Tag Archive for 'GRE math'

Sample GRE Multiple Choice Math Problem – Way Too Many Variables

bond_villain

“I hope you’re not … under-prepared … Mr. Bond.”

On every GRE Math section, the test makers try to come up with a few extremely difficult problems that will leave even the cleverest students scratching their heads. The really evil part, though, is that even these problems can be solved in under a minute without a calculator – if you know what to do. This means that once you “figure out the trick,” these difficult problems become easy. So, while those test makers are busy cackling with sadistic glee, let’s see if we can’t beat them at their own game.

Consider the following problem:

In the correctly solved subtraction problems above, each of the seven letters represents a different single digit integer. What is the average (arithmetic mean) of DE and BG?

A) 32

B) 50

C) 68

D) 100

E) 137

DNJamesBond41

“Actually, I did take that Test Masters course.”

“How on Earth am I supposed to figure out the average of DE and BG if I have no idea what any of those seven numbers are?” Well, you could try the guess and check method, but that would waste a lot of time. If you think about it, you will realize that you don’t actually need to know what all seven integers are in order to solve this problem: you only need to figure out one.

What we have in the first subtraction problem is a three digit number that is the sum of two two digit numbers. A two digit number must be less than 100 by definition. 100 + 100 = 200, so the two two digit numbers added together must be less than 200, since each of them alone is less than 100. Thus, the three digit number must be between 99 and 200. Therefore, the letter A must be the number 1.

If A represents 1, then we can rewrite ABC as:

ABC = 100 + BC

The one other thing you need to realize is that we can get rid of DF. Write out each problem horizontally:

ABC – DE = DF

DF – BG = BC

In the second one, isolate DF:

DF = BC + BG

Now, it becomes clear that you can set the two equations equal to each other:

ABC – DE = DF = BC + BG

ABC – DE = BC + BG

Now substitute 100 + BC for ABC:

(100 + BC) – DE = BC +BG

100 + BC = BC + BG + DE

100 = BG + DE

The BC appears on both sides of the equation, so it cancels out, and you are left with the realization that the sum of BG and DE is 100. Finding their average is now elementary (as Sherlock Holmes might say):

Average of BG and DE = (BG + DE)/2 = 100/2 = 50

The answer must be 50, or choice B. You didn’t have to know the other variables at all. If you’re curious as to what they were, here is a solution (I believe the only solution) that works:

However, figuring that out would take way too much time on test day. If you know what to do, it takes only about 30 seconds to solve this problem. So you see, with practice, even the hardest problems on the GRE become easy. Check back here each week for more extra hard problems and the tricks you need to solve them! Also, remember that you can find out all the tricks from experts like me with a Test Masters course or private tutoring. Until  then, keep up the good work and happy studying!

Sample GRE Multiple Choice Math Problem – FOIL or Factor?

GRE test writers are always trying to find new ways to discombobulate students.

GRE test writers are always trying to find new ways to discombobulate students.

On every GRE Math section, the test makers try to come up with a few extremely difficult problems that will leave even the cleverest students scratching their heads. The really evil part, though, is that even these problems can be solved in under a minute without a calculator – if you know what to do. This means that once you “figure out the trick,” these difficult problems become easy. So, while those test makers are busy cackling with sadistic glee, let’s see if we can’t beat them at their own game.

Consider the following problem:

For all x and y where ,

A) 2(x – 2y)

B) 2y – x

C) 1

D) 0

E) -2

Well, as you know, you can’t add fractions unless their denominators are the same, so if you want to add these two fractions, then you would have to multiply the numerator and denominator of the first fraction by the denominator of the second fraction and the numerator and denominator of the second fraction by the denominator of the first fraction, like so:

So, the answer is E. But that’s the long way. By the time you get to the third line of all that awful math, you should realize that the numerator and denominator of each fraction are essentially the same, except that one has been multiplied by -1. In line 3 above, we factor out the -1 and realize that these fractions are just a glorified way of writing -1 + -1, but if we had stopped to think about it at the beginning, we could have realized it without multiplying all those binomials:

The canny student, however, can usually spot their tricks.

