop's suggestion -- so which base would be the best "starter base" to introduce our students.

Well, consider that my knowledge of number base terminology (cube alpha, SPD, {c} default dozenal, etc.) comes from the Dozens Online website. The members of that forum would obviously recommend that dozenal be the base we introduce to students. The only problem with dozenal is that the two extra digits might confuse students. If we were to choose dozenal, I'd recommend using the digits introduced in the Schoolhouse Rock video "Little Twelvetoes" -- dek (X), el (E), doe (10). I might as well post the video here:

{c} (default dozenal)

Here are the relevant divisibility rules similar to the ones Bishop uses for decimal. The divisor rule applies to 2, 3, 4, 6, and omega applies to el (E). For Bishop's cube alpha (1001) rule, we find that 1001 = 7 * 11 * 17 in dozenal, and so this rule also applies to three primes, 7, 11, and 17. If we're going to have cube alpha, then we might as well have square alpha (SPD), which gives us two more primes since 101 = 5 * 25. So the resolvable primes are 2, 3, 5, 7, E, 11, 17, 25. But 17 and 25 are quite large, with two skipped primes 15 and 1E.

But should dozenal indeed be the base we introduce to students? If we want to avoid transdecimal digits (digits greater than 9) yet have a base as close to decimal as possible, then we should choose base 9 (nonary). If we wish to avoid odd bases, then base 8 (octal) is the best choice. Furthermore, some might consider teaching number bases to be a waste of time, but if we choose octal, then we can at least tie this to computer programming.

Another computer-related base, hexadecimal, suffers from needing more digits than dozenal. Binary, the pure computer base, is well below the human scale. Base 2 (binary) has several interesting divisibility rules, but if the purpose of teaching the number base is to compare it to decimal, then binary becomes a poor choice.

I'm actually in a base 4 (quaternary) mood right now because of the recent Pappas question. Indeed, there was a second Pappas question this month that I didn't mention:

?10 (base four) = 52 (base ten)

The correct digit is 3 -- and of course, this problem is from July 3rd.

{4} (default quaternary)

Imagine that we're teaching a math class, and that following Bishop's suggestion, we've chosen quaternary as the starter base to introduce to our students. Here are some questions we might ask:

1. In decimal, the divisor (last-digit) rule applies to two, five, and ten. To what numbers does the divisor (last-digit) rule apply in quaternary?

This one should be easy -- in quaternary, the divisor rule applies only to two (2) and four (10).

2. Notice that Bishop doesn't mention the rules for regular numbers, like four in decimal. Regular numbers have divisibility rules where we must check the last several digits (for four in decimal, we must check the last two digits for divisibility by four).

Even the regular test for four is a bit tricky. The next regular number, eight, requires checking the last three digits, but it might be doable if we piggyback on the rule for four. Sixteen, unfortunately, is too difficult to resolve in decimal.

On the other hand, regular tests in a base like quaternary are more important and much easier. To test for divisibility by eight (20), the last two digits must be -00 or -20. To test for divisibility by sixteen (100), the last two digits must be -00. For the next two powers of two (200 and 1000), we check the last three digits. If they're -000 or -200, the number is divisible by 200, and in the former case -000, the number is also divisible by 1000, and so on. Compare this to the difficult task for checking the fifth and sixth powers of two in decimal.

3. In decimal, the omega (sum-of-digits) rule applies to three and nine. Does the omega rule apply to three (3) in quaternary? Does it apply to nine (21)?

Students should be able to check that the omega rule applies to three but not nine. A counterexample for nine is nine itself -- nine is written as 21, and 2 + 1 = 3, which clearly isn't a multiple of nine.

10 (since we're still counting in quaternary). Why or why not?

As for "why," it's said that regular rules "extend upward," while omega rules "extend downward." So the regular rule in quaternary (and theoretically in decimal) works for any power of two. But the omega rule only works for the omega number (one less than the base) and any of its factors. In decimal, nine is the omega, and three inherits nine's omega properties. But in quaternary, three is the omega, so no other number inherits three's omega properties. The fact that nine is 3^2 doesn't mean anything in quaternary, just as 3^3 doesn't have a simple omega rule in decimal.

