The last time we looked at a short verbal problem, it was of the “ridiculously, stupidly difficult” variety. You might find it interesting to see how the GRE test makers use challenging vocabulary, but on a medium-difficulty problem. Here’s a two-blank sentence completion we cover in our GRE Bootcamp event:

The social reformer proposed locating juvenile detention facilities close to parks, playing fields, and greenbelts, theorizing that (i) _____ to such sites could have a favorable effect on the troubled youths, as opposed to the (ii) _____ influence associated with locations in the heart of high-crime slum areas.

In a multi-blank sentence completion, there are three choices for each blank. But I’m not even going to show the choices to you yet, because you shouldn’t look at them on Test Day until you figure out the meaning of the blanks yourself.

One nice trick on these problems is to start with the blank that’s easiest to figure out. I’ll start with the first blank here because, in my opinion anyway, it’s the clearest one. But never start with the first blank just because it’s first!

The reformer wants these parks and fields to be “close to” the detention facilities, so the first blank needs to mean “closeness” or something.

Now look at your options for that blank:

ubiquity
diversity
propinquity

Uh oh, vocabulary! But contrary to the last problem, this one has a backdoor. “Diversity” is a word you know, and it definitely doesn’t mean “closeness,” so knock it out. And I’ve found that ubiquity is a pretty commonly studied word; you might very well know that it means “universality,” which also doesn’t make any sense. So even if you have no idea what propinquity is, you can click it confidently on Test Day: it’s gotta be the answer.

Let’s look at the second blank now. “As opposed to” is an excellent clue phrase; it signals that the second blank is the opposite of “favorable.” In addition, the phrase “high-crime slum areas” is a rather explosive clue that whatever “influence” the second blank describes is negative.

Quick tip: don’t get overly fancy with your predictions. If you happen to think of a word like “unfavorable” or “negative,” that’s great. But if not, don’t sweat it. Your prediction doesn’t even have to be a single word; it can be a descriptive phrase, like “not the greatest!” The point is that you make a prediction, not that it be exquisite. We’re making predictions to guard ourselves against the insidious influence of the wrong choices, not to guess the exact right choice.

Now look at your options for blank two:

timorous
malevolent
propitious

More vocabulary! But, even if you didn’t know that malevolent means “evil” or “harmful,” you should recognize the prefix mal-, which means “bad.” (Think of every word you can that starts with “mal-”: malignant, malady, malice, malign … none of them mean happy things!) So malevolent has to be the word you seek. This is another case when, even if you aren’t sure of the vocabulary, you can be sure of the choice you’re clicking.

Don’t get frightened just because a GRE short verbal problem has vocabulary you don’t know. It’s true that a hard problem will force you to know some ridiculous words, or guess; but a medium-level problem like this one features several varieties of back entrances that let you bypass most of the words. Notice how, to get this problem right, you don’t even have to care what propinquity, timorous, and propitious are! This is the kind of point you should expect to score on Test Day.

GRE teachers have a love/hate relationship with the calculator that testers get on their quantitative section. It can be useful in certain situations – if you have to multiply 372 by 754, for example, then by all means take advantage of the calculator to get a quick answer. By and large, though, we find that the majority of our students, even the ones who use a lot of math as part of their jobs or classes, overuse the calculator on problems that they could solve much faster by hand. It’s not because they’re unintelligent or lazy – given how ubiquitous calculators and spreadsheets are in our daily lives now, a lot of people just don’t remember how to do quick scratch-paper arithmetic. (You’ll notice that I didn’t say “mental” arithmetic – do not do any steps in your head! Write your work down. But that’s a topic for another day.)

So, to help remedy this problem, here are some quick arithmetic tricks that you can use to save yourself time and energy on the GRE:

- To divide by 4: Divide the number by 2, and then divide by 2 again.

- To multiply by 4: Flip the last trick. Double the number, and then double again.

- To multiply by 5: Multiply the number by 10, and divide by 2. (For integers, this just means adding a 0 to the end of the number and taking half of the result.)

- To divide by 5: Again, flip the previous tip. Divide the number by 10 (take away a 0 or move the decimal point one unit to the left), and double it.

- Once you’ve multiplying and dividing by 4 and 5 down, working with other numbers becomes simpler as well: Multiplying by 6, for example, just involves doubling a number and then multiplying that by 3.

- To quickly find percentages, multiply the number by the integer value of the percent, and use logic to determine where the decimal point goes. For example: If you need to find 12% of 300, first multiply 12 and 300: That gives us 3600. But we’re looking for a value that’s just a bit bigger than 30 (which is 10% of 300), so our answer must be just 36.

