The post Mean, Median and Mode on the GRE appeared first on Magoosh GRE Blog.

]]>Over the last six months, I’ve gone through quite a few Revised GRE prep books. One thing I’ve noticed missing is difficult questions relating to mean, median, and mode. Sure, most books describe how to find the average, and what the difference between the mean and the median. Many already know the above, but mean, median and mode questions on the actual Revised GRE are going to require much more practice than you are going to get from simply applying the basic formulas.

Below are five questions that should hopefully give you a good mental workout. If you whip through them, answering all correctly, then you are very well prepared for the concepts you’ll find on the GRE. If you stumble, don’t despair. Remember, we have plenty more questions like these—some that are every more difficult—waiting for you at gre.magoosh.com.

1. Set S is comprised of six distinct positive integers less than 10. Which of the following must be true?

I: The median is an integer

II: The median is less than the average

III: The range of digits in Set S is less than 8

(A) I only

(B) I & II

(C) II & III

(D) III only

(E) None of the above.

2. A list is comprised of five positive integers: 4, 4, x, 7, y. What is the range of the possible values of the medians?

(A) 2

(B) 3

(C) 6

(D) 7

(E) Cannot be determined by information provided.

3. The average of five positive integers is less than 20. What is the smallest possible median of this set?

(A) 19

(B) 10

(C) 4

(D) 3

(E) 1

4. Set S is comprised of 37 integers

Column A | Column B |
---|---|

The median of Set S | The mean of the lowest and the highest term |

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

1. A good idea is to choose numbers. For instance, just using 1, 2, 3, 4, 5, 6 we can see that the first condition does not need to be true. The median of this range of numbers is 3.5. For the second condition, we can also use numbers and determine that the average and the median are equal. Therefore the second condition also does not have to hold true. Finally, for the third condition, the range of digits is 9 – 1 = 8. The third condition says less than 8. Therefore the answer is (E) none of the above.

2. Here we want to find the lowest possible median and the great possible median. Then we want to subtract the least possible median from the greatest possible median to find the range of the medians. If x and y are less than or equal to 4, then the median is 4. If x and y are greater than or equal to 7, the median is 7. Therefore the range of median is 3, Answer (B).

3. Choosing numbers will help us on this one. If we choose 1, 1, 1, 1, 16, the sum is equal to 20. The median is 1, Answer (E).

4. This is a conceptual problem. Nonetheless we can still pick numbers. Imagine all 37 numbers are the number ‘1’. The mean is ‘1’ and the average of the least and greatest is 1. n this case both columns are equal (C).Now, just change the last number to a 3. That is you have the number ‘1’ 36 times and the number three. The median is still 1. But now the average of the least and greatest is greater than ‘1.’ In this case column (B) is larger. If the answer switches depending on the numbers we use, the answer is (D).

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]]>The post GRE Math: Solving Quantitative Comparisons appeared first on Magoosh GRE Blog.

]]>First off, here’s a tricky algebraic QC:

1) x > -3

If you find this puzzling, this article discusses an efficient method for handling such problems. Before we talk about that, a quick refresher on a few basic ideas dealing with equations and inequalities.

Suppose you have an equation, P = Q. You probably remember that you can add, subtract, multiply, or divide one side of an equation by just about anything, as long as you remember to do the same operation to the other side. The only restriction is — it’s illegal to divide by zero — all other adding & subtracting & multiplying & dividing is perfectly legal.

Now, suppose you have an inequality, P > Q or R < S. It is still 100% legal to add or subtract anything you want to both sides. If you multiply or divide both sides by a *positive* number, that’s perfectly legal as well. If you multiply or divide by a *negative* sign, this is also allowed, but it reverses the direction of the inequality —- if P > Q, then (–5)*P < (–5)*Q. Of course, we never can divide by zero, and it’s also problematic to multiply by zero, because that would obliterate the inequality, making both sides equal.

