Monthly Archives: December 2016

GRE Solutions Manual, Problem 5.6

Exam Prep, GRE, Quantitative Reasoning

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!

5.6: Absolute Value

This problem is an excellent candidate for plugging in. It’s also a great opportunity to discuss what to plug in on Quantitative Comparison problems. When a QC problem involves variables, our goal is to figure out which relationships between the two quantities are possible, given the constraints on those variables.

Often, only one relationship is possible:

  • If quantity A is always greater, we choose answer A.
  • If quantity B is always greater, we choose answer B.
  • If the two quantities are always equal, we choose answer C.

Sometimes, however, the relationship varies depending on the numbers we choose:

  • If quantity A is sometimes greater, but sometimes quantity B is greater, we choose answer D.
  • If quantities A and B are sometimes equal, and sometimes not, we choose answer D.

Plugging in is our way of figuring out whether the relationship between the quantities is fixed (answer A, B, or C) or not (answer D). To do this, we plug in multiple different sets of numbers, solve for the two quantities, and note the relationship that results.

But we don’t just plug in numbers at random. Instead, it helps to try out numbers that have special properties:

  • Zero. Multiplying anything by 0 gives a result of 0. Adding 0 to anything leaves the sum unchanged.
  • One. Multiplying anything by 1 leaves the product unchanged.
  • Negative numbers. These are useful in problems that involve absolute value (like the one on this page!) or exponents.
  • Extremely large and small numbers. These are especially helpful in problems involving fractions: when the denominator is small, the fraction is large, and vice versa.

You can think of these four types of numbers as being “in the ZONE,” and in fact, I recommend trying them out in that order. 0 and 1 in particular are such useful plug-ins that the problem will sometimes rule them out preemptively, using phrases like “x is a positive number” or “y > 1″.

Side note: every test prep company seems to have their own magic list of numbers to try. The Princeton Review textbooks, for example, promote a somewhat longer list, which they call the FROZEN numbers. (Their “extra” numbers include Fractions and Repeats.) In my experience, this is overkill, for a couple of reasons. Plugging in fractions makes calculations unnecessarily difficult: “1/2” is much harder to manipulate on the GRE calculator than “0.5”. For “repeats,” as for pairs of different numbers, it makes the most sense to start with 0 and 1, which are already on our list.

Now, back to problem 5.6. We’re told up front that x > y, so we have to pick two different numbers for x and y. Keeping our ZONE list in mind, let’s try setting x = 1, y = 0.


Plug-In Attempt #1

Let x = 1 and y = 0. Then

When x = 1 and y = 0, quantity A = 1.
When x = 1 and y = 0, quantity B = 1.

So in this case, the two quantities are equal. At this point, we’re left with two options:

  • Quantities A and B are always equal. (If so, we choose answer C.)
  • Quantities A and B are sometimes equal, sometimes not. (If so, we choose D.)

In order to figure out what’s going on, we need more information, which means we need to plug in another set of numbers and see if the A/B relationship changes.


Plug-In Attempt #2

Now, let x = 1 and y = -1. Then

When x = 1 and y = -1, quantity A = 0.
When x = 1 and y = -1, quantity B = 2.

This time, quantity B is greater. This means that depending on the values of x and y, the relationship between quantities A and B can change. The corresponding answer choice is (D), which stands for “Depends on the variables” or “Don’t know.”

We’ll get lots more plug-in practice in subsequent problems.


Math Review Reference

For more on this topic, see the following section of the GRE Math Review:

  • 1.5: Real Numbers (p. 8)

GRE Solutions Manual, Problem 5.5

Exam Prep, GRE, Quantitative Reasoning

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!

5.5: Inequalities, Part 1

This problem is a fairly straightforward check on your knowledge of inequalities. (For a more advanced problem on the same theme, see 5.15.) Just like an equation, an inequality can be manipulated via addition, subtraction, multiplication, or division — as long as you apply the same operation to both sides. There’s one important difference, however:

When you multiply (or divide) both sides of an inequality by a negative number, you have to flip the signs.

Equivalently, we could say that when you multiply both sides of an inequality by a negative number, the direction of the inequality changes. For example, the statement

Three is less than five.

means that 3 is to the left of 5 on the number line. But if we multiply both sides by -1, we get:

The quantity "negative 1 times 3" is greater than the quantity "negative 1 times 5."
Negative three is greater than negative five.

On the number line, -3 is to the right of -5.

