Tag Archive for 'GRE math'

GRE Example Problem – Order of Operations

Published by Bill on May 18, 2013 in Miscellaneous, Preparation and Revised GRE.
GRE Math isn't so scary; just try this GRE example problem.

GRE Math isn’t so scary; just try this GRE example problem.

Learn more about what you need to know to do well on GRE Math by taking some time to complete this GRE Math example problem.

If L = (a – b) – c and R = a – (b – c), then L – R = ?

This example problem is an exercise in basic mathematical principles, particularly the Order of Operations and your understanding of the Commutative, Associative, and Distributive Laws of mathematics. Let’s do a brief review:

The Commutative Law essentially states that, when you add or multiply, you can swap the order of numbers and get the same answer. So, for example:

Addition:             X + Y = Y + X

Multiplication:  A x B = B x A

The Associative Law states pretty much the same as the Commutative Law with the additional declaration that when you are multiplying and adding groups of numbers, the grouping of those numbers is irrelevant. So, for example:

Addition:             (X + Y) + Z = (Z + Y) + X

Multiplication:  (A x B) x C = (C x B) x A

The Distributive Law says you get the same answer when you multiply a number by a group of numbers added together or multiply each number separately and then add them together. So, for example:

A x (B + C) = AB + AC

This might be easier to understand with actual numbers:

3 x (4 +5) = 3(4) + 3(5)

3 x 9 = 12 + 15

27 = 27

The Order of Operations determines the order in which certain mathematical operations act. The actual order of operations is Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction. A particularly useful mnemonic device to remembering this (rather than memorizing the acronym PEMDAS) is “Please Excuse My Dear Aunt Sally.”

Okay, now duly armed, let’s return to the question above:

L – R = [(a – b) – c] – [a – (b – c)]

Notice that each equation has been bracketed off from the other. This is not because you cannot add or subtract these equations; it is only to signify and help you recognize that you are, in fact, beginning with and looking at the two different variables, L and R. Mainly, in this problem, brackets will help you keep track of which numbers are positive and negative.

In order to solve this problem, the first thing you should do is distribute the negative in front of the equation R represents, a – (b – c). The reason for this is that this equation includes two subtractions; so, when you subtract R from L, you will inevitably subtract a negative. Subtracting a negative turns that negative into a positive number. Observe:

L – R = [(a – b) – c] – [a – (b – c)]

L – R = [(a – b) – c] – [a – b + c]

L – R = [(a – b) – c] – a + b – c

After having successfully distributed the negative, the Commutative and Associative Laws, and the Order of Operations, tells us that we are free to solve this problem with no more hang ups:

L – R = a – b – c – a + b – c

You can reorganize for coherency:

L – R = a – a + b – b – c – c

L – R = 0 + 0 – c – c

L – R = -c – c

L – R = -2c

Thus, the answer is -2c.

Find more GRE example problems here. Have a question about GRE Math or graduate school admissions? Ask the experts at Test Masters!

 

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Sample Math Problem: It’s hip to be a square…or a cube!

Published by Calvin on December 18, 2012 in Example Problems and Preparation.

How can you get more hip than this?

Geometry is heavily tested on the GRE Math section, and a thorough review of geometrical concepts is essential to a high score. Consider the following problem:

“If the length of an edge of a cube X is twice the length of an edge of cube Y, what is the ratio of the volume of cube Y to the volume of cube X?”

The easiest way to solve this is to pick a number for the initial edge length and plug it into the problem. For instance, let’s say cube X is a 4x4x4 cube. Cube X would have a volume of 64. Cube Y would have to be a 2x2x2 cube, since 2 is half of 4, and it would have a volume of 8. The ratio of the volume of cube Y to the volume of cube X would thus be 8 to 64, or 1/8.

