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 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 GRE Math Multi-subject Challenge Question appeared first on Magoosh GRE Blog.

]]>Once in a while it is fun to take a concept in an official GRE question and run with it. The question below is inspired by a question from the 2^{nd} edition Official Guide. That is not to say it is that similar; I have only extracted a couple of concepts from that problem.

I’ve also made my question far more difficult. The first question was already difficult enough, so it is safe to say that a question of this difficulty would never appear on the GRE (though it could appear on the GMAT). Another of way expressing this: only attempt this question if you are already very strong in Quant. If not, just have fun with it, remembering that questions on the actual GRE will be easier.

1. A semicircle with area of xπ is marked by seven points equally spaced along the half arc of the semicircle, such that two of the seven points form the endpoints of the diameter. What is the probability of forming a triangle with an area less than x from the total number of triangles formed by combining two of the seven points and the center of the diameter?

- 4/5
- 6/7
- 17/21
- 19/21
- 31/35

Again this is a very difficult question, one that requires many steps, and one that has numerous twists. The question is without a doubt expansive, covering in its sweep combinatorics, probability, basic geometry, advanced triangle theory, counting properties and much more. So hang on for the ride!

In order to deal with this whole area of the triangle business, we first have to figure out the area of the circle. That way, we can also derive the radius, a number that be instrumental in helping us in Step #3.

First though we find the area of the circle, which is twice that of the semi-circle . To solve for the radius, we use .

So we have the radius. But how does that help in figuring out the resulting area of any of the inscribed triangles? Well, remember the seven points? If two points are different sides of the diameter and the other five points are evenly spaced along the arc, then the semicircle is broken into equal sixths. Therefore the central angle of each will be 30 degrees (the arc of a semicircle corresponds to 180 degrees).

** **

Using the information above, we can start to play around with triangles, keeping in mind that any two points along the arc will form radii with the center of the diameter (remember that the original question asked us to make triangles using the middle of the diameter).

A good idea is to come up with a triangle that both meets the criteria and is easy to solve area-wise. That way you can see how close you are to an area of x. For instance, you could assume that the triangle is an equilateral, the two points on the arc a distance of 60 degrees away (and thus the central angle will be 60).

A radius of will form the sides of the equilateral. Using the formula for the area of the equilateral triangle, , where s equals the side of the triangle, or in this case the radius, we get . Because equals approximately 1.7, we can see that such an equilateral triangle has an area less than x.

But don’t give up just yet. In trying to increase the area you should keep in mind that skinnier triangles, say with points on the opposite side of the semicircle, will not have an area greater than that of an equilateral. So we want to increase the angle but not too much.

The next most sensible triangle—that is one in which we can use simple formulas to determine the area—is a 45:45:90 triangle, which means that the two points on the semicircle will have to be 90 degrees apart (that way the central angle will be 90 degrees. And with two equal sides (remember all radii are equal), we know have a 45:45:90 triangle with sides of . Using the formula for the area of a triangle, we get .

An important insight: if we bring the points together, we end up with an equilateral, which we know to have a smaller area. Therefore, moving the points on the arc together results in a triangle with less area. Moving the points apart makes the triangle skinnier, something that also results in a triangle with less area.

Therefore, the isosceles right triangle represents the largest possible triangle, area-wise, that you can inscribe given the conditions of the problem (pretty advanced stuff, right?).

Voila! Finally, we have it. The only possible triangle to yield x will be an isosceles right triangle, or a 45-45-90, with a central angle of 90. If we find the number of these triangles and subtract that from 1, we have the probability of choosing a triangle with an area less than 1.

The total number of isosceles right triangle is four, which we can find by choosing points along the circle that are 90 arc degrees apart.

Next we want to find the total number of triangles. Because one point is already fixed (the center of the diameter), we have to determine how many triangles can be formed by using two of the seven points. The ordering of the vertices is not important. Meaning that a triangle with points A, B, and C, is the same as triangle with points B, C, A. Therefore we use 7C2 = 21.