11. Describe an alpha rule for five (11) in quaternary.

Notice that Bishop doesn't give the alpha rule for eleven in decimal -- instead, eleven is one of three primes available via cube alpha. But five is a more important prime than eleven. Thus in quaternary, we should resolve five (11) using just alpha, not cube alpha.

The alpha rule involves adding and subtracting alternate digits. For example:

21: 2 - 1 = 1, therefore 21 isn't a multiple of 11.
313: 3 - 1 + 3 = 11, therefore 313 is a multiple of 11.
3001: 3 - 0 + 0 - 1 = 2, therefore 3001 isn't a multiple of 11.

12. To what numbers does the cube alpha (1001) rule apply?

We factor 1001 is quaternary: 1001 = 11 * 31. Thus cube alpha applies to the primes five (11) as well as thirteen (31). In practice, cube alpha will be used for 31 only, since 11 is resolved by alpha itself (and it's not worth using cube alpha for an extra prime much larger than the alpha, if all we want to resolve is the alpha).

13. To what numbers does the cube omega rule apply?

In quaternary, cube omega is 333, one less than the cube of the base 1000. In decimal, cube omega only helps to resolve two large numbers, but cube omega is more useful in quaternary. We notice this by factoring 333 = 3 * 3 * 13. So not only does cube omega provide us an additional prime seven (13), but we can also do nine (3^2 = 21) -- which, as you recall, is unavailable with just omega. (In decimal, cube omega likewise gives us an extra power of three, 3^3, and a prime larger than 3^3.)

Bonus: Is there a divisibility rule for eleven (23) in quaternary?

The answer is no, if we allow divisibility tests up to cube alpha (as Bishop does). The number eleven is resistant to intuitive divisibility tests -- we can show that unless eleven is resolved by divisor, omega, or alpha, it's resolved only by fifth-power omega/alpha. Fifth powers are too large to work with in any base except 2 (binary).

But as I mentioned in my last post, quaternary is easily convertible to binary -- one quaternary digit equals two binary digits. Then we apply fifth-power alpha to the binary number to resolve eleven.

I consider this to be a "bonus" question -- I wouldn't mention a rule for eleven with students unless it's an honors class. If we do give a rule for eleven, then we have divisibility rules for all two-digit numbers in quaternary. Of course, 100 is regular, 101 is square alpha, and 102 = 2 * 21, which we resolve as even + 21 (from cube omega). So the first prime we can't resolve is nineteen (103).

{a} (default decimal)

When exactly would we teach these number bases anyway? Since Bishop mentions number bases in connection with Garelick's word problems, perhaps there should be a unit on traditional word problems and number bases. Students learn about divisibility rules, then use these rules in some  calculations similar to those mentioned in this post, then reinforce the divisibility rules by comparing decimal to other number bases. In Algebra I, such a word problem unit can occur late in the first or early in the second semester. This lesson on number bases might be an interesting one to teach just before/after Thanksgiving/winter break.

Another possible time to teach it is during the lesson on factoring. An interesting question is:

There are square alpha, cube omega, and cube alpha rules? Why isn't there a square omega rule?

The reason is factoring. Square omega would be one less than the square of the base -- that is, in base b, square omega is b^2 -1. But this factors as (b - 1)(b + 1), which is omega * alpha. So any prime available in square omega would be available in omega or alpha.

Sometimes, though, alpha rules are given as square omega -- for example, a frequently given rule for 11 in decimal is to add the last two digits to the rest of the number. The assumption is that anything omega is easier than alpha (where we must remember to alternate signs).

There is one case where square omega is relevant -- odd bases. If b is an odd base, then both b - 1 and b + 1 are even. One of these is a multiple of four (and possibly of a higher power of two as well). If without loss of generality, say this is b - 1. Then b + 1, which is also even, provides one more factor of two than is available using omega only:

• In quinary, omega resolves 4, while square omega resolves 8.
• In nonary, omega resolves 8, while square omega resolves 16.
• In pentadecimal (base 15), alpha resolves 16, while square omega resolves 32.

