Regents Recap — June 2013: Erroneous Questions

Published by MrHonner on

Here is another installment in my series reviewing the NY State Regents exams in mathematics.

It’s probably unavoidable that large-scale standardized exams will contain errors.  But some errors are more serious than others.

The New York math Regents exams consistently contain errors that demonstrate a lack of mathematical understanding on the part of the test-makers.  These aren’t just “typos”, as administrators and politicians often suggest;  they are serious conceptual errors, and they call into question the validity of these assessments.

Consider number 24 from the Geometry exam demonstrates several different kinds of errors.

2013 June G 24

The test-makers indicated that (2) is the correct answer.  Presumably, they believe the acronym SAS stands for the Side-Angle-Side Similarity Theorem.  However, SAS typically stands for the Side-Angle-Side Congruence Theorem.  If a student interpreted SAS, SSS, and ASA in the usual way, i.e. as congruence theorems, then AA would be the only possible way to prove two (non-congruent) triangles similar, and thus (1) would be the correct answer.

Perhaps the test-makers might claim that the context of the problem, proving similarity, should have led students to assume the acronyms stood for similarity theorems.  But, alas, there is no ASA similarity theorem.  What were they thinking here?

More generally, “Which method could be used” is not a good mathematical question.  Lots of different methods could be used.  It’s not inconceivable that AA could be used to prove these triangle are similar, so how could that possibly be an incorrect answer to this question?  Ultimately this question was thrown out, but not before thousands of students across the state had already taken their final exam.  And even as they tossed the problem out, the state still refused to accept responsibility for publishing an erroneous question, hiding behind the old alternative methods defense:

Since there are alternative methods to prove that the two triangles given in Question 24 are similar, all students should be awarded credit for this question. (link)

Unfortunately, this is just the latest example of serious mathematical errors in NY State Regents exams.

[Update:  An earlier version of this post criticized #25 on the Algebra 2 / Trig Exam.  Thoughtful comments provoked me to re-examine my criticism, and also pointed to a different issue with this question, which can be found in a separate post.]

Categories: Teaching


Anonymous · July 10, 2013 at 10:27 am

You’re right about Question 24, which is at best confusing. (Arguably, there could be an ASA similarity theorem, but the S is redundant.) This sort of question penalizes students who think carefully, while rewarding anyone who pattern-matches SAS without thinking.

On the other hand, I wouldn’t say Question 25 is mathematically incorrect in the way you claim. Rationalizing the denominator of an algebraic function is perfectly reasonable, with “rational” meaning “rational function” and not “rational number”. Admittedly, they never specify whether this expression is meant to represent a function of x or a specific number. They certainly intended the former, since they say the answer is (1) rather than (4): technically, (1) and (4) define slightly different functions (since 0 is in the domain of (4) but not (1)), while if you interpreted the original problem as a number, then (1) and (4) would define exactly the same number as you started with.

    MrHonner · July 10, 2013 at 12:03 pm

    You are right about #25. My criticism is off-the-mark. Indeed, “Rationalize denominators involving algebraic radical expressions” is a standard for the course ( I have updated the post based on your thoughtful and polite feedback; thank you.

Graeme McRae · July 10, 2013 at 11:21 am

I agree with Anonymous’s comments about question 25. I also see a different way they could have fixed the question. They could have added the assumption (which may have been in the mind of the person writing the question) that x is a rational number that is not the square of a rational number. If that assumption were stated, then even viewing the expressions as functions of x, only (1) and (4) are both equal to the original expression and have rational denominators. (4) would be the correct answer in this case, since it is arguably simpler than (1), and (1) and (4) don’t define different functions since neither domain includes x=0 with the “x is rational nonsquare” assumption.

Graeme McRae · July 10, 2013 at 11:39 am

On second thought, the way to fix question 25, and the assumption that must have been in the mind of the question-writer is as follows:

“25. Expressed with a rational denominator and in simplest form, assuming x is an integer, x/(x-sqrt(x)) is…”

In this case, the only way to ensure the denominator is rational is to multiply the numerator and denominator by x+sqrt(x), and this does in fact ensure the denominator is rational if we assume x is an integer. It also makes response (1) the only correct one, for the reason Anonymous mentioned: x=0 can’t be in the domain.

    MrHonner · July 10, 2013 at 12:08 pm

    I also agree with Anonymous’s comments about question 25, and the accompanying refutation of my criticism. As a result, I have removed that portion and updated the post.

    The question presumably refers to rationalizing “denominators involving algebraic radical expressions” , and as such it seems ok.

CCSSIMath · July 10, 2013 at 4:09 pm

Since we don’t live in the contrived and arcane world of Regents math, we can still take exception to Algebra2/Trig question #25.

Because eliminating radicals in (and imaginary parts of) a denominator are oft-practiced high school exercises, stronger students will likely understand what they’re being asked to do regardless of how the question is posed. However, in the outside world, and for weaker students, the three possible interpretations of the word “rational” make the wording of this question both grammatically weak and ambiguous.

The original expression is already “rational” in one mathematical sense of the word because it’s in the form f(x)/g(x), whether the entire expression be part of a function or not. That aspect comes in the “rational functions” part of a typical Alg. 2 syllabus, but might be mistakenly invoked here.

As for the usage of the word in “rational denominator”, notwithstanding A2.A.15, more commonly the notion of “rationalizing a denominator” refers to a denominator that’s the sum of an integer and a radical (or just a radical alone):


Lastly, we agree with the other posters that even with the expression rewritten as answer (1) or (4), the denominator is not necessarily a “rational number” (the third possible meaning of the word), and we can extend that line of thinking to argue that the original expression could already have a rational denominator if x were, say, 4.

The domain issue seems to be well covered already, but questions should always be clarified by including simplifying conditions such as “If x ≠ 0 and x ≠ 1…”

Agreeing with the ongoing theme of this series of postings, Regents exams questions’ writers really don’t seem to understand the mathematics all that well.

This question unnecessarily penalizes weaker students who may have the algebraic skills, but won’t understand the directive. If it’s going to reasonably test A2.A.15, the question might possibly be improved, but we think as tested, the question should be tossed.

MrHonner · July 11, 2013 at 9:48 am

The thoughtful remarks about the subtle domain issues at play in A2T #25 led me to double check the scoring guide: it turns out that (1) is not the “official” correct answer; (4) is.

In light of this, I have written a follow-up post:

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