Just when you thought we were finally done with this monstrosity of the problem, there is one final, and subtle twist. We have to discount one of these twenty-one triangles because it is actually not a triangle. If we choose both end points of the diameter and the center of the diameter, we have a straight line (this line is the diameter of the semicircle).

Therefore, there are twenty triangles, sixteen of which have an area less than x. so the answer is (A) 4/5.

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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 Data Interpretation Strategies appeared first on Magoosh GRE Blog.

]]>If I had to count the number of questions/concerns regarding combinations or permutations, I’d have to use some pretty mighty factorials to do so. On the other hand, if I had to count the number of times students have expressed the same misgivings regarding Data Interpretation, I’d simply use my hand.

What’s truly surprising is there may be one combination/permutation question or probability question on the test. That’s right—for all those hours students slave away trying to tease out the nuances of these question types, not a single one of these concepts could be covered test day.

What is 100% sure, however, is that you will see a Data Interpretation question. Indeed, over the two math sections you can see as many as eight questions. So make sure you spend a lot of time practicing this question. Below are some helpful tips to get you started.

All of sudden, you are tossed into an unruly world of facts and figures, bars and charts, and pies and lines. The first thing to do—before throwing up your hands in defeat—is to get your bearings. If you see bar graphs, read what is to the side of the bar graphs (the y-axis). Then read the information below the bars. Now you should know what each bar stands for and what it means for one bar to be higher than the other.

At the same time, do not scrutinize the graph. You want to get a sense of the big picture. Once you’ve done so move on to the question. Answering the question will allow you to focus on the relevant information in the passage. Knowing the big picture will allow you to correctly interpret this information.

If you have waded through all the information, determining that you have to find the total dollar amount of exports for Company B, you may be stymied yet. Often ETS gives us unpleasant numbers, like 149,000 and 41%. In fact, we can say the first figure is the dollar amount in thousands of the exports of Companies A – E, and that 39% is the percent of that share attributed to Company B.

Do not whip out the calculator to perform the tedious calculation. Instead, round up 149,000 to 150,000 and round 41% down to 40%. Quick math should give you 60,000. Then find an answer close to 60,000. If in the off chance that there are happen to be two answers very close to 60,000, then use the calculator. However, there will most likely be only one answer close to 60,000.

Sometimes, a graph may not give you an exact dollar amount. Instead, you may only have a bar or a line. Unlike Quantitative Comparison, everything you see in Data Interpretation is drawn according to scale. So if they are asking you for the difference between two bars or points on lines, then use your eyeball. It will save you a lot of time.

Almost every Data Interpretation Set will try to catch you mixing up percent with the actual total. To illustrate: Let’s say that Company X has increased by 40% from 1998 to 2000. Company Y has decreased by 30%. Without knowing the total amount of either company, we cannot say that Company X has more than Company Y.

Don’t just read the tips above—apply them. Hundreds of Data Interpretation questions abound, from those found in Magoosh to those created by ETS.

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]]>The post Compound Interest vs. Simple Interest on the GRE appeared first on Magoosh GRE Blog.

]]>For some this is the most “interest”ing concept on the Revised GRE (at least for those with a fondness for bad puns). For most, compound vs simple section can be a nuisance. Many think, what is the difference between the two, and/or how did that formula go again?

But remember, this concept involves money, and for many that means its practical (especially if you invest money yourself). But enough rambling…

Principal: The Amount of Money initially invested

Interest Rate: The amount return on an investment expressed as a percent of the principal.

Time: The length of time in which a principal is invested

1. John invests 100 dollars in account that yields 8% simple interest annually. How much money will John have in his account after one year?

(A) $4

(B) $8

(C) $104

(D) $108

(E) $110

2. Bob invests 100 dollars in a fund that yields 15% simple interest monthly. If Bob invests the principal in the middle of January, which is the first month will he have more than $200 total?

(A) June

(B) July

(C) August

(D) February

(E) March

3. In 2001, John invests x dollars in a special account that yields y% simple interest annually. If he has $250 in his account in 2006 and in 2008 he has $270 in his account, what is x + y?