OK, probably all of that was review, but we need that as our ground rules.

For some folks, especially folks with good mathematical intuition, what I am going to explain here will be patently obvious — probably you have already thought of it. But, for folks for whom math is a constant struggle and not particularly intuitive, this new method might bring a landslide of insights.

You see, one way to think about an ordinary variable like x is as a “holder for a number.” In any algebraic equation, such as 2x + 5 = 13, the variable x is holding some as yet unknown number, and of course the goal of most standard algebra is to find what number it is holding.

As you know, the GRE Quantitative Comparisons set a very different task. We are not trying to solve for a number in the ordinary sense. Instead, we are trying to figure out the relationship between columns A & B —- either A > B, A < B, A = B, or we can’t determine the relationship. What is unknown is not a number but a relationship — what mathematicians call a **relation**. At the outset of a typical GRE QC, the unknown relation could be < or > or =.

I will suggest introducing a “mystery symbol”, which I will denote by the arbitrary combination **$\!|$**. This is a holder, not for an unknown number, but for an unknown relation. We could begin any GRE QC by sticking this mystery symbol between the two columns.

What’s the point of that? Now, we just have a funky symbol between the two columns! How does this help us? Well, think about it. Whether **$\!|$** stands for < or > or =, there are certain mathematical transactions that are 100% legal. Here is a summary of what we can do to both sides of **$\!|$**:

a. We can add or subtract any number to both sides of **$\!|$**.

b. We can multiply or divide both sides of **$\!|$** by any positive number.

This opens up a panoply of ways of handling the two columns.

I will demonstrate a solution to the QC at the top of the page using this “mystery symbol” method. Begin with the mystery symbol between the content of the columns.

2x – 4 **$\!|$** 5x + 7

First, because I can subtract any number from both sides, I will subtract 2x from both sides, to get all the x’s on one side.

–4 **$\!|$** 3x + 7

Now, I will subtract 7 from both sides.

–11 **$\!|$** 3x

I am allowed to divide by any positive number, so I will divide by 3.

–11/3 **$\!|$** x

Well, we were told that x > –3, and –3 > –11/3, so the “value” of **$\!|$** must be <, and column **B** is bigger.

Here, I made up the arbitrary symbol **$\!|$** to demonstrate the idea of handling an unknown relationship, but of course, you don’t actually need this symbol once you understand the idea. You can leave the space between the columns blank and simply follow guidelines (a) & (b) above, doing simultaneous operations on both columns until they are simplified to a form that allows you to decide directly.

The problem at the top of this article was a relative easy problem — folks with good number sense might have been able to solve it by inspection. Here’s a somewhat more challenging problem in which the same principle can be used with powerful effect.

2) http://gre.magoosh.com/questions/165

Questions about this? Do you have thoughts you would like to share? Let me know in the comments section below! 🙂

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]]>The post How to Approach Complicated Math Problems appeared first on Magoosh GRE Blog.

]]>Every GRE test is going to have a few math questions that are very difficult. Upon seeing them, you may become completely flustered and believe that there is no way you are able to solve such a question. Oftentimes this is not the case. Rather, you fall into any number of traps that prevent you from solving a problem that is well within your ability.

Below are five important points to keep in mind when you are dealing with a difficult GRE math problem.

It is very easy to become flustered on difficult math problems, especially word problems. One reason our normal breathing quickly changes to short, agitated breaths is we start reading and re-reading the same question over again. Hope that by the fourth read we will finally get it. At this point, re-reading is clearly an example of diminishing—and frustrating—returns. What to do?…

It often may take about 30-seconds—and a couple (and only a couple) of calm rereads of the question—to decipher a complicated math question. After all, it is complicated. Deciphering the question means understanding what the question is asking. Next, find the solution path. To do so, think—or even write down—the necessary steps to get to the solution.