Now let’s turn back to problem 5.5. We need to find out whether t is greater than 0, equal to 0, less than 0, or whether this relationship changes for different values of t. The clue is in that other variable, r.

Looking at the leftmost term of the inequality, we can see that the product rt is negative. This tells us two things:

  • Neither r nor t can be zero, since then the product would be zero. (This lets us rule out answer C.)
  • Either r or t, but not both, must be negative. (If both were negative, their product would be positive.)

Fortunately, we get some additional information about r in the third term of the inequality:

Zero is less than negative r.

If -r is positive, then r itself must be negative. This means that t cannot be negative, and we already determined that t can’t be zero. So t must be positive: quantity A (t) is greater than quantity B (0), and the correct answer is (A).


Math Review Reference

For more on this topic, see the following section of the GRE Math Review:

  • 1.5: Real Numbers (pp. 7-8)

GRE Solutions Manual, Problem 5.4

Uncategorized

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!

5.4: Three Circles

In the previous question, I touched on two complementary strategies for GRE math:

  • plugging in
  • solving algebraically

It’s easiest to see the distinction between these two approaches in algebra questions, since those have specific, named variables that you can plug into. But the “plug in” approach can also work on geometry, data analysis, and even arithmetic problems. Knowing when to plug in and when not to is a matter of experience, and it depends to a large extent on your personal strengths and weaknesses. Thus, whenever possible, I’ll be showing both approaches, so that you can decide which is faster and more intuitive for a particular problem type.

One more caveat before we begin: except for data graphics and coordinate planes, diagrams on the GRE are not drawn to scale. (The Paper Practice Book explains this at the beginning of each section — see, for example, p. 74.) So in this problem, we can’t assume that PR is larger than RQ just because the diagram is drawn that way.


The Algebraic Route

In order to solve this problem, we have to recognize that PR + RQ = PQ. In other words, the diameters of the smaller circles add up to the diameter of the larger circle. Once we realize that, we can break out the circumference formula

C = πd

and start figuring out quantities A and B. The circumference of the larger circle is

C1 = π(PQ)

and the circumferences of the two smaller circles are

C2 = π(PR)
C3 = π(RQ)

But since we know that PQ = PR + RQ, we can rewrite the circumference of the larger circle as

C1 = π(PR + RQ) = π(PR) + π(RQ)

which is the sum of the two smaller circumferences. Thus, quantities A and B are equal, and the correct answer choice is (C).


The Plug-in Route

We could also select values for PR, RQ, and PQ, as long as these obey the constraint that we noted above: the two smaller diameters have to add up to the larger diameter. For example, let PQ = 3, PR = 2, and RQ = 1. Then the circumferences are

C1 = 3π
C2 = 2π
C3 = 1π

This way, it’s even easier to see that C1 (quantity A) is equal to C2 + C3 (quantity B). Note that there’s nothing special about the numbers 3, 2, and 1. We could have picked

PQ = 8, PR = 6, RQ = 2
PQ = 15, PR = 14, RQ = 1

or any other three numbers that satisfy the original rules of the problem. Any valid triple will get us to the same conclusion.


Challenge Question

How does the area of the smaller circles relate to that of the larger circle? Is the ratio fixed, as it is for circumference, or does it vary? If it does vary, why does area behave differently than circumference?


Math Review Reference

For more on this topic, see the following section of the GRE Math Review:

  • 3.5: Circles (p. 52)

GRE Solutions Manual, Problem 5.3

Uncategorized

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!

5.3: Percents and Proportions

This problem illustrates one of the basic rules of GRE math: whenever you see percentages, turn them into decimals if at all possible. Decimals are much easier to work with and, in my experience, less likely to yield errors in calculation. In word problems (which we’ll see later), turning percentages into decimals also forces you to be clear about the role of any markups, discounts, or other percent changes.

We could think of this problem as a “word problem lite,” where our first task is to translate the written phrases into mathematical expressions. “4 percent of s” becomes

0.04s

and “3 percent of t” becomes

0.03t

Putting the two together, we get

0.04s = 0.03t

From here, we can divide out the coefficient of 0.04 from both sides of the equation:

s = (0.03/0.04)t
s = (3/4)t
s = 0.75t

Because both s and t are positive, we now have enough information to conclude that quantity A (meaning s) is less than quantity B (that is, t). So our answer is (B).

To see this more clearly, try plugging in a number for t. If t = 4, then

s = 0.75(4) = 3

It doesn’t matter what value of t you plug in — as long as you choose a positive number (as the problem instructs), you’ll find that s is less than t. You could even start your solution by plugging in for t, then simplify from there. For this problem, though, I find it easier to simplify the expression first.