The canny student, however, can usually spot their tricks.

Granted, if you remember the FOIL method, multiplying the binomials shouldn’t take that long, but it’s still nice if you can do a 2 minute problem in 10 seconds, since that gives you more time for other questions you may need to think about. A method with fewer steps also leaves less room for careless errors. But how are you going to find these shortcuts on the day of the test? Ironically, if you are rushing through the test as fast as you can, you’re more likely to do problems the long way and miss the shortcuts. Before you dive into a problem and break out your calculator, pause and reflect: every GRE math problem can be solved in under a minute without a calculator. This means that for the harder problems, there is always some trick. If you ever find yourself scribbling down lines and lines of scratch work, that probably means that you’re doing a problem the long way. For this problem, just look at it. You will notice a lot of repetition: everything is made up of x and 2y. This is a red flag that there’s a trick to solving this problem. If everything is kind of the same, then if you arrange the expressions properly chances are things will start to cancel out. In this case, you have to factor out a -1 from either the numerators or the denominators. After that, it becomes clear that each fraction is equal to -1, and the problem becomes, as they say, easy as π!

So you see, with patience and practice, even the hardest problems on the GRE become easy. As you do more practice problems you will get better and better at spotting these shortcuts – the test makers tend to use the same tricks over and over again. Check back here each week for more extra hard problems and the tricks you need to solve them! Also, remember that you can find out all the tricks from experts like me with a Test Masters course or private tutoring. Until  then, keep up the good work and happy studying!

Sample GRE Multiple Choice Math Problem – Combinations

timcurrycardinalrichelieuOn every GRE Math section, the test makers try to come up with a few extremely difficult problems that will leave even the cleverest students scratching their heads. The really evil part, though, is that even these problems can be solved in under a minute without a calculator – if you know what to do. This means that once you “figure out the trick,” these difficult problems become easy. So, while those test makers are busy cackling with sadistic glee, let’s see if we can’t beat them at their own game.

Consider the following problem:

King Louis XIII must pick a team of 5 musketeers to investigate one of Cardinal Richelieu’s nefarious schemes. If there are 10 musketeers to choose from, what is the probability that four of them (Athos, Porthos, Aramis, and d’Artangan, of course) will be selected?

A) 1/2

B) 1/10

C) 3/5

D) 1/42

E) 5/252

To solve this problem, we must first remember that the probability of any event is calculated by taking the number of desired outcomes over the number of possible outcomes. In this case, figuring out the number of desired outcomes is not too difficult. We know who four of the five musketeers should be, so the only variable is the remaining musketeer. We have already used 4 out of the 10 possible musketeers, so there are 6 possibilities left for the remaining musketeer. If we let P, Q, R, S, T, and U represent the unknown musketeers, then we could represent the desired outcomes like so:

Athos, Porthos, Aramis, d’Artangan, and P

Athos, Porthos, Aramis, d’Artangan, and Q

Athos, Porthos, Aramis, d’Artangan, and R

Athos, Porthos, Aramis, d’Artangan, and S

Athos, Porthos, Aramis, d’Artangan, and T

Athos, Porthos, Aramis, d’Artangan, and U

That leaves figuring out the total number of possible outcomes. You could try to write down all the possible combinations of five musketeers, but with 10 musketeers to choose from that’s going to take a long time, and there would be many opportunities for making mistakes. What we are trying to figure out here is how many possible combinations of 5 musketeers we could make from a group of 10. To calculate this, all we need is a little formula that you might remember from math class:

Where n is the number of items to choose from and r is the number of items to be selected. Combinations and permutations are occasionally tested on the GRE, so you would do well to memorize this formula and other relevant formulas before test day. Using the formula, we find that the total number of ways to select a group of 5 from a group of 10 is:

Thus, the number of desired outcomes over the number of possible outcomes is:

the-three-musketeers-6-1Thus 1/42, choice D, is correct. If you know what to do, it takes only about 30 seconds to solve this problem. So you see, with practice, even the hardest problems on the GRE become easy. Check back here each week for more extra hard problems and the tricks you need to solve them! Also, remember that you can find out all the tricks from experts like me with a Test Masters course or private tutoring. Until  then, keep up the good work and happy studying!