A few pieces of good news: you can take longer on complicated questions. After all, there aren’t too many of them. Just make sure to solve the easy and medium questions quickly and accurately. Secondly, you can always come back to a question. Sometimes, it is easier to decode the second time around.

Sometimes a problem seems much more complicated than it actually is. The reason is we are misreading a word, or injecting or own word into the question. We spend several minutes laboring through difficult equations only to realize that none of the answer choices matches up with our answer. To avoid this **make you sure you don’t rush through the question**. Instead, read carefully, and know what the question is asking before attempting the question.

The great thing about the new GRE vs. the old format is that you can come back to questions. Sadly, many students do not take advantage of this and are unable to pull themselves from a question once they’ve bitten their teeth into it. But knowing when to unclench that jaw is very important. If a minute has gone by and you are unable to make sense of the question, move on! That question is worth the same number of points, so there is no point in wasting precious minutes on it.

The best thing about being able to come back to questions is your brain is able to make sense of the question a lot more easily the second time around. What seemed cryptic and inscrutable now seems much clearer.

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]]>The post Quantitative Comparison Pacing Drill appeared first on Magoosh GRE Blog.

]]>The very first problem you will see test day will be a Quantitative Comparison one (QC). Indeed, the first seven problems will be QC, unless you decide to skip them (which in the case of the last couple may, for some, be a wise strategy).

Therefore, while prepping, you should make sure not to neglect QC, as it makes up over a 1/3 of the questions (remember, each question is worth the same points). To assess your ability on QC tackle the seven questions below. I’ve tried to make them similar in difficulty to what you’ll see test day.

The first question is the easiest, and the last the most difficult. That is, theoretically, the higher the number the more difficult the problem. I say theoretically because sometimes a question #6 turns out to be more difficult then question #7. Though, a question #3 will never be more difficult than a question #7.

What this means in terms of pacing is that you should not spend too much time at the beginning of the QC section. Yet, many do just that, thinking that they want to make sure to get the first few questions correct. So instead of checking your work several times on the first QC question, try to go through these questions quickly–but accurately–so you can spend more time focusing on the medium level questions.

That said, if you are struggling on the last couple QC (the difficult ones), then you should skip to other parts of the section and spend your time answering easier and medium level questions. With the questions below, you only have seven QC–as I didn’t want to write an entire GRE math section! So do you best, and see how many you can answer correctly in 12 minutes.

1. The average price of a home County X is 250,000. The average price of a home in County Y is 300,000. The average price of a home in both County X and Y is 265,000.

A

The number homes in County X

B

The number homes in County Y

2.

A

X

B

3

3. x + 1 = |x – 1|

A

x

B

0

4. Event X and event Y are independent. The probability of event X is 40%. The probability of events X and Y both occurring is 16%.

A

The probability of X occurring

B

The probability of Y occurring

5. A

The number of factors of

B

The number of factors of

6. The area of a square is doubled.

A

The percent increase in one side of the resulting square

B

50%

7.

A

The number of distinct numbers that can be the units digit of n

B

4

Answers:

1. A

2. B

3. C

4. C

5. A

6. B

7. B

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]]>The post GRE Math: Percentiles and Quartiles appeared first on Magoosh GRE Blog.

]]>Fact: An 8 year old boy who is 4’5″ (53 inches) tall is in the 86th percentile for height for his age.

What on earth does that mean? Well, the percentile of an individual tells you what percent of the population has a value of a variable is below that individual’s value of the variable. For example, to say that a 4’5″ 8 year-old boy is in the 86th percentile for height for his age, we are saying: gather together all 8 year-old boys on Earth, and measure their heights; if you sort out all the 8 year-old boys who have a height less than 4’5″, they will comprise approximately 86% of the population. That boy is taller than 86% of other boys his age – that means he’s in the 86% percentile.

Percentiles is a relatively unlikely topic to see on the GRE, but if it does show up, here are a few handy facts to have up your sleeve.