Math Review Reference

For more on this topic, see the following section of the GRE Math Review:

  • 1.7: Percent (pp. 9-12)

GRE Solutions Manual, Problem 5.2

Exam Prep, GRE, Quantitative Reasoning

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!

5.2: Repeating Decimals

In order to see which of these two quantities is greater, we can expand the two decimals by a few place values:

0.717 repeating, expanded to six digits, is 0.717717.
0.71 repeating, expanded to six digits, is 0.717171.

These quantities are the same down to the thousandths place (the third place value after the decimal point), but they differ at the ten-thousandths place, which is underlined in the above equations. Specifically, Quantity A has a greater ten-thousandths digit than quantity B. Because this is the most significant digit in which the quantities differ, quantity A must be greater (answer A).


Math Review Reference

For more on this topic, see the following section of the GRE Math Review:

  • 1.4: Decimals (p. 6)

GRE Solutions Manual, Problem 5.1

Exam Prep, GRE, Quantitative Reasoning

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!

5.1: Transversals and Triangles

Depending on how much you like geometry, this first problem could be nerve-wracking or reassuring. Triangles are hands-down the GRE’s most favorite shape, so it behooves you to get comfortable with them if you aren’t already. There are at least two approaches we can use to solve this one.


Method 1: Transversals

The key to this method is that k and m are parallel. When two parallel lines are intersected by a third line, we wind up with a situation called a transversal of parallel lines. This arrangement has a special property that will help us out: corresponding angles are congruent. This means that for every angle at a given intersection, there is a congruent angle in the corresponding position at the other intersection. Think of it in terms of streets: the “northwest” corner of each intersection is the same. The “southwest” corners also match, and so forth.

Add to this the fact that vertical angles are congruent (this is true for any intersection of two lines), and you get the following:

A transversal of parallel lines forms two intersections.

But how does this fit into the diagram from the problem? Well, let’s forget about angles y and w for a moment and just focus on angles x and z. Imagine extending the left leg of the triangle upward so that it crosses line k, and downward so that it crosses line m. What do we get? A transversal of parallel lines.  The angle that is vertical to x must be congruent to x, and that angle corresponds to z, so all three of these angles must be congruent:

In this transversal, vertical angles and corresponding angles are congruent.

So we’ve shown that x = z. Using the same process, we can also show that angles y and w are congruent. If x = z and y = w, then

x + y = z + w

and the two quantities are equal (answer C).


Method 2: Triangles

The other route of solution depends on that middle angle between x and y. Let’s call it angle t. We know that x + y + t forms a straight angle, which is 180 degrees:

x + y + t = 180

But t is also an interior angle of a triangle, together with angles z and w. These three must also add up to 180:

z + w + t = 180

Because equality is transitive, we can combine the two previous equations to get the following:

x + y + t = z + w + t

Then we subtract t out from both sides to find that

x + y = z + w

which, reassuringly, is the same answer we got with the first method.


Why Two Solutions?

You may wonder why I’m bothering to show multiple methods of solution, when you only need to get the problem right once. With GRE math, it’s always a good idea to have more than one way to solve a problem, in case they surprise you with a variation where one solution method works but the other one doesn’t. In this problem, for example, what if lines k and m weren’t parallel? Method 1 (transversals) wouldn’t work anymore, but Method 2 (triangles) would work just fine.


Math Review Reference

For more on these topics, see the following sections of the GRE Math Review:

  • 3.1: Lines and Angles (pp. 45-46)
  • 3.3: Triangles (p. 47)

GRE Solutions Manual, Problem 3.25

Exam Prep, GRE, Verbal Reasoning

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 3 a shot before reading this!

3.25: Dolphins and Boat Paint

It’s time for another round of “Same, Worse, or Better.” The way to approach this problem type (as I explained in my solution to 3.14) is to separate the answer choices into three piles: those that weaken the argument, those that strengthen the argument, and those that have no clear effect. Here, we’re only concerned with those answers that would strengthen the argument; thus, our first goal is to identify and reject the “Weaken” and “No Effect” answers so that we can focus on the most promising answer choices.

A good preliminary step is to identify the argument and try rephrasing it in your own words. In this case, you might come up with something like:

“Compounds in boat paint are contributing to increased dolphin mortality rates. Therefore, banning the paints will cause the mortality rate to drop quickly.”