Applying these tricks may feel like a poor use of time at first, but if you practice by doing just a couple of calculations a day this way, by Test Day you’ll be a scratch-paper math whiz – able to outpace both the calculator and everyone else in the testing room!

Join Kaplan’s Director of Graduate Programs, Lee Weiss, to get answers to some of our most frequently asked questions from GRE students like you. This video discusses GRE FAQs including the exciting new GRE ScoreSelect option for submitting your best scores to graduate schools. We also address the common question “What is a good score?” It comes down to the best score for you, and for the specific programs you are applying to. Your GRE scores are good for 5 years from the day you test, and you can now retest every 30 days if needed.

If you have additional questions, please visit us on Facebook or Twitter. For more info on the New GRE, visit our GRE Test Info Center. If your GRE Test Taking skills need a boost, check out Kaplan’s GRE Advantage course, specifically designed to focus on essential tools for success on the GRE.

My fiancé and I have five pets: A dog, a kitten, and a trio of guinea pigs.  We are normally excellent animal parents, but the one thing that we consistently disagree on is whether it’s okay to feed them outside of their mealtimes and occasional treats.  I taught the dog that begging from the table gets her nowhere with me, but my fiancé is the king of slipping her bites of steak or “dropping” scraps of bacon onto the ground (or into her already-open mouth, as is more often the case).  Our last conversation on the topic began when he, instead of putting his leftover pork into the fridge, put it into the dog’s bowl.  It went something like this:

Me:  “If you keep doing that, she’s going to get fat!”

Him:  “Oh, please – how much can a few bites possibly matter?  That much meat wouldn’t affect my weight at all.”

He made two crucial mistakes in that moment:  He both triggered my inner GRE instructor, who loves percentages and data analysis, and made me determined to prove him wrong once and for all.  So I decided to figure out the percentage by which that extra pork increases the dog’s caloric intake, and compare that to how much it increases his.

My fiancé eats anywhere from 2,500 calories to 2,800 calories per day, with the upper end being during football season – not because he plays, but because he and his friends spend every Sunday feasting on buffalo wings and sliders while watching NFL games.  So what percentage of his daily intake does a 4-ounce pork loin comprise?  We can use the percentage formula to find out.  The percentage formula is (y/x) * 100%, where x is the total value and y is the smaller portion of the whole.  In this case, since 4 ounces of pork has roughly 160 calories, our percentage would be (160/2,800) * 100%.

Now, you could very reasonably be asked to estimate this percentage on the GRE, so let’s simplify it without using the calculator.  Let’s ignore the * 100% at the end of the equation for now, and just focus on the fraction.  We can start reducing 160/2,800 by cancelling out one factor of 10 in the numerator and the denominator, and then factoring out a 4 from each term.  That would look like this:

160/2,800  =  16/280  =  4/70

Now, if we add the * 100% back in, we have (4/70) * 100%, which gives us 400/70 when we multiply 100 times the 4 in the fraction’s numerator.  Once again, the numerator and the denominator each have a factor of 10, and when we cancel it out we’re left with 40/7.  We know that 42/7 equals 6, so 40/7 equals something just a bit smaller – we can estimate it to be 5.8%.

By comparison, our 30-pound dog only needs 900 calories per day to maintain her weight.  So every time someone makes her day by giving her 160 calories worth of meat, he increases her intake by (160/900) * 100%.  The fraction in this case doesn’t reduce quite as easily as the previous one did, so let’s look at the 100% and the 900 in the denominator – that reduces to 1/9.  So we’re really dealing with 160/9, which again is not very easy to reduce.  Let’s take a quick quiz to test your strategic estimation skills:  Without using the calculator, can you determine which of these percentages is roughly equivalent to 160/9?

(A) 8%

(B) 10%

(C) 18%

(D) 34%

(E) 56%

Again, this is another very GRE-like question, and your ability to quickly approximate will serve you well.  160/10 would give us 16%, and 160/9 is going to be just a bit larger than that – the only answer that can possibly match is (C), 18%.

So while that piece of pork would only increase my fiancé’s calories by less than 6%, it would increase our dog’s calories by nearly 20% – a significant difference that definitively proved me right.  I presented my findings to my fiancé, who had no choice but to cede his case.  When dinnertime rolled around, however, one of the ham steaks went missing – apparently tired of listening to us debate, the dog had taken matters into her own hands and stolen it off of the counter.