I think back to my former middle school classes. It might have been interesting to replace one of the Illinois State projects with a different number base for each grade. (And unlike Illinois State projects, number bases would be Bishop-approved). One idea might be to teach the same number base as the grade level -- senary in sixth grade, septimal in seventh grade, octal in eighth grade. Senary might be a good starter base for sixth grade (with divisor rules for 2, 3, omega rules for 5, alpha rules for 7, regular rules for 4, 8, 9, forget about 11). Septimal adds the challenges of an odd base plus a more complicated rule required for 5. In eighth grade, we can discuss the relationship between octal and computers/binary (and perhaps even mention converting to binary for 11).

If we continue this pattern, then we teach nonary in Algebra I. Since nonary is an odd base, we might discuss the connection to difference of squares (divisibility rule for 16). And according to this pattern, the number base for Geometry would be decimal -- but Geometry is all about shapes and figures, not calculations, so there'd be no need for number bases at all. By Algebra II, students would be ready for transdecimals with undecimal (base 11), but by this point it's better just to skip to dozenal. (Of course, traditionalists might prefer octal in Algebra I, since that's their recommended class for eighth grade.)

By the way, I was continuing to think about number bases and Mocha/Atari music. You might think -- is anyone really going to perform the divisibility test for 11 right in the middle of a song just to determine whether a note's Degree is amber (11-limit) or not?

If we stick to the 7-limit, we notice that among the bases that are powers of two, octal resolves 7 much more easily (in the omega) than quaternary or hexadecimal can. (It's also the only base that resolves 19, via the cube alpha, for khaki notes.) The problem is that octal doesn't resolve 5 as easily as the other bases. But an argument can be made that 7 is more important than 5, even in 5-limit music (on the Atari, since we found the importance of 28, and hence 7, for the notes in 16-bit).

Anna on Homeschooling

The most prolific commenter in this thread for once isn't SteveH -- it's Anna. Let's see what Anna has to say in this thread:

Anna:

Excellent question, Mr. Garelick. My daughter actually had a question similar to this for her algebra class this year. Once your students get to ratio and rates, they could tackle one like this:

“A plane flies from Alphaville to Betaville and then back to Alphaville. When there is no wind, the round trip takes 4 hours and 48 minutes, but when there is a wind blowing from Alphaville to Betaville at 100 miles per hour, the trip takes 5 hours. How many miles is the distance from Alphaville to Betaville?”

I don't need to post all of Anna's work -- her answer is 1200 miles. Here's her second comment:

Anna:
It was fun to listen to my son (who is in pre-Algebra) discuss and argue concepts with her on this question. He couldn’t see why the tail wind/head wind didn’t cancel each other out and my daughter had to reason through the problem herself to explain it.

The way I think about this is, since traveling into the wind slows us down, we must spend more time in the headwind than in the tailwind -- which only slows us down even more. After all, if the wind speed had been 250 mph then it would have taken twice as long to get from Betaville to Alphaville as it would in still air. In other words, that one way trip would have taken just as long as the round trip in still air (that is, 4:48). There's no way that the tailwind can make up that time (since even with a tail wind, the trip from Alphaville to Betaville takes more than 0:00). And if the wind speed were more than 250 mph, the one-way trip would take longer than the round trip in still air.

On Square One TV, there was a Mathnet case where George Frankly, one of the main characters, is on trial for robbing a bank. The case against him hinged on whether a plane could fly from the mainland to the island where Frankly is staying, to bring him back to the mainland in time to rob the bank. But it's a windy day. In still air the trip would be 30 minutes each way, and with a tailwind, the plane arrives on the island in 15 minutes. But Kate Monday, Frankly's partner, determines that the return trip to the mainland would take close to an hour -- too late for Frankly to have robbed the bank, thus clearing his name. (They eventually reveal that an impostor robbed the bank.)