GRE Multiple Choice Math Problem – Speed Trap

Do you know how fast you were going on that problem?

Do you know how fast you were going on that problem?

On every GRE Math section, the test makers try to come up with a few extremely difficult problems that will leave even the cleverest students scratching their heads. The really evil part, though, is that even these problems can be solved in under a minute without a calculator – if you know what to do. This means that once you “figure out the trick,” these difficult problems become easy. So, while those test makers are busy cackling with sadistic glee, let’s see if we can’t beat them at their own game.

Consider the following problem:

If x and y are integers and y < 20, for exactly how many ordered pairs (xy) will x^2 = y?

A) 4

B) 5

C) 7

D) 8

E) 9

This one actually doesn’t seem so bad, does it?

1^2 = 1

2^2 = 4

3^2 = 9

4^2 = 16

5^2 = 25 > 20

So we’ve got (1,1), (2,4), (3,9), and (4,16). Answer choice A, right? Not so fast! You forgot that the square of a negative number is also positive, so for every y, there are two x values: one positive and one negative. So really our list should look like this:

(1,1) and (-1,1)

(2,4) and (-2,4)

(3,9) and (-3,9)

(4,16) and (-4,16)

So the answer is D, right? Wrong again! There’s one last square you forgot:

0^2 = 0

Thus, there are in fact 9 pairs: the eight already mentioned, plus (0,0). Thus, the correct answer is actually choice E.

On the GRE, sometimes slow and steady does win the race.

On the GRE, sometimes slow and steady does win the race.

Was there actually anything hard about this question? Not really. However, if you were going fast and running out of time, you might have easily made one of the careless errors above. Note that 4 and 8 are traps set for students who see this problem, think it’s easy, and then blow through it too fast without thinking carefully (if you forgot the negatives but remembered 0, there’s also choice B, 5). If you get toward the end of a math section and see a problem that looks really easy, be careful – there’s probably more to it than meets the eye. Sometimes it’s just as bad to spend too little time on a problem as it is to spend too much, so make sure you don’t go too fast through any “easy” problems at the end of a math section.

Check back here each week for more problems and the tricks you need to solve them! Also, remember that you can find out all the tricks from experts like me with a Test Masters course or private tutoring. Until  then, keep up the good work and happy studying!

GRE Multiple Choice Math Problem – Prime Factorization

Math problems that appear hard but are secretly easy? Most excellent!

Math problems that appear hard but are secretly easy? Most excellent!

On every GRE Math section, the test makers try to come up with a few extremely difficult problems that will leave even the cleverest students scratching their heads. The really evil part, though, is that even these problems can be solved in under a minute without a calculator – if you know what to do. This means that once you “figure out the trick,” these difficult problems become easy. So, while those test makers are busy cackling with sadistic glee, let’s see if we can’t beat them at their own game.

Consider the following brain teaser:

What is the least common multiple of 18 and 14 that is also a perfect square?

A simple enough question. But how do you figure it out? Could we just multiply 18 by 14?

18*14 = 252

That’s definitely a common multiple of 18 and 14, but is it a perfect square?

252^(1/2) = 15.87450787…

Should’ve known it wouldn’t be that easy. This IS a hard problem after all. Maybe we should multiply 18^2 by 14^2:

(18^2)*(14^2) = 63504

Well, that works, but is it the LEAST common multiple? Don’t look at me! How should I know? Hmm…something tells me it probably isn’t. Maybe we could just go through the multiples of 18 trial and error style?