Note that this argument is a prediction about what will happen if we take a certain course of action, not an ethical judgment about what we ought to do.

Now for the answers. Our first choice, (A), actually weakens the argument, if it has any effect at all. If concentrations of these harmful compounds are dropping, but mortality rates are rising, how can we be sure that the compounds are causing the die-off, or even contributing to it? And if we’re not sure that the compounds are causing the die-off, why should we believe that banning the paints will help?

Answers (B) and (D) are carefully-crafted distractors that blur the difference between a logical argument and an appeal to our emotions. If, as (B) suggests, other sea creatures are harmed by the paints too, then that gives us an added incentive for banning them (assuming we care about sea creatures). But this new information doesn’t affect the validity of the original argument, which only concerned the compounds’ effects on dolphins. The fact that manatees or seahorses are poisoned by the paints doesn’t tell us whether we are correct in blaming them for the dolphin die-off as well, just as the fact that chocolate is poisonous to dogs doesn’t imply that it’s harmful to humans. (D), likewise, attempts to cloud the issue: we can’t use other “marine animals” as the basis for predictions about dolphins. In a way, it’s the reverse of (B): the fact that anemones and clownfish can tolerate the compounds doesn’t mean that the same is true of dolphins. (After all, many types of berries are safe for birds but toxic to humans.) Both of these answers belong in the “No Effect” pile.

Answer (C), in contrast, strengthens the argument by answering a possible objection that someone might raise: “What about all the boat paint compounds that have already been released? Won’t those keep harming the dolphins?” (C) anticipates this line of reasoning and suggests that the compounds already in the ocean (and therefore exposed to water) will pose a threat only for “a few months.” If this is true, and we stop introducing these compounds via boat paint, then the problem should quickly resolve itself. Just as raising questions about an argument weakens it (answer A), anticipating and answering questions is one way to strengthen an argument.

Finally, answer (E) fails to address the original argument that banning the boat paints will lower the mortality rate. If the compounds are getting into the water in the first place, it’s fair to assume that the “manufacturer’s directions” are being disregarded, and banning the paints remains a reasonable course of action. In other words, (E) goes into the “No Effect” pile as well, leaving us with (C) as the only answer that strengthens the argument.

GRE Solutions Manual, Problem 3.24

Exam Prep, GRE, Verbal Reasoning

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 3 a shot before reading this!

3.24: Renaissance Prints, pt. 2

This is yet another recurring problem type. I call it “Vocab in Context,” for reasons that will soon become apparent. You’ll see these about once or twice per section, usually as one of multiple questions attached to a medium-sized reading passage.

With “Vocab in Context,” the trick is to focus first on the vocab, then on the context. This particular problem could be paraphrased as follows:

Which of the following five words means the same thing as “passive,” in the context of the passage?

There are two parts to this question, which suggests a two-step approach. In Step 1, we consider each of the answer choices and ask: is this a valid meaning for the target word? In this case, our target word is passive, which our learned friend Dr. Google defines as “accepting or allowing what happens.” With that definition in mind, here’s how the process plays out:

  • (A) Disinterested means “detached, uninvolved, or impartial.” This is a close match for the meaning of passive, so we keep this word.
  • (B) Submissive means “ready to yield to the will of others,” which is close in meaning to passive. Put this one in the keep pile too.
  • (C) Flaccid literally means “limp,” but figuratively it means “dull” and “uninspiring.” This is not a close match, so we discard this answer.
  • (D) Supine literally means “lying face up.” Figuratively, as you might expect, it means “weak” or “yielding” — a good match for passive and therefore a word to keep.
  • (E) Finally, unreceptive means, well, “not receptive,” as in “not open to new ideas.” This is not the same thing as passive, so we can discard this one.

At this point, without even looking at the passage, we’ve managed to eliminate two of the five answer choices. Now, in Step 2, we take all the answers that survived Step 1 and plug them into the sentence to see which one fits the best. Before we proceed, though, look at the definitions of submissive (B) and supine (D) once more. Notice that they’re virtually the same. Just like emerge and coalesce in question 3.5, these two words are too close in meaning for just one of them to be correct. Since they can’t both be right, they must both be wrong.

Even without that consideration, though, disinterested (A) is our winner here. We’re told that the printmakers “reliably record” what’s going on around them — not that they bend or yield (B, D) to rulers, the Church, or any other kind of authority.