A few details to clarify. The individual with the lowest value of the variable, with the minimum value, is not bigger than anyone, so the lowest percentile, the percentile of the rock-bottom minimum, is the 0th percentile. If my score is in the 0th percentile, then I am not higher than anyone.

What’s trickier is the maximum score. If my score is the highest score, I am higher than everybody else, but that’s ** not** the 100th percentile, because in order to be higher than 100% of the population, higher than everyone, I would have to have a score higher than my own score: a paradox! In fact, for this very reason, there’s no such thing as a 100th percentile. The person with the highest score is higher than everybody else, but not higher than herself, so she’s in the 99th percentile. If we are sticking with whole numbers, the 99th percentile is the highest possible percentile. If we go to decimals, we can get higher with the 99.9th percentile (1 out of a 1000), the 99.99th percentile (1 out of 10000), etc.

The median is the middle of a list: the median divides a list into an “upper half” and a “lower half.” This means, the median is higher than the lower half of the population, higher than 50%, so the median is the 50th percentile. Now, we have to be careful here. On a list with only three members — e.g. {2, 4, 7} — the median is the middle number, here 4, but that number is higher than only one number out of three — so 4 is the 33rd percentile of that list. In a technical sense, the median is not always the 50th percentile.

In some sense, though, that’s a specious objection. When there are only 3 members on a list, nobody in their right mind talks about percentiles. When the total number is less than a few hundred, there’s seldom talk of a percentile. Percentiles, by their very nature, are a way to make sense of tens of thousands, even millions of individuals. How many 8 year-old boys are there on Earth? Who knows, but it’s certainly a very very large number. That’s where percentiles are used in practice.

When the number of folks in the group is that large, then for all intents and purposes,the median is the 50th percentile. If you are familiar with the idea of quartiles, then the first quartile is the 25th percentile and the third quartile is the 75th percentile, again, when the group sizes are truly huge.

1) Sasha took a nationwide standardized test that is graded on a scale from 20 to 60. Sasha got one of the best scores recorded on that this test.

**Column A Column B**

Sasha’s score the percentile of Sasha’s score

(A) The quantity in Column A is greater.

(B) The quantity in Column B is greater.

(C) The two quantities are equal.

(D) The relationship cannot be determined from the information given.

2) Alice took nationwide standardize test that is graded on a scale from 0 to 100. Alice scored the highest score recorded on this test.

**Column A Column B**

Alice’s score the percentile of Alice’s score

(A) The quantity in Column A is greater.

(B) The quantity in Column B is greater.

(C) The two quantities are equal.

(D) The relationship cannot be determined from the information given.

3) A large distribution of score is normally distributed

**Column A**

score that’s one standard deviation above the mean

**Column B**

score that has the 80th percentile

(A) The quantity in Column A is greater.

(B) The quantity in Column B is greater.

(C) The two quantities are equal.

(D) The relationship cannot be determined from the information given.

(1) **B**; (2) **D**; (3) **A**;

1) We know that Sasha is near the top of the scoring distribution, so that would mean a score with a percentile close to the 99th percentile. Because of the scoring scale, the score is not going to be above 60, so the percentile is clearly bigger. Answer = **B**.

2) Alice got the highest score, so by definition, that’s the 99th percentile. What we don’t know is: how hard was this test? What score was the highest score? If it was a particularly challenging test, it could be that the highest score anyone achieved was only, say, a 73. In that case, the percentile would be greater. If, on the other hand, it was possible to get a perfect score, and Alice did in fact do that, then her score of a 100 would be greater than the percentile. We don’t have enough information to decide. Answer = **D**.

3) Here, it might be helpful to brush up on Normal Distribution. On a normal distribution, it’s always true that 68% of the populations lies within one standard deviation of the mean. That means, half of that, 34%, lie between the mean and one standard deviation above the mean. The score that is one standard deviation is higher than the 34% between the mean and one standard deviation above the mean, as well as than the 50% below the mean. That means, a score that lies one standard deviation above the mean is the 50 + 34 = 84th percentile. Thus, it’s higher than a score in the 80th percentile. Answer = **A**.

