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Thursday, November 6, 2008

Solving the WASL Questions of Doom

posted by on November 6 at 18:58 PM

Winding down—still at work—from my day, I decided to solve the WASL questions of doom.

Question 1:

The easiest was the Tom, Dick and Harry one.

Harry, earning more than the manager, cannot be the manager, nor the teller (that earns the least.) Therefore Harry is the Cashier.

Tom cannot be the teller, as he has a sister. Since Harry is the Cashier, that makes Tom the manager.

Dick is the teller by default. Sucks to be him, being poorest paid.

Total Time: 2 minutes, using a neato match-chart.

Question 2:

65 sides, by my calculation.

5 on each short side x two short sides = 10
25 on each long side x two long sides = 50
5 on the top x 1 top = 5

Total = 65

I did this by making a little engineering plan style drawing.

Total time: 2 min.

Question 3:
This fucking compactor/cube problem was the worst.

x is the edge size of the older compactor, and y is the edge size of the newer compactor.

We’re told x > y and x and y are integers.

The volume of each cube is x^3 and y^3.

Each cube has twelve edges, so the total length of all the edges on both cubes is 12x + 12y.

Some asshole observed the following:
x3 + y3 = 12x + 12y

I spent much misery in algebra, attempting to get this down to something edible. My misery was unconstrained.

Among many others, I can make it all of the following:
x(x2 -12) + y(y2 -12) = 0

x3 / (x+y) + y3 / (x+y) = 12

and my personal favorite manipulation:

x4 + y4 + x^3y + xy3 -12 = 0

at this point, I became stuck, and brute forced it. (I have a feeling the last equation above is readily solvable. I got fed up, and realized that there aren’t that many integral candidates that even potentially would work.)

x = 4, y = 2 satisfies the initial equation.
4^3 + 2^3 = 12(4+2)
64 + 8 = 72 = 12*6

Total time? 30 fucking minutes.

Who the fuck “notices… that the combined volumes of one cube from each compactor was numerically the same as the combined length of all their edges.” I work in a lab, a real scientific lab, and can guarantee such a situation never happens. Nor will it ever happen. I mean, really, this has no business being a word problem. Fucking stupid. I’m glad she lost.

RSS icon Comments


I think the best question would be where would I ever need to use these types of problem-solving in my daily life?

Posted by Mr. Poe | November 6, 2008 7:26 PM

Srinivasa Ramanujan, that's who.

Posted by elenchos | November 6, 2008 7:28 PM

*any of these

Posted by Mr. Poe | November 6, 2008 7:30 PM

Jonathan - Question 3 is not solvable by conventional means, other than plugging in integers until it works. That question serves no purpose on a high school level test other than frustrating the students who actually know how to do algebra.

Posted by Mahtli69 | November 6, 2008 7:43 PM

Woah, I was stuck by the second one. I can tell you these kinds of questions have never up in my daily life so far.

What age students are taking this test?

An aside--My son was terrified to start 3rd grade this year due to horror stories told to him by older kids.

Posted by beatgrl | November 6, 2008 7:47 PM

Grades 3, 7 and 10 respectively.

For those interested, here is Jonah's original post on Bergeson's failure on all three. I was pleased to see the sample answers matched my own.

Posted by Jonathan Golob | November 6, 2008 7:56 PM

Question 3 is indeed incredibly dumb. Not only is the wording not particular clear, but the relationship described is unit-dependent: if it is true when the sides are measured in cm (2^3 + 4^3 = 12*2+12*4), it is not true when the sides are measured in mm (20^3 + 40^3 != 12*20 + 12*40). This kind of crap drives physicists crazy.

That said, as a pure mathematics problem, x^3 + y^3 = 12x + 12y is solvable by analytic means, although the method is usually not taught in high school. Factoring (x+y) out of the RHS reduces it to x^2 - xy + y^2 = 12. This is a quadratic diophantine equation, for which a standard algorithm exists. This calculator will give you the solution and show you the steps.

Replacing the WASL with another standardized test is fine with me, as long as we continue to use a standardized test and not grant diplomas to students who can't pass it.

Posted by David Wright | November 6, 2008 8:02 PM

David and Mahtli69--

Part of what took me so long, is my refusal to recognize the third question as the Kobayashi Maru of high school algebra math tests.

Posted by Jonathan Golob | November 6, 2008 8:08 PM

You could start by realizing that since y^3-12y = -(x3-12x), you need one example where 12n>n^3 and also an example where n^3>12n.

Since 12n>n^3 for only three integer values of n (1,2, and 3),
And n^3 gets enormously larger than 12n for n>4,
one number must be 4, and the other is easy to figure out from there.

Posted by F | November 6, 2008 8:09 PM

Jonathan @ 8: That's such a great scene! Thanks for the link.