(A) 5

(B) 25

(C) 200

(D) 205

(E) 210

Answers:

1. D

2. C

3. D

Okay, that was the easy part. Now for compound interest. In compound interest things become complicated. We no longer have a nice, clean linear increase. To illustrate:

If Mike invests $100 at 10% simple annual interest, he will have $110. After two years he will have $120. That is his money grows by $10 every year. After 10 years, Mike will have doubled his money.

Now, let’s say Mike’s friend Thomas invests $100 at a 10% rate that is compounded annually. After one year, Thomas will have made the same amount as Mike. But then things start diverging. Remember how Mike always gets 10% of the original 100 (the original 100 is called the principal)? Well, Thomas – because things are getting compounded annually – gets 10% of whatever the value of the account is at the end of the year. Let’s see how this plays out over time.

1^{st} year: 10% of 100 = 110

2^{nd} year: 10% of 110 = 121

3^{rd} year: 10% of 121 = 133.10

4^{th} year: 10% of 133.00 (rounding down) = 146.30…

After 10 years, Thomas will have made $260, which is $60 more than Mike.

Okay, that may all seem like chump change, but the same percent increase applies to numbers with a few more zeroes thrown in. How would $260,000 vs. 160,000 sound?

Of course the point of this lesson is to understand the conceptual difference between the two forms of interest—and not to have you running to the nearest ban, since the numbers above are very generous.

Now for the fun part: Notice how, in the case of Thomas, I seemed to be doing mathematical wizardry. After all, how did I know that 10% compounded annually at 10 years is going to yield 160% of the principal? Well, let’s meet the formula:

V = Total Value

P = Principal

r = annual interest rate

n = number of times per year invested

t = number of years

Pretty unpleasant, no? Well, let’s try to put the formula to the test. And you may want to get your calculators out (this is the Revised GRE after all!)

If $10,000 is invested at 10%, compounded semi-annually, how much will the investment be worth after 18 months?

(A) 11,500

(B) 11,505

(C) 11,576.25

(D) 11,625.30

(E) 12,000.50

Now don’t worry about the semi-annual bit—it just means twice a year. And remember the *n* from the scary little formula above: the number of times per year invested. And that 18 months? That corresponds to t, the number of years, which translates to 1.5.

.

That was easy—once you know where to put everything (and provided you remember the formula)!

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]]>The post GRE Exponents: Practice Question Set appeared first on Magoosh GRE Blog.

]]>Each of the math questions below is directly inspired by a question in the on-line Revised GRE test. I’ve provided an easier version of the question (#1) and a more difficult version of the question (#2).

My recommendation is to try the easier version first. Then, if you answer it correctly, click on the link, and take a stab at the actual Revised GRE question.

If you are able to answer that question correctly, then as prize – you get a fiendishly difficult question (#3). Okay, maybe that’s not a prize – but it is great practice for those aiming for the 90% on quant.

The good news is I have explanations. For the Revised GRE question, I have recorded an explanation video you can watch. Finally, it is a good idea to try the easy question before the medium one, and the medium question before the difficult one.

Good luck!

If , where n is a non-negative integer, what is the greatest value of ?

- ½
- 1
- 5
- 32
- 64

**Explanation:** Don’t think big – think small. That is the smaller n becomes the greater ½^n becomes. So what is the smallest value? You may be tempted to say 1, which would give us ½. But remember n = 0, because . Therefore **Answer: B.**

The “hidden zero,” as I like to call it, is a classic GRE math trick. So always keep your eyes open, especially when you see “non-negative integer,” which includes zero.

(5/4)^{-n} < 16^{-1}

What is the least integer value of n?

**Explanation:**

The best place to start here is by getting rid of the unseemly negative signs and translating the equation as follows:

(4/5)^{n} < 1/16

A good little trick to learn using 4/5 taken to some power is that (4/5)^{3} = 64/125, which is slightly — but only slightly — greater than ½. Therefore, we can translate (4/5)^{3} to ½.

(1/2)^{4} = 1/16

That would make (4/5)^{12} a tad larger than 1/16. To make it less than 1/16 we would multiply by the final 4/5, giving us n = 13.