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]]>The post Quantitative Comparison: “The Relationship Cannot Be Determined from the Information Given” Answer Choice appeared first on Magoosh GRE Blog.

]]>- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

Many people dread choosing answer choice (D) on Quantitative Comparison (QC) Some feel it may be conceding defeat. Others think that the GRE is trying to trick them by making them pick (D). After all, they think, there must be some pattern that I’m not getting.

The truth is answer (D) comes up often. And to determine whether an answer cannot be determined is actually not too difficult.

#1 Determine a relationship

Say you find an instance, in which the answer is *(A) the information in column A is greater*. If that is the case, then the next step is to disprove that.

#2 Disprove that relationship

Meaning, see if you can come up with an instance, either through plugging in different variables, manipulating algebra, or manipulating a geometric figure, in which the answer is not (A). As soon you do that, you can stop. The answer is (D).

If you can’t disprove your answer, then it must be correct: it must be (A), (B) or (C).

1.

Column A | Column B |
---|---|

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

Question #1

After choosing a few numbers you should note something: that Column A will always be greater than Column B.

Why? Will, whenever, you have an even exponent, positive or negative, that exponent will always yield a positive number.

Odd exponents, on the other hand, give you a negative output if the base (the number below the exponent) is negative. Remember x has to be negative. So no matter what number you plug in Column A will always be positive, Column B negative. This is a definite not (D). The answer is (A).

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]]>The post GRE Perfect Math Score Challenge appeared first on Magoosh GRE Blog.

]]>The Revised GRE is a long, grueling test. To get a perfect score on the quantitative section (a 170) you will not only have to answer the difficult questions correctly, but you will also have to answer all of the easy ones. Nonetheless, I’ve focused on the difficult questions for the 170-challenge. I can’t promise a perfect correlation, but something tells me that if you can answer all five of these questions correctly in less than 10 minutes, you are well on your way to a perfect score.

Of course missing a few hardly precludes a perfect score. But make sure you learn from the ones you miss to make sure you don’t make similar mistakes in the future. And if time is a problem, remember the Magoosh GRE product has plently of questions to help you hone your chops—both from a pacing and a conceptual standpoint.

Good luck, and feel free to post your answers. Let’s see who the first perfect score will be!

1. If s and t are both primes, how many positive divisors of v are greater than 1, if v is an integer?

(A) two

(B) three

(C) five

(D) six

(E) eight

2. A quadrilateral has a perimeter of 16. Which of the following alone would provide sufficient information to determine the area of the quadrilateral. Choose ALL that apply.

[A] The quadrilateral contains equal sides

[B] The quadrilateral is formed by combining two isosceles right triangles

[C] Two pairs of congruent angles are in a 2:1 ratio

[D] The width is 4o% of the length and all angles are of equal measure

[E] If the perimeter was decreased by 50%, the area would decrease to 25% of the original

3. Product Question: Triangle in a Parabola

http://gre.magoosh.com/questions/2192

4. If x is an integer, and 169 < < 324, which of the following is the sum of all values of x?

(A) 61

(B) 62

(C) 75

(D) 93

(E) None of the above.

5. x = 350,000

y = 45,000

Column A | Column B |
---|---|

The total number of positive divisors of x | The total number of positive divisors of y |

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

[Note from our intern, Dylan: “I can do multivariable calculus, and got 2/5 on these in 10 minutes. These guys are tricky. :P”]

https://www.youtube.com/watch?v=BxkB8R7esB4

https://www.youtube.com/watch?v=jZdxqajkZtw

http://gre.magoosh.com/questions/2192

https://www.youtube.com/watch?v=pwzlclrvqWU

https://www.youtube.com/watch?v=–U8aDmLNzE

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]]>The post GRE Quantitative Comparison: The Devil is in the Details appeared first on Magoosh GRE Blog.