However, you really should have known that to begin with. Imagine that cube X had edges that were three times as long as those of Cube Y. Then Cube X would now be a 6x6x6 cube if Cube Y remains a 2x2x2 cube, and the volume ratio would be 8 to 216, or 1/27. Notice something? 8 is 2 ^3, and 27 is 3^3. If the ratio of the sides is 1:4, the ratio of the volumes will be 1:64. If the ratio of the sides is 1:5, the ratio of the volumes will be 1:125. Since these are cubes, you just cube the ratios. 1^3 is 1, and 4^3 is 64; 5^3 is 125. If you know this simple property of the relationship between length and volume, it will take a problem that would take 30 seconds to solve and turn it into a problem that takes 5 seconds to solve. On a timed exam, that could be the difference between getting another, harder question right or wrong. Memorizing these kinds of mathematical facts is something that the GRE test writers expect top scorers to do, and they write the questions so that they can be solved quickly if you know them. It also pays to memorize the squares and cubes of the numbers 1 through 12.

Sometimes a picture is worth a thousand words.

So with cubes, you cube the ratio of the sides. What about squares? If you guessed that you square the ratio of the side lengths in order to get the ratio of the areas, you’d be right, as you can see from a quick demonstration. If the original square has side lengths of 1 and the new square has side lengths of 2, the side ratio is 1:2 and the area ratio is 1:4. If the new square has side lengths of 3, then the side ratio is 1:3 and the area ratio is 1:9. If the new square has side lengths of 4, then the side ratio is 1:4 and the area ratio is 1:16, and so on. Sure enough, you just square the original ratio.

The classic children’s science fiction novel, A Wrinkle in Time by Madeleine L’Engle, featured tesseracts as wormholes used to travel vast distances through space.

So now you know about cubes and squares, but what about tesseracts? “Tessawhats?” you say? A tesseract is to a cube as a cube is to a square, just as a cube is to a square what a square is to a line. Still confused? Let me explain it this way: say you draw a line a foot long running from east to west. This line only exists in one dimension: east-west. Then, you decide to square it by adding three more lines: two perpendicular to it running north to south and one parallel to it running east to west. This square exists in two dimensions: east-west and north-south. Now you decide to turn the square into a cube by adding lines in the up-down dimension, so that each edge of the original square is now the edge of another square emanating from it. This cube exists in three spatial dimensions: east-west, north-south, and up-down. Now you take this cube you’ve made and decide to square it…in a fourth spacial dimension.

What is this fourth dimension? Who knows. We live in a world in which we experience only three spacial dimensions, so it is impossible for us to imagine what a four dimensional object would look like. That hasn’t stopped mathematicians from naming four-dimensional objects, and this hypercube I’ve just described to you is called a tesseract. As you know, even though a cube is a three dimensional object, it is possible to draw a cube on a piece of paper in only two dimensions by using perspective and all those other artistic illusions. Likewise, some have attempted to render tesseracts in three dimensions in order to give some approximation of what they might look like. Having never seen an actual tesseract, though, you might still find these representations confusing.

A tesseract!

In terms of doing calculations, though, tesseracts are simple as can be. For a square with side lengths of 1 and another square with side lengths of 2, the ratio of side lengths is 1:2^1 (since sides are 1 dimensional), or 1:2, and the ratio of areas will be 1:2^2 (since squares are 2 dimensional) or 1:4. For a cube with side lengths of 1 and another cube with side lengths of 2, the ratio of volumes is 1:2^3 (since cubes are 3 dimensional), or 1:8. So, for a tesseract with side lengths of 1 and another tesseract with side lengths of 2, the ratio of hypervolumes(?) is 1:2^4 (since tesseracts are 4 dimensional), or 1:16. It just follows the pattern. Try not to think about it too much.

If you’re having trouble with tesseracts, don’t worry. They’re not on the test. I just wrote about them to mess with your head.

Remember, if you ever want extra help getting ready for the GRE, you can always study with experts like me through Test Masters. Until then, happy studying!

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GRE Sample Math Problem: Pattern Problems

Published by Calvin on November 22, 2012 in Example Problems.

Doing GRE problems is kind of like lifting weights, except less sexy.

“It’s not GREek!” will present you with question types you are likely to see on the GRE, as well as a brief explanation on how to arrive at the answer for each question. This week we will turn our attention toward a sample GRE Math problem.

Remainder problems are word problems that always involve some sort of repeating pattern. Consider the following example:

“Arnold decides to start a new exercise regimen at his gym. He will devote the first day of his plan to exercising his upper body, the second day to his lower body, the third day to cardio, the fourth day to resting, and then the pattern will repeat: upper body, lower body, cardio, rest, repeat, ad infinitum.  Arnold continues this until he pulls a muscle and is forced to rest while it heals. If he pulled a muscle on lower body day, then which of the following could be the number of days he spent following his regimen? Select all possible answers.”