The equation is true for how many unique integer values of n, where n is a prime number?

- 7
- 4
- 2
- 1
- None of the above

This problem can be difficult, indeed downright inscrutable, unless you take your time and process one piece of information at a time. Once you understand what the problem is saying, you should be able to solve the question relatively quickly.

**Explanation: **

The most important piece of info is n is a prime number. So do not start by plugging in zero or one. Neither is a prime. The lowest prime is 2. When we plug in ‘2’ we get:

This is clearly true. Thus we have one instance.

As soon as we plug in other prime numbers a pattern emerges.

is always a negative number if n is odd. Because all of the primes greater than 2 are odd, the number in the middle will always be negative:

Because in each case n is a positive number we can never have the middle of the dual inequality be positive, if n is an odd prime.

Thus the only instance in which the inequality holds true is if we plug in ‘2’, the answer is (D).

If you got that right – congratulate yourself. It’s a toughie.

For even more GRE questions, check out our GRE Quant problems with answers and explanations! And to find out where exponents sit in the “big picture” of GRE Quant, and what other Quant concepts you should study, check out our post entitled: What Kind of Math is on the GRE? Breakdown of Quant Concepts by Frequency.

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]]>The post Systems of Equations on the GRE appeared first on Magoosh GRE Blog.

]]>How would you solve the system of linear equations below?

** x – y = 5**

** 2 x + y = 13**

There are two primary approaches for solving systems of linear equations:

1) Substitution Method

2) Elimination Method

With this method, we take one of the equations and solve for a certain variable. For example, we might take *x* – *y* = 5 and add *y* to both sides to get *x *= *y* + 5.

Then we take the second equation (2*x* + *y* = 13) and replace *x *with *y* + 5 to get: 2(*y* + 5)+ *y* = 13

From here, we have an equation we can solve for *y*: *y* = 1

Now that we know the value of *y*, we can take one of the equations and replace *y* with 1 to find the value of *x*: *x *= 6

So, the solution is* x *= 6 and *y* = 1.

With this method, we notice that, if we add the two original equations (*x* – *y* = 5 and 2*x* + *y* = 13), the *y*’s cancel out (i.e., they are eliminated), leaving us with: 3*x* = 18.

From here, when we divide both sides by 3, we get: *x* = 6, and from here we can find the value of *y*: *y *= 1.

Okay, so that’s how the two methods work. What’s my point?

The point I want to make is that, although both methods get the job done, the Elimination method is superior to the Substitution method. And by “superior,” I mean “faster.”

First, the Elimination method can often help us avoid using fractions. Consider this system:

** 5 x – 2y = 7**

** 3 x + 2y = 17**

To use the Substitution method here, we’d have to deal with messy fractions. For example, if we take the equation 5*x* – 2*y* = 7 and solve for *x*, we get *x* = (2/5)*y* + 7/5. Then when we take the second equation (3*x* + 2*y* = 17) and replace *x* with (2/5)*y* + 7/5, we get: 3[(2/5)*y* + 7/5] + 2*y* = 17. Yikes!!

Alternatively, we can use the Elimination method and add the two original equations (5*x* – 2*y* = 7 and 3*x* + 2*y* = 17). When we do this, the *y*’s cancel, leaving us with: 8*x* = 24, which means *x* = 3. No messy fractions.

It has been my experience that many students rely solely on the Substitution method to solve systems of equations, and this can potentially eat up a lot of time on test day. So, be sure to learn the Elimination method soon. In fact, if I were you, I’d drop the Substitution method from my repertoire; it isn’t very useful.

If you still believe that the Substitution method is just as good as the Elimination method, try solving this question using the Substitution method.

**If 5x – 8y = 11, and 4x – 9y = 4, what is the value of x + y?**

**3****4****5****6****7**

To view the solution to the above question, see my post, The Reasonable Test-Maker.

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]]>The post Three GRE Challenge Combinations and Permutations Problems appeared first on Magoosh GRE Blog.