]]>Many quantitative comparison questions come with parameters. Parameters are basically a few ground rules that are listed above Column A and Column B.

x is a positive integer

Column A | Column B |
---|---|

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

In this question *x is a positive integer* is a parameter. The parameter is essential information. That is you must take into account if you want to get the correct answer.

The answer to the question above is (D). If x = 1, then the columns are equal. For all other values (B) is bigger.

To see how a simple tweak in a parameter can change a question have a look at a similar problem.

x is an integer greater than 1

Column A | Column B |
---|---|

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

Let’s now look at two questions that are very similar save for a word or two. Both questions are much more difficult so you may want to grab some scratch paper.

1. Isosceles right triangle ABC and Square EFGH have the same area.

Column A | Column B |
---|---|

Perimeter of ABC | Perimeter of EFGH |

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

2. Quadrilateral EFGH and right triangle ABC have the same perimeter

Column A | Column B |
---|---|

Area of ABC | Area of EFGH |

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

Hint: The answers are different.

In this question, we have an isosceles right triangle, aka a 45-45-90 triangle. This is fixed triangle—meaning you can’t endlessly manipulate it. Two sides are equal and it is a right triangle, so the sides must be in the ratio of x: x: x√2.

But let’s back up a moment. A square has the same area as our 45-45-90 triangle. At this point, I like to use a convenient number (but if you are predisposed to using ‘x’ then go ahead and do so). Let’s say the area of the square is 8. That means each side of the triangle is 4 (Area of triangle: (4)(4)/2 = 8).

The three sides of the triangle added together will be 4 + 4 + 4√2 = 8 + 4√2.

If the square has an area of 8, each side is equal to √8 = 2√2. The perimeter of the square is 4 x 2√2 = 8√2.

Now we can compare these two numbers:

Column A | Column B |
---|---|

Be very careful here! First off do not add 8 + 4√2. You cannot add an integer to a number expressed as a radical. One solution:

√2 = 1.4 (approximately)

4 x 1.4 = 5.6

5.6 + 8 = 13.6

8√2 = 8 x 1.4 = 11.2

Column A | Column B |
---|---|

Therefore (A).

Another solution:

8√2 = 4√2 + 4√2

4√2 + 8 vs. 4√2 + 4√2 (Subtract 4√2 from both sides)

8 vs. 4√2

(4)(2) vs (4)(√2) so (A).

A slight (but devilish) change in the details gives us question #2. Note that the square has changed to a quadrilateral and the triangle is now simply a right triangle. Does this change anything?

Imagine that quadrilateral is a square. Imagine that the triangle is an isosceles. We have the first question. How does this help us? Well, in the first question (A) was bigger. If we find an instance in which (A) is not bigger, then the answer is (D).

Imagine the quadrilateral is very skinny, let’s say length is 9 and width is 1. The area would only be 9, but the perimeter would be 20. A triangle with area nine could have sides 3 and 6. Just to equal 20, the triangle would have to have a hypotenuse of 11, which would be impossible: longest possible side of triangle with legs 3 and 6 is < 9.

And like that we’ve found a case where (B) is larger. So the answer is (D).

On Quantitative Comparison a simple word can change an entire answer.

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]]>The post Common GRE Math Mistakes appeared first on Magoosh GRE Blog.

]]>We all make mistakes. That being said, you should strive to never make the same mistake twice. If you understand some of the most common math mistakes, you can use your awareness to actively avoid falling into these traps.

Under timed conditions, you may feel compelled to rush. But remember, by misreading a word (or not reading it entirely), you can make a relatively straightforward problem seem intractable. You may flail about the answer choices, picking one—usually the incorrect one—that happens to be somewhat close to your answer.

Worse yet, you may get a numeric entry question and blithely enter in the wrong answer—something you could easily have avoided doing had you read the question carefully.