A) 243

B) 122

C) 567

D) 84

E) 370

F) 284

To solve this, we just need to think a little bit about the nature of the pattern. We know the last day he worked out was a lower body day, because that’s when he pulled the muscle, so the number of days he spent following the regimen would have to be such that the last day would be a lower body day. The first day that was a lower body day was the second day; the next was the sixth day, then the tenth, and so on. Because the pattern is four days long, every fourth day the pattern repeats. We can thus represent the lower body days like this:

LBD = 2 + 4N

Where N is the number of times the pattern has repeated and an LBD is a lower body day. We have to add two because the lower body day is the second day in the pattern. Thus, the first lower body day would be:

LBD = 2 + 4(0)

LBD = 2 + 0

LBD = day 2

The second lower body day would be:

LBD = 2 + 4(1)

LBD = day 6

The third would be:

LBD = 2 + 4(2)

LBD = day 10

And so on. The formula fits our earlier predictions, so we should be able to use it to work backwards to solve the problem; to answer the question, we just need to plug in each answer choice for LBD and solve for N. If N comes out as an integer, then that LBD was in fact a lower body day.

Consider choice A:

243 = 2 + 4N

241 = 4N

60.25 = N

N is not an integer, so A is incorrect. Consider B:

122 = 2 + 4N

120 = 4N

30 = N

N is an integer, so B works. Consider C:

567 = 2 + 4N

565 = 4N

141.25 = N

N is not an integer, so C is incorrect. Consider D:

84 = 2 + 4N

82 = 4N

20.5 = N

N is not an integer, so D is incorrect. Consider E:

370 = 2 + 4N

368 = 4N

92 = N

N is an integer, so E works. Consider F:

284 = 2 + 4N

282 = 4N

70.5 = N

N is not an integer, so F is incorrect. Thus, the two correct choices are B and E.

Remember, if you want, you can always get extra help studying for the GRE from the experts at Test Masters. Good luck!

Want more GRE example problems? Click here! Want more information on GRE test preparation or on GRE courses in you area? Click here!

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Sample Math Problem: Two trains leave the station…

Published by Calvin on November 12, 2012 in Example Problems.

If a nineteenth-century Russian adulteress goes to a train station…

“It’s not GREek!” will present you with question types you are likely to see on the GRE, as well as a brief explanation on how to arrive at the answer for each question. This week we will turn our attention toward a sample GRE Math problem.

Ah, the dreaded train problem. Surely these kinds of questions must be the the most infamous of all inane word problems. They can haunt the mathematically disinclined for years after leaving school, causing people undue anxiety waiting in traffic for a locomotive to pass. You probably thought you left these behind long ago, but they’re back. Who cares about some stupid trains, you ask? The GRE, that’s who.

Never fear though – all GRE math questions are written so that they they can be solved in less than two minutes, if you know what to do. This means that they aren’t going to require going through a lot of complicated steps to solve, and remember, the GRE doesn’t test anything beyond high school math. It just asks questions in unfamiliar ways that may require you to read carefully, and if you’re more of a verbal person than a math person, that shouldn’t be so bad, right? With some practice, the test makers’ tricks become familiar and recognizable, and problems that once seemed confusing become plain as day. Today, we’ll banish your siderodromophobia (fear of trains) for good.

Consider the following GRE math problem:

“At 10:00 AM train A left the station and an hour later train B left the same station on a parallel track. If train A traveled at a constant speed of 60 miles per hour and train B at 80 miles per hour, then at what time did train B pass train A?”

Monet’s “The Arival of the Train” at the Gare St. Lazare in Paris.

The first step to solving this problem is understanding what the question is really asking. What is this question asking? Well, “at what time did train B pass train A.” Yes, but what does that mean? When will train B pass train A? When they have traveled the same distance.