Posted by David Wright | November 6, 2008 8:14 PM

Golob, the problem is you are thinking about it in a way too mathematical way. You HAVE to define at least one variable to solve for the other instead of chasing your tail round and round.

step 1; Identify the smallest possible value (and since they state it is an integer you can start at 1 and move upwards) Y can be.
step 2; When you write the equation x^3 - 12x = 12y - y^3 you can observe that at a certain point when X is sufficiently large y will be a negative number.
step 3; Factor. You soon realize that it can't be any larger thatn 2root3 or the but has to be larger than zero while still being a positive integer. That leaves 3 possible solutions; 1, 2, 3. when plugging those into the initial part of Y, you can then solve for X.

The problem is horrific though or takes for granted kids will logically go through the problem based on constraints provided in the outset of the problem. For a lot of people they won't try and substitute a value into Y because thats not how they were taught to approach the problem.

Posted by Bellevue Ave | November 6, 2008 8:21 PM

fuck, in the time it took me to write that on my phone other people kicked my ass to it.

Posted by Bellevue Ave | November 6, 2008 8:23 PM

Question #2 is ambiguous. Although it says that she made 5 stacks, it actually doesn't say they were lined up in a wall. So an alternative solution could look like:



In that case, the number of stamps would be 55.
Alternately, if the two columns were centered on the edges of the other three columns with no gaps -- hard to render in a fixed font, something like:

 * *

* * *

then the solution would drop to 50.

Who the hell wrote these questions?

Posted by A | November 6, 2008 8:32 PM

@13 Nor does it say that the stacks are of the same height.

I can also see zero as the correct answer to a certain interpretation of the question.

Posted by Matt | November 6, 2008 8:40 PM

The second question is terrible. There are any number of configurations that fit the actual description in the question.

The third question wasn't that bad, though, I admit, you pretty much have to do plug and chug. I rearranged to an equation of (x^3+y^3)/(x+y)=12. At that point, you know that both numbers have to be even (because all multiples of 12 are even). So, I just started with 4 and 2, and there was the answer.

Posted by Julie in Chicago | November 6, 2008 8:41 PM

It also doesn't specify that you need to stamp the top, which may or may not count as a "side" depending on your definition.

Posted by F | November 6, 2008 8:42 PM

I need to amend my comment. They are either both even or both odd. But, I skipped 1 and 3, because that obviously wouldn't produce a large enough difference.

Posted by Julie in Chicago | November 6, 2008 8:42 PM

Where's that fucking psychotic Blomstrom guy?

Posted by Mr. Poe | November 6, 2008 8:55 PM

#3 isn't that bad if you just start with the obvious equation and don't manipulate it: x^3 + y^3 = 12(x+y) (okay, obvious after getting through the tangled stupid mess of the phrasing, which is a separate problem). as julie said, they're either both even or both odd. by the same token, they're either both multiples of 3 or both not multiples of 3, so the smallest numbers that are even possible are 2 and 4. it's a dumb problem, and i'm not really sure what it's trying to test, but i think it's a little harsh to say it penalizes the students who "actually know algebra." part of knowing algebra is knowing when to use it.

Posted by BL | November 6, 2008 9:15 PM

i did pretty really good in algebra because i knew how to figure things out without memorizing some stock equations. in the real world, people solve math problems like that all the time. you don't pull out the quadratic trinary equation or whatever. i always did awful on tests where they were strict about you "showing your work" so that they can see you use whatever formula they're testing. :[

that said, it's still a dumb test.

Posted by jrrrl | November 6, 2008 9:36 PM

ps i wish david blomstrom got elected. :[

Posted by jrrrl | November 6, 2008 9:51 PM

1. if prob is bad questions -- fix the questions.

Fails the logic test to say there are bad questions, therefore throw out the test.

2. Slog must have political hangover today, skipping impt. news:

Item one--Emanuel is chief of staff, a former Clinton staffer who is a hard knuckled pol. I.e., nothing about him is "new politics" at all. Not a tall.

P-I: "the party's liberals and labor leaders are wary of Emanuel. He helped Clinton push the North American Free Trade Agreement through the House, angering the left."

Um-hmm. Not only has Obama walked back that stuff about NAFTA that he used to distinguish himself from the "failed" policies of Clintons, not only has he seized their focus on economics, not only has he vocalized the economic benefit we got from the Clinton years, not only did he wrap his arm around Bill Clinton and praise him to the skies, he is now hiring the Clinton staffer as his chief of staff.

And it doesn't stop their as Podesta is head of transition team.

Comment: This "change" mantra as applied against HRC was brilliant, it was one of those fake this way then move the other way moves that Slog commenters famously admired, not realizing that of course it meant much of his critique of HRC was primary fodder, to get elected.

Like faking left on NAFTA then walking that back.

This of course is exactly what a great politician of the old style does.

"I'll never send your boys to war in Europe"--FDR, 1940, election.

I am sure that Julius Caesar promised a new politics, too, helping draw folks to his side in his struggles in ancient Rome. The image of a new kind of politics is perhaps the oldest kind of politics, no?

--closed door session between Reid and traitor Lueberman. Hmmm, just exactly how much punishment of Lieberman is proper? He deserves a good swatting.

How else do you deal with traitors?


Posted by PC | November 6, 2008 10:07 PM

@22 - PC, do you just pick threads at random, or do you go out of your way to find one that has nothing to do with whatever it is you're talking about?