Before you begin attempting a solution, you need to know *exactly* what the question is asking you to solve. Sometimes, students will see that a variable like ‘x’ is involved, and they will immediately start solving for ‘x’. They hastily input an answer, only to realize later that the prompt was actually asking for the value of ‘1/x’.

Never start blindly solving a problem. Always acknowledge what you are specifically seeking.

Many common math mistakes result from forgetting something as minor as writing a negative sign. Other times, simple mathematical errors—like thinking that 16 x 5 = 90—can be very costly. Math is about precision, so use your prep time to become an efficient and unerring human calculator.

When presented with geometric diagrams, students often erroneously assume the figure accurately represents the information given by the prompt. However, there’s no guarantee that a figure will be drawn to scale on the GRE. If we see the following on the GRE:

Beware—although this figure *looks* like a square, there’s no information to prove it. We need to have data that irrefutably shows ∠P = ∠Q = ∠R = ∠S = 90º, and that PQ = QR = RS = PS to confirm this shape is a square. Otherwise, the shape above could actually be a rhombus, or a rectangle, or even an irregular quadrilateral. So be wary of diagrams—all assumptions need to be backed up by geometrical reasoning.

It’s important to keep your units straight, especially when handling work rate problems. For example:

- If a problem is asking you
*how long*a machine takes to make an order, make sure your answer reflects hours, or seconds, or some metric of time. - Or if a problem is asking about a
*rate*, then your answer needs to be in a form like widgets per hour, or widgets per second. - Sometimes, tricky problems will ask you to find the
*inverse rate*, which reverses the above: seconds per widget, or hours per widget.

In any case, you’ll want to read the question carefully and know exactly what you’re solving for.

This is one of the common math mistakes that can trip you up on the GRE test. Especially in Quantitative Comparison, you always want to make sure to plug in 0 and 1 if the constraints permit doing so. Oftentimes plugging in a 0 or 1 will prove the exception, thus making the answer (D).

Since this is a common GRE mistake, we’ve provided a practice problem so you can try this technique for yourself.

*x* is a non-negative integer.

Column A | Column B |
---|---|

2x^{2} | 3x^{3} |

A) The quantity in Column A is greater.

B) The quantity in Column B is greater.

C) The two quantities are equal.

D) The relationship cannot be determined from the information given.

Remember to plug in 0 and 1. If you don’t, you will choose the obvious—but actually incorrect—answer (B). However, if *x* is zero the two columns are equal. Because of this one instance, we cannot say for sure which side is bigger. Thus (D) is the correct answer.

Remember, 2 is the smallest prime number. It is the only *even* prime. 1 is NOT a prime.

There is a subtle, but important difference here. If a question is phrased “must be,” then the answer you choose must always hold true for the conditions stated in the problem. “Could be” means that the answer only holds true in certain instances (i.e., for certain numbers).

All of this makes a lot more sense when in the context of the problem. So, let’s take a look!

*c* and *d* are prime numbers. If *c – d* is an odd prime, then which of the following must be true?

A) *c*is even

B) *d* is odd

C) *c x d* is odd

D) *d* is even

E) *c x d – c* is even

First off, don’t let the variables throw you. There is an answer, so there must be some pattern that you have to discern.

If you remember, I mentioned that 2 is the only even prime. Thus, the rest are all odd. The question says that *c – d* is an odd prime. The only way to get an odd number when we subtract two numbers is that one number must be odd, and one must be even.

Since 2 is the only even prime, we know that 2 must be *d*. (*c* cannot equal ‘2’ because *c – d* would end up being a negative number, and primes can’t be negative.)

We don’t have to know what exact number *c* equals, as long as *c – d* equals an odd prime. *c = 5* is perfect. We plug in those values into the question.

Only (D) works as the correct answer. And we know that *d* must be even, because *d* must equal 2, an even number.