This is key to understanding how to solve the problem. We are going to need to know how to use the information we have been given, the speeds of the trains and the times at which they left the station, to calculate the distance they have traveled. As you know, the distance formula is usually written as:

speed = distance/time

If we want to find distance, we rearrange this familiar equation like this:

distance = speed(time)

So, if we want to calculate train A’s distance after a given length of time, we would multiply train A’s speed times the length of time it has been traveling. We know train A’s speed is 60 mph, so if we let the variable t represent the number of hours it has been since 10:00, we could write this as:

If you like the Monet painting, consider visiting the Musee d’Orsay. It’s full of them.

train A’s distance = 60(t)

Now, for train B, it’s slightly more complicated. How far has train A gone by 11:00, one hour after leaving the station? 60 miles, of course. But how long has train B gone? Zero, because train B doesn’t start traveling until 11:00. If we were to write this mathematically, we would have to express the distance traveled by train B as:

train B’s distance = 80(t – 1)

We have to write (t – 1) because train B starts an hour later than train A. This makes sense, because if we let t be one, that is, one hour after 10:00, then train B has gone zero miles:

80(1 – 1) = 80(0) = 0

Now, what were we trying to find again? The time when train A and train B have traveled the same distance.In other words, we want to know when:

train A’s distance = train B’s distance

If train A’s distance is equal to 60(t) and train B’s distance is equal to 80(t – 1), then we can just set those termsd equal to each other and solve for t:

60(t) = 80(t – 1)

60t = 80t – 80

80 = 20t

4 = t

So, four hours after 10:00 is when train A and train B have traveled the same distance. So that’s 2:00 PM. Was that so bad?

All you need to do is break it down step by step and practice. Try this one on your own and post the answer as a comment if you think you got it right:

“Train A leaves Paris at noon and travels at a constant speed of 75 mph toward Berlin. At the same time, train B leaves Berlin headed toward Paris at a constant speed of 50 mph. If Paris and Berlin are 500 miles apart, then at what time will the two trains pass each other?”

Remember, if you want, you can always get extra help studying for the GRE from the experts at Test Masters. Good luck!

Want more GRE example problems? Click here! Want more information on GRE test preparation or on GRE courses in you area? Click here!

 

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New GRE Math Problems – More Tips and Techniques

Published by admin on October 12, 2011 in Example Problems, Preparation, Revised GRE and Test Day.

You won't see calculus on the New GRE!

In the last math post we talked about how the new GRE is putting more emphasis on word problems. However, there are plenty of pure algebra problems on the new test. Here’s a couple of good examples of the types of algebra problems you might see.

 

1. This first problem involves two equations and two variables. You may have learned how to do a problem like this in school with your graphing calculator. Unfortunately, you won’t have that available to you for the GRE. Continue reading “New GRE Math Problems – More Tips and Techniques” »

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New GRE Math Tips and Tricks

Published by Jason on September 8, 2011 in Example Problems and Preparation.

This is a fraction, not a word problem

The new GRE Quantitative section places a greater emphasis on word problems. Here are a couple of good examples of the types of word problems that will be more common on the new GRE.

Problem #1 –   This first problem is a numeric entry problem, one of the new problem types. For these problems, you will not be given multiple choices.  Here is the problem: Continue reading “New GRE Math Tips and Tricks” »

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Grad School vs. the Working World: Should I Continue My Education or Start Working?

Published by Jason on April 5, 2011 in Graduate School.

Grad school: It seemed better than getting a real job

Now that it’s April, and seniors everywhere are looking forward to graduation and beyond, the question, “should I go to graduate school or find a job right out of college,” is probably floating around in a lot of minds. The answer, as it is for every important, life-altering decision, is: it depends. Luckily, there’s this neat thing called the Internet, where people can find answers to all of their questions!

Continue reading “Grad School vs. the Working World: Should I Continue My Education or Start Working?” »

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I Want To Go To Graduate School To Study Stuff!

Published by Jason on March 22, 2011 in Admissions and Graduate School.

What field do you intend to specialize in? All of it.

Far be it from me to discourage anyone from going to graduate school, but this Xtranormal video, which has been floating around for quite a while, demonstrates exactly the kind of vague and unfocused thinking that is exactly wrong for graduate school.

Watch the video after the jump.

Continue reading “I Want To Go To Graduate School To Study Stuff!” »

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