OK, that's enough on that. Let's get to Anna's third and final comment:

Anna:
We finally pulled the plug on the public school and decided to go the homeschool route. They jettisoned Geometry in 8th grade 2-3 years ago and in the fall, they will no longer offer separate math courses. It will be 7th grade math, 8th grade math. Middle schoolers will be bussed to the HS if they want Algebra.

She isn't the first traditionalist (again, I assume that everyone who comments on Garelick's traditional math blog is a traditionalist) who is a former public school parent before homeschooling. And the implication is obvious -- if only the public schools would teach math (or other subjects) the way she'd prefer it to be taught, her children would still be at the public school.

For example, she intimates that if the district would make it easy for students to take Algebra I and Geometry in eighth grade, then her children would still attend. (Notice that a compromise presents itself to me immediately -- teach Algebra I to some eighth graders and bus only the Geometry eighth graders to high school, since there would be much fewer of them.)

No, her district isn't San Francisco. She explains that her family lives in Washington State. But Washington, like California, is an SBAC state. This strongly suggests a reason for her district dropping eighth grade Algebra I -- the SBAC that all eighth graders must take at the end of the year is based on Common Core 8, and so it discourages eighth grade Algebra I.

I've mentioned a solution to this many times before -- just as ninth graders in Algebra I don't have to take the math SBAC, neither should eighth (or seventh) graders have to take it. We can even come up with an incentive for schools to place students in Algebra I -- for school accountability purposes, eighth graders in Algebra I are counted as taking the SBAC and earning the same score as they did on the seventh grade test. And so it's good for schools to place students who earn a 4 (or possibly a 3) on the seventh grade SBAC in eighth grade Algebra I, where they don't have to take the test. They're guaranteed to be counted as earning another 4 (or 3), with no possibility of the score dropping.

In fact, I was thinking about state accountability incentives. Before we had the Common Core, California had the CST, and schools were given API scores. The API was determined by assigning each student a score and then taking the average over all students:

• Advanced = 1000
• Proficient = 875
• Basic = 700
• Below Basic = 475
• Far Below Basic = 200

Notice that raising a student from FBB to BB was worth 275 points, while raising a student from Proficient to Advanced was worth only 125 points. And so schools had an incentive to help the lower students more than the higher students. (And this was even in the era when California encouraged schools to place eighth graders in Algebra I.)

Most traditionalists prefer that schools focus on the smart students. And so they'd reverse the differences between the five levels:

• Advanced = 1000
• Proficient = 725
• Basic = 500
• Below Basic = 325
• Far Below Basic = 200

In the old days, the goal for all schools is to attain an API of 800 (majority Proficient). With this scale, we might set the goal at 600, which can be attained by either having a near-majority Proficient or a significant number of Advanced students.

Anna continues to describe another problem with her district -- the textbooks:

Anna:

For middle school, they’ve just adopted Big Ideas by Larson and will change the HS curriculum to that series I think. I previewed the text and it is very heavy on linear algebra to the detriment of other concepts and has a strong focus on discovery. No partial fraction decomposition is presented, it is weak on ratios and rates, barely covers completing squares and factoring, and doesn’t touch optimization.

It's ironic that Anna would mention the Big Ideas text here. My district uses the Big Ideas text as an eighth grade Algebra I text, since it includes eighth grade SBAC material as well. Thus it's strange that her district would adopt the publisher of the ideal eighth grade Algebra I text and yet not offer eighth grade Algebra I class.

Anna reveals her ideal Algebra I text -- Dolciani. OK, that's settles it -- she really is a traditionalist:

Anna:
My oldest started with the 1988 Dolciani Algebra text and then moved to Art of Problem Solving Algebra. I think the AoPS material is great BUT students may need extra psets for review and practice. The online AoPS courses moved very fast. They covered the basics plus complex numbers, exponentials and logs, arithmetic and geometric series (finite and infinite), optimization with linear and quadratic inequalities, and special functions.

Recall that I, as a young student, used a Dolciani Algebra I textbook. This is back in 1993-94, so it's possible that our class used the 1988 text. Meanwhile, I've linked to the Art of Problem Solving website, but that was for college math contests with local Professor Kent Merryfield. Apparently, AoPS is also used by traditionalist parents disillusioned with public schools.