I know that I’m not the only one who uses GRE-like percentage, ratio, and arithmetic calculations in my everyday life – how do these types of math factor into your day-to-day activities?  Let us know in the comments!

GRE Geometry problems with multiple figures can often seem intimidating – the components of each figure overlap and intersect in sometimes-unclear ways, and you just know that they’re going to be time-consuming – rarely can GRE multiple figures problems be solved with just a few calculations.  Take a look at this problem, for example:

It’s easy to picture the GRE test-maker laughing fiendishly as he inscribed an equilateral triangle into a circle, and then added a square in just to make your head explode.  However, that portrait is likely inaccurate – each problem on the GRE is designed to test your critical thinking skills, and multiple figures problems are no exception.  Let’s take this one apart:  We need to find the area of polygon ABCDE, which is not a regular polygon.  The language in the question stem gives us an important clue, though – if we find the areas of triangle ABE and triangle BCDE independently, then we can add them together to get the total area of the larger figure.

The only figure for which we have any real values is the circle, so let’s start there.  The area is 36π, and if we plug that into the equation for the area of a circle, we get .  Once we divide both sides by π and take the square root of 36, we see that the circle has a radius of 6.  It would be very convenient if the radius or diameter of the circle corresponded to any side length of the triangle, but we’re not quite so lucky here.  Instead, we’ll have to rely on a lesser-known geometry rule in order to continue solving.

Since we don’t have any radii drawn into the figure, let’s call the center of the circle P and add two radii, as shown:

Now, here’s that not-so-common geometry rule:  If an angle is formed by two radii creating an arc (minor arc AE, in this case), and another angle starts on the opposite side of the circle and ends at the same arc-endpoints, then the former angle has twice the measure of the latter angle.  If we apply that rule to this figure, we know that angle APE is twice the measure of angle ABE.  And since angle ABE is an angle in an equilateral triangle, its measure is 60 degrees.  When we double that, we find that angle APE is 120 degrees.

With me so far?  I hope so, because we haven’t even found the triangle’s side length yet.  We’re getting there, though!  Now that we know the angle measure of APE, let’s cut it in half by dropping a height down to line AE, and creating two identical triangles.  Since we’ve divided angle APE in half, each of the top angles measures 60 degrees.  And since the newly-formed angles along line AE are both right angles, that means that the remaining angles in each triangle must measure 30 degrees each.  Do these angles sound familiar?  Knowing the ratio of side lengths in a 30:60:90 right triangle will put you far ahead of the pack on GRE Test Day.  The ratio of sides in triangles with these angle measures are always x:x√3:2x.  Since the known side in this case, the radius of the circle, is opposite the 90 degree angle, we know that 6 = 2x, which means that x = 3 and x√3 = 3√3.  Here’s all of the information we currently have, drawn into the figure:

If we double the 3√3, we have the side length of the equilateral triangle: We now know that it’s 6√3.  Now that we have the length of the triangle’s base, all we need is its height in order to calculate its area.  Let’s take the triangle out of the figure for a moment:  If we add a height, we create another pair of 30:60:90 right triangles.  The side length opposite the 90 degree angle is 6√3, which means that the base is 3√3 and the height is 3√3 * √3, which equals 9:

Since we finally have the base and height of the equilateral triangle, we can calculate its area:

½ (base)(height) = ½ (6√3)(9), which gives us 27√3.

Looking at the answer choices, which ones now jump out at you?  Answers (A) and (E) both have 27√3, and look to be the most promising choices.  If you were tackling this problem on Test Day and were running out of time, now would be a perfect time to take a strategic guess and move on.

Calculating the area of square BCDE is now a snap:  The equilateral triangle and square share a side, so we know that the square also has a side length of 6√3.  Therefore, the square’s area is (6√3)², or 108.  The area of polygon ABCDE, then, is 108 + 27√3, which matches answer choice (A).

That was a marathon of a problem, by anyone’s standards.  Let’s review all of the concepts that you needed to know and recognize in order to solve that problem:

1. The formulae for areas of circles, triangles, and squares,
2. The special geometry rule governing relative angle measures in circles,
3. The ratio of sides of a 30:60:90 right triangle, and
4. Recognition of the shared sides between the figures that allowed us to transfer information from one figure to another.

Putting all of those pieces together is a test of your critical thinking skills, if I’ve ever seen one!  With practice, though, applying all of these geometry concepts to one problem will become easier and faster, and will allow you to rack up points on GRE Test Day.