Posted by Mahtli69 | November 6, 2008 10:39 PM

#7 is the perfect answer and within what a high school algebra class' curriculum would least in Brother Macauley's class at St. Ann's Academy (nee Archbishop Molloy H.S.).

Posted by John Bailo | November 6, 2008 11:33 PM

Mr. Poe for most insightful.

I for one am glad Terry's fired.


Posted by Will in Seattle | November 7, 2008 12:49 AM

Solving a cubic function analytically is hard stuff, not the kind of thing you would expect a high school (or grad student for that matter) to be able to do off the top of their head.

Posted by matt | November 7, 2008 1:38 AM

It's funny how the WASL doesn't work compared to real life. And the fact that everything on that test is a muddle (poorly worded, etc) for the average student to work out.

Years ago I failed the WASL math portion (passing everything else) - I then went to college and got straight As in everything from economics to calculus I-II. The WASL doesn't judge abilities worth shit.

Posted by dkstar | November 7, 2008 7:38 AM

Okay, one last gripe about #2. Obviously the wording in the 4th sentence should have been "Penny put the packages into 5 stacks with an equal number of packages". I assumed that's what they were getting at, but still.

However, I would have missed that question because of the wording in 2nd sentence "It is Penny's job to stamp the sides of packages that are not touching the floor...". I actually read that as, none of the boxes on the floor need stamps at all. So, my original answer was 12 less than the right answer.

This is bringing me back to my days as a Mathlete. That's right. I said it. I was a Mathlete.

Posted by Julie in Chicago | November 7, 2008 7:53 AM

Well, I, personally, do this type of problems all the time at math parties, and have gotten laid on many an occasion just because of this type of integer-plugging.

Posted by Jude Fawley | November 7, 2008 8:10 AM


Yes! Mathcounts! The second problem could be interpreted any number of ways. The stacks could be of uneven numbers of boxes. They could have been arranged so there was a 2 x 2 square and one extra, or in an L shape. And I did not count the tops as a side that needed to be stamped. In short, what a terrible test.

Posted by kebabs | November 7, 2008 8:41 AM

@22 point 1 re "fix the questions; don't throw out the test" - when I went down to Seattle School district last Dec to review my 5th grader's answer booklets (because he didn't meet standard in math), the proctor who sat with me while I reviewed them (no cell phones! no notes!) told me that when it's all said and done, each and every test question costs $20,000 (to develop, vet, make a rubric for scoring, etc).

I saw so many ambiguous questions and instructions in the math section (which, by the way, isn't aligned with the standard math curriculum recently chosen by the district called Everyday Math, and so uses different nomenclature), that if you were to throw them out and start over, the cost would be more prohibitive than it already is.

I actually appreciate standardized (or norm-based, as the WASL is) testing, but what I saw in the my son's WASL tests was pretty concerning.

Not to mention that you don't find out how your child did until 6 months have passed and a new school year (with new teacher) has begun - and even then the teachers see only cryptic summary results by section, e.g., how he did in "number sense", "geometric sense", "algebraic sense". Not a problem if the child has met or exceeded standard - but not sure what you do with that as a teacher (or parent) if he hasn't - and aren't those the ones this test is designed to assure "aren't left behind?"

Posted by momster | November 7, 2008 8:58 AM

The WASL is crap. I was in Math Olympiad from 5th grade all the way through senior year. Most of the problems were designed to trick you - several were ridiculously hard to solve unless you knew that you could use trigonometry or calculus as a shortcut. All of them were better written, clearer, easier to understand, less ambiguous than the math problems on the WASL.

The whole state test felt like an insult - this poorly organized crap was supposed to be representative of the curriculum?

I should probably also mention that my mom, who taught math at the junior high and high school levels for 30 years, is not a fan of the WASL.

Posted by Greg | November 7, 2008 9:49 AM

And what exactly is a Cashier doing in a bank? I mean, there's a teller but a cashier? And who would believe that the manager earns LESS than a teller or a cashier? Really!?

#2 is way too ambiguous, as others have pointed out.

What they should do is add the Monty Hall question to the test. That would get people thinking!

Posted by slugbiker | November 7, 2008 11:35 AM

1. was quick and fun
2. for the least stamping I get

(top view)
6 6 1
6 6

that is 6 x 6 + 5 + 3 = 44 stamps

3. was easiest is you skipped the algebra,
wrote a little table and noticed
that the only even solution was
4, 2

if you read this deep in comments, you are
really, really bored.

I R Engineer

Posted by Jack | November 7, 2008 11:37 AM

I've helped my kids with WASL in the past. The trick is not to write closed form solutions, but to play with numbers. #3 solved this way:

1-side   all-sides   volume
1          12          1
2          24          8
3          36          36
4          48          64

The trick is to notice 24+48 = 8+64.

My time: ~5 minutes (2 false starts)

Posted by butterw | November 7, 2008 4:15 PM

Oops. Line 3 volume s/b 27, but no matter.

Posted by butterw | November 7, 2008 4:18 PM

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