1*18 = 18. Not divisible by 14, not a perfect square.

2*18 = 36. It is a perfect square, but not divisible by 14.

3*18 = 54. Not divisible by 14, not a perfect square.

4*18 =

Time is running out, my pretty!

Time is running out, my pretty!

This is clearly going to take too long. We’re running out of time on this section – whatever shall we do? Never fear – prime factorization is here! Remember, all GRE math problems can be solved in under a minute without a calculator. If you know what to do, hard problems like this one become fast and easy. In this case, we should begin by finding the prime factors of both 18 and 14:

18 = 9*2 = 3*3*2

14 = 7*2

Remember, to find the prime factors of a number means to write out the numbers as products of only prime numbers like 2, 3, 5, 7, 11, 13, etc. In this case, the prime factors of 18 are 3, 3, and 2, and the prime factors of 14 are just 7 and 2. Now, if we just wanted to find the least common multiple of 18 and 14, we would multiply all the prime factors of 18 by all the prime factors of 14, but leave out any duplicates that appear in both lists. In this case you would multiply:

3*3*2*7 = 126

This works, because

(3*3*2)*7 = (18)*7 = 126

3*3*(2*7) = 3*3*(14) = 126

Note that you need only one 2 in 3*3*2*7, since both the prime factors of 18 (3*3*2) and 14 (7*2) include a 2. This is all well and good, except that 126 is not a perfect square:

126^(1/2) = 11.22497216…

So, what are we to do? Note that a perfect square multiplied by a perfect square is also a perfect square:

4*25 = 100

This means that we can pair up all of the unique prime factors of 126 so that we have a bunch of squares multiplied by each other. Thus,

3*3*2*7 = 126

becomes

(3*3)*(2*2)*(7*7) = 1764

We have to add an extra 2 and an extra 7 in order to make sure those numbers are part of perfect square pairs (we don’t need to add any extra 3s because there are already two of them). Sure enough, 1764 is a perfect square:

1764^(1/2) = 42

and 1764 is a multiple of both 18 and 14:

1764/18 = 98

1764/14 = 126

But, how can we be 100% sure that 1764 is the least common multiple? Well, think about it. Every perfect square that isn’t the square of a prime number will always be prime factorized into pairs like the ones above ((3*3)*(2*2)*(7*7) = 1764). Consider the following examples:

16 = 4*4 = (2*2)*(2*2)

36 = 6*6 = (3*2)*(3*2) = (3*3)*(2*2)

144 = 12*12 = (3*2*2)*(3*2*2) = (3*3)*(2*2)*(2*2)

You may have figured out this problem, but I'll get you next time! Next time!

You may have figured out this problem, but I’ll get you next time! Next time!

 

Et cetera. It’s inevitable. If the prime factors of a perfect square aren’t two prime numbers, then they will consist of multiple pairs of other prime numbers. Now, is there any way we could remove any pairs or make any of these factors smaller without breaking the conditions of finding the least common multiple of 18 and 14 that is also a perfect square?

(3*3)*(2*2)*(7*7) = 1764

I don’t think so. Remove or change one number and the product will either no longer be a perfect square or no longer be a multiple of both 18 and 14. Thus, 1764 must be the least common multiple of both 18 and 14 that is also a perfect square.

To sum up, the “trick” to solving a problem like this is to:

-find the prime factors of the two numbers in question

-pair them up

-multiply

If you know what to do, it takes about 30 seconds to solve this problem. So you see, with practice, even the hardest problems on the GRE become easy. Check back here each week for more extra hard problems and the tricks you need to solve them! Also, remember that you can find out all the tricks from experts like me with a Test Masters course or private tutoring. Until  then, keep up the good work and happy studying!