Vocab Notes

Struggling to remember the meaning of supine? Med students use this mnemonic: the hand is in supine position when you can scoop soup with it (i.e., when the palm is upward). The opposite of supine is prone, which means “lying face down.”

By the way, disinterested is not the same thing as uninterested, at least for GRE purposes. If you’re disinterested, you’re impartial — you don’t have a direct stake (an interest) in the outcome. Uninterested means you don’t care — you lack interest in what’s going on.

GRE Solutions Manual, Problem 3.23

Exam Prep, GRE, Verbal Reasoning

This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 3 a shot before reading this!

3.23: Renaissance Prints, pt. 1

Here we have an example of another common Reading Comprehension format. I call this one the “triple true/false” because it presents three answer choices, each of which has to be evaluated separately. Usually, as in this case, the goal is to determine whether each statement is supported by the passage.

The key here is to distinguish between the “orthodox position” and the other positions described in the passage. The “orthodox position” is explained in ll. 4-8. Later on, we learn about a different position — the one taken by art historians Scribner and Moxey. We know that their work is unorthodox for two reasons: it’s contrasted with the orthodox position (l. 9), and the author calls it “pioneering” (also l. 9), which means “novel” or “innovative.”

Now let’s look at the answers. We’re in luck: (A) quotes directly from l. 4, which is part of the description of the “orthodox” view. Similarly, answer (C) just restates ll. 5-7, which also belong to the orthodox position. So both of these answer choices are correct.

What about (B), though? It too quotes from the text (ll. 17-18), but if we read through those last lines carefully, we see that (B) is the opinion of Scribner — the “pioneering” or unorthodox researcher — rather than the orthodox view. This is a devious little move on the GRE’s part, and we’ll see it again in future questions. To avoid falling for it, remember: if a passage introduces multiple viewpoints, we need to be clear about who said what.

Book Review: One-on-One 101

For Instructors, Review

Ahdoot, Robert. One-on-One 101: The Art of Inspired & Effective Individualized Instruction. New York: Morgan James, 2016.

Weighing in at a tidy 120 pages, One-on-One 101 is equal parts how-to manual and manifesto. The author, Robert Ahdoot, is best known as the creator of Yay Math, a highly popular video series in which zany costumed characters introduce concepts from high-school mathematics. (Here’s the “Mathemagyptian” explaining inscribed angles.) Ahdoot’s prose, like his YouTube persona, exudes a warmth and enthusiasm that are downright contagious. His unconventional path to a teaching career (“my job found me,” he explains in the preface) has given him a unique perspective on today’s educational system — one that will resonate with tutors, coaches, and others working outside the traditional classroom format.

The basic premise of One-on-One 101 is that teaching individual students is an art, one for which subject matter expertise is necessary but not sufficient. Individualized instruction, as Ahdoot notes, is on the rise, filling a void created by ever-increasing class sizes and passively consumed curricular media. But simply knowing the material isn’t enough to make one a good tutor, or, to use the author’s favored phrase, “One-on-One artist.” Much of the advice in One-on-One 101 seems designed to close the gap between “well-meaning expert” and “professional educator.”

Another commonality between Ahdoot’s videos and his book is his fondness for bold, quirky imagery. For the most part, Ahdoot’s snappy metaphors work well; I found the seeds-soil-sun paradigm introduced in Chapter 8 to be especially … well, illuminating. On occasion, though, I wondered whether Ahdoot’s idiosyncratic vocabulary was embellishing his argument rather than clarifying it; the unconventional use of “tangible” and “intangible,” for example, left me a bit puzzled. I also found myself shaking my head at the suggestion to drop buzzwords like “swag” and “dope” into interactions with students. (To be fair, Ahdoot points out that this technique won’t work for everyone.)

These are small quibbles, though. On the whole, One-on-One 101 presents an excellent balance of practical tips and high-level perspective. The former include ideas on communicating clearly and empathetically, dealing with difficult students, and modeling good organizational habits (hint: hole-punch anything you want your students to file). As for perspective: Ahdoot describes the student-teacher relationship using a host of analogies, but the central concept is that of a “sacred apprenticeship bond.” That may seem a bit lofty for someone whose students just want to boost their GRE score, but it remains an admirable ideal.

In fact, that may be the most important thing that One-on-One 101 offers: a sense of the tremendous value that the best tutoring can have. In an age of MOOCs, YouTube, and Khan Academy, it can sometimes be hard to make the case for involving one more professional in a student’s learning process. One-on-One 101 reminds tutors of the life-changing potential of their work — then challenges them to live up to it.