You’ll definitely make some common math mistakes as you study, and that’s okay. Any time an error comes up, treat it as an opportunity to correct a misguided approach. Keep an error log, and make it your goal to never repeat a mistake.

*Editor’s Note: This post was originally published in February 2012 and has been updated for freshness, accuracy, and comprehensiveness.*

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]]>Here is a quiz based on math concepts covered over the last month on the Magoosh GRE blog. See how many of the following five questions you can answer correctly in less than 8 minutes. Go!

1. Which of the following is a multiple of ?

(A) 36,045

(B) 43,623

(C) 52,224

(D) 89,892

(E) 92,044

2. To graduate John needs to complete eight courses. Of these eight, he must take only three science courses, only two math courses, and only one history course. If the college offers five science courses, six math courses, and four history courses, how many different class schedules can John have if the college offers a total of twenty courses?

(A) 150

(B) 600

(C) 3000

(D) 6000

(E) 12450

3.

Column A | Column B |

The sum of the multiples of 4 less than 100 | The sum of the multiples of 5 less than 100 |

(A) The quantity in Column A is greater

(B) The quantity in Column B is greater

(C) The two quantities are equal

(D) The relationship cannot be determined from the information given

4.

Column A | Column B |

The number of pentagons that can be inscribed in an octagon | The number of triangles that can be inscribed in an octagon |

(A) The quantity in Column A is greater

(B) The quantity in Column B is greater

(C) The two quantities are equal

(D) The relationship cannot be determined from the information given

5. A “trio-prime” is any number made up of three factors, not including itself and 1, that are consecutive primes. How many “trio-primes” are less than 1,000?

(A) 2

(B) 3

(C) 4

(D) 7

(E) 11

**Answers:**

** ****1. ****D**

**2. ****D**

**3. ****A**

**4. ****C**

**5. ****B**

** **

**Explanations:**

1. I never talked about divisibility rules of 36. No test prep strategy would ever address such a random concern. Based on what you know, you have to be able to figure out how to answer this question.

We know divisibility rules of 6. But, that doesn’t really help us here. We have to think about this problem from a different angle: 36 can also be broken down into 9 and 4. In a previous post, I covered divisibility rules for both of these numbers. To determine which one of the answer choices is divisible by 36, I have to make sure that it passes the divisibility test for 9 and 4.

The rule for 9 is, if the sum of the digits of the number is a multiple of 9. The rule for 4 is, the last two digits should be divisible by 4. Only answer (D) fits both these requirements:

8+9+8+9+2 = 36

92/4 = 23

2. We must use the combination formula here for the three different classes. For science, John has 5C3 = 10, for math 6C2 = 15, for history 4C1 = 4. We multiply these results together- 10 x 15 x 4 = 600. The twist is, John has two other courses to take. If the college offers a total of 20 courses, John already has chosen from a total of 5+6+4 = 15. So, he only has 5 classes left or 5C2 = 10. Therefore, he has 600 x 10 = 6000 possible schedules. Answer (D).

3. Your first instinct may be to think that 5 is greater than 4, so the sum must be greater. However, there are more multiples of 4 under 100 than there are multiples of 5. You can write out the first few multiples for each, and see that 4 is already bigger than 5. Extrapolating this all the way to the multiple closest to 100, you can see that the sum of the multiples of 4 will be much bigger than the sum of the multiples of 5.

4. This question may look intimidating, but it is actually very straightforward. When we are figuring out how many smaller shapes (i.e. shapes with fewer sides) can be inscribed in large shapes (i.e. shape with more sides), we only need to use the combinations formula. For Column A, we get 8C5; for Column B, we get 8C3. These two values are equal. Therefore, the answer is C.

5. The only trio-primes are 2 x 3 x 5 = 30, 3 x 5 x 7 = 105, and 5 x 7 x 11 = 385. Answer B. If you chose answer C, remember that 7 x 11 x 13 yields 1001, which is too great by a hair.

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