Notice that Anna doesn't distinguish between Algebra I and II here. So Anna criticizes Big Ideas because it "doesn't touch optimization." I assume that by "optimization," she means what the U of Chicago Algebra II text calls "linear programming" (Lessons 5-8 and 5-9 of that text), especially since she mentions optimization in connection with inequalities later on (but some of those were quadratic inequalities, which explains why she avoids the phrase "linear programming"). But I don't recall seeing "optimization" in the Dolciani text either -- the first time I'd seen linear programming is in that very U of Chicago text (which indeed was the Algebra II text I used as a young student).

And "partial fractions" don't appear in Dolciani either. The first time I've ever seen partial fraction decomposition was in AP Calculus BC (since the sole purpose of partial fractions is to make certain integrals easier). And so unless I'm misunderstanding what she's saying, Anna criticizes Big Ideas for lacking lessons that don't appear in her beloved Dolciani Algebra I text (based on my own memories of that same text).

Here is Garelick's reply to Anna:

Garelick:
I didn’t care for the Big Ideas series; I skipped the discovery lessons and just focused on the direct instruction lessons. So I gave up on it and use Dolciani’s algebra textbook from 1962. I was able to get them for very low cost via Amazon. The students like them a lot better than Big Ideas, which I find interesting.

Once again, Garelick likes to point out that his students actually prefer the direct instruction lessons of Dolciani to the discovery lessons of Big Ideas. Again, this sounds incredible, but I must respect the fact that Garelick is referring to his real students. Sometimes I wonder whether or not he's talking about real students. (Well, he did say that they kept reaching for a calculator -- OK, that definitely sounds like real students!)

SteveH on Governing Equations

Well, I guess I saved the "best" for last. Let's see what SteveH has to say:

SteveH:

Back when I taught college math and computer science, I realized that when I prepared to explain how to do a problem, I consolidated all of my analysis into a final (and clear) process to explain. My error was thinking that my final solution process was helpful to them when they approached a new problem – as if they should be able to get immediately to that level of problem understanding. They might have understood exactly what I did, but that didn’t mean that it would be helpful to them on a new problem. I evolved into giving them a more general process that starts with finding the governing equation, like D=RT, and then defining unique and simple equations, like

550 = R1 * T1
and
425 = R2 * T2

This isn't the first time that SteveH has mentioned "governing equations," but this is the first time that I notice his full explanation.

I've heard of setting up tables or charts containing such information, but SteveH likes to jump directly into setting up equations.

He's also criticized the Common Core for trying to avoid equations. For some problems, such as the Pythagorean Theorem, even Common Core texts recommend using a^2 + b^2 = c^2 -- the governing equation -- since there is no other reasonable way to solve such a problem. But for other problems, particularly proportions, the Core goes out of its way to avoid equations.

SteveH:
I told them to define proper variables and write down legal equations. Let the math do the thinking for them. This is also psychologically good because they can quickly get partial credit even if they somehow get confused about wind being completely additive and subtractive to rate. Is it exactly? I can imagine my son questioning that. They might cancel out, but what cancels out might not be the free wind speed.

SteveH claims that writing equations means not having to think about it. But if this is true, why does the appearance of an equation in a book cut its sales in half? I'd argue is because readers don't want to think about equations -- that is, equations force readers to think more, not less.

SteveH:
Math will explain it all to you, but that doesn’t seem to be the “understanding” many educators are looking for. They seem to think that it’s a concept/gut level approach – which can get you into big trouble.

And of course by "many educators," SteveH includes proponents of the Common Core. To SteveH, writing equations should replace having to explain one's own answers on the Common Core.

Conclusion

Earlier in today's post, I asked the question:

Why is Garelick's distance problem considered "rich," but not the SBAC price problem? Conversely, why does the Common Core approve of this price problem, but not Garelick's distance problem?

Maybe that's the reason -- Garelick's distance problem is best explained by equations, while the Core's price problem is explained by discourse. Each type of word problems reflects one of the two views of what "understanding" really is.