]]>*These challenge questions are great for extra practice. However, be warned: while I provide the answers, I don’t provide the explanations. If you’d like to try practice problems with explanations, check out Magoosh’s 15 GRE math practice problems and my biweekly Brain Twisters!*

Over the last few weeks, I have gone through GRE combinations and permutations. My goal is to make this formidable concept less intimidating. If you’ve watched the videos and feel comfortable with the math, I would say continue practicing but do not worry too much about this problem type. You will probably see one, though you will unlikely see two, on the actual test. So your time will better be spent on other concepts, such as work rate, sets, series, and coordinate geometry. (I will be covering these concepts soon!). Unless, that is, you are aiming for the 160+ category/the top 90% (it’s great that I can finally speak of the Revised GRE in terms of the Revised GRE scoring). The combinations and permutations I covered, while tough, may not be as difficult as a question you may see on the test, if ETS decides to make one of its most difficult questions a combinations and permutations one. Thus I have decided to devise some pretty diabolical questions. In fact, the last two are so convoluted and tricky that they wouldn’t even show up on the test. Again, this is only for those who are already comfortable with combinations and permutations, and who are going for the near perfect score. Otherwise, this could be a waste of your time (of course the less practical side of me thinks that this level of problems is fun!).

**Warming Up** An equal number of juniors and seniors are trying out for six spots on the university debating team. If the team must consist of at least four seniors, then how many different possible debating teams can result if five juniors try out?

(A) 50 (B) 55 (C) 75 (D) 100 (E) 250

**Difficulty Level: Hard** A bibliophile plans to put a total of seven books on her marble shelf. She can choose these seven books from a mixture of works from Antiquity and works on Post-modernism, of which there are seven each. If the shelf must contain at least four works from Antiquity, and one on Post-Modernism, then how many ways can he select seven books to go on the shelf?

(A) 441 (B) 1225 (C) 1666 (D) 1715 (E) 1820

**Difficulty Level: Are you kidding me?!?** An artist is planning on mixing together any number of different colors from her palette. A mixture results as long as the artist combines at least two colors. If the number of possible mixtures is less than 500, what is the greatest number of colors the artist could have in her palette? (A) 8 (B) 9 (C) 11 (D) 12 (E) 13

**Check out the answers:** 1. B 2. D 3. A

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]]>The post Strategies for Difficult GRE Questions appeared first on Magoosh GRE Blog.

]]>Below is a challenging GRE question. However, instead of just having you take a blind hack at it, I want to provide a little bit of context.

First off, if you sense that a question is very difficult on the test – be it a verbal or quantitative one – then skip it. Because the new GRE allows you to skip, you should take advantage of this option when you encounter a question that you either have no idea how to approach, or sense that you can probably get the answer, but doing so will be time consuming.

The good news is, each question is worth the same number of points, so even if you blindly guess on a really difficult question, the worst that can happen is you guess incorrectly, and simply only lose one point. The worst that can happen is, you become so caught up in solving one problem that you miss attacking any number of easy questions that you could have answered correctly, had you not burned your precious minutes on the difficult question.

The key is learning to gauge when a hard problem is truly a difficult one, and when you are simply stressed because of the nature of the situation (you are taking the GRE, after all). The best way to prepare for this is by doing practice problems, and practice tests, so you get a sense of where it is advantageous to skip, and where it is better to see a problem through till the end (usually, if you spend a minute staring at a problem without any noticeable progress, it is time to move on).

Now that I’ve finished prefacing challenging questions, try the following one. And, in this case, don’t simply give up after a minute – this isn’t the real test, so you can spend a few minutes. The important part is how you approach the problem. Remember, taking an approach and following through on it is better than simply staring at the screen, and rereading the problem over and over again. Good luck!

*A “descending number” is a three-digit number, such that the units digit is less than the tens digits and the tens digit is less than the hundreds digit. What is the probability that a three-digit number chosen at random is a “descending number”?*

*(A) **3/25*

*(B) **2/9*

*(C) **2/15*

*(D) **1/9*

*(E) **1/10*

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