Quantitiative Comparison Angles of a Triangle

The Quantitative Comparison question type on the GRE can be very challenging. Essentially, you are given some information (usually in the form of a sentence, equation, or picture) and then two quantities. You are then asked to determine:

A if the quantity in Column A is greater;

B if the quantity in Column B is greater;

C if the two quantities are equal;

D is the relationship cannot be determined from the information given.

Like most questions on the GRE Quantitative section, the concepts involved in solving this question type are not of themselves very advanced, you just have to be careful in their application. Let’s go through an example problem:

First, look at the triangle formed by the points T, U, and V. We are given two angles of this triangle. Remember that the sum of the angles for a triangle will always equal 180°. This means we can subtract the measures of the two marked angles from 180° to find the third angle.

180° – 120° – 25° = 35°

This means that the angle at point U has a measure of 35°.

Now, the key to this problem is that the triangle formed by points R, S, and V has the same angle measurements as the triangle formed by points T, U, and V. The angle at point S is equal to the angle at point U. These two triangles are what are known as similar triangles. Similar triangles are triangles that have equal angles, and whose sides are proportional to each other. In other words, similar triangles will have the same shape, though not necessarily be the same size.

It may seem obvious from the picture that these triangles are similar, but how can we be certain? Generally, you will not want to only rely on the shape of the figure in the picture. Figures on the exam may not be drawn to scale.

We can be certain because of a rule of triangles known as the Side-Angle-Side rule. Here is the rule:

  • First, the two triangles must have an equal angle.
  • Next, the sides that form the angle on one triangle are equal to the sides that form the angle on the other triangle. Either this, or there is a common ratio between the sides of the two triangles.
  • If these two conditions are true then the triangles are similar.

In this case, the two triangles share the 120° angle at point V. Also, the sides forming that angle are in proportion to each other. Notice that line segment RT is equal to segment TV. This means that line segment RV is exactly twice the length of line segment TV. Since line segments VU and US are equal to each other, line segment SV is exactly twice the length of line segment VU. These two triangles pass the Side-Angle-Side rule and must be similar.

Since the triangles are similar, the angle at point S will be equal to the angle at point U. This means that the value of x is 35°. This is greater than 30, which means that the answer is A.

Quantitative Comparison Example Problem

The Quantitative Comparison question type on the GRE can be very challenging. Essentially, you are given some information (usually in the form of a sentence, equation, or picture) and then two quantities. You are then asked to determine:

A if the quantity in Column A is greater;

B if the quantity in Column B is greater;

C if the two quantities are equal;

D if the relationship cannot be determined from the information given.

Like most questions on the GRE Quantitative section, the concepts involved in solving this question type are not of themselves very advanced, you just have to be careful in their application. In fact, sometimes there is little or no need at all to do any “real” math; let’s take a look at the following GRE Quantitative Comparison Example Problem:

With geometry problems, it always helps to start with a formula. In this case, the problem is involving the volume of a cylinder. The formula for this is pi times the radius squared times the height.

Volume of a cylinder = πr2h

Therefore, to find the volume of a cylinder you must find both the radius of the cylinder and the height.

In this problem, two cylinders are given. Both cylinders have a different radius, and so it might be assumed that they must have a different volume. However, no information about the height is given. It is possible that the cylinder with the smaller radius has a larger height and might be the larger cylinder. Since there is no way of knowing how these cylinders compare to each other, the answer is D.

Test Masters offers the most comprehensive and successful GRE course available; every Test Masters GRE course, whether it is online or in-class, comes with a 10 point Score Increase Guarantee.

Quantitiative Comparison Word Problem

The Quantitative Comparison question type on the GRE can be very challenging. Essentially, you are given some information (usually in the form of a sentence, equation, or picture) and then two quantities. You are then asked to determine:

A if the quantity in Column A is greater;

B if the quantity in Column B is greater;

C if the two quantities are equal;

D is the relationship cannot be determined from the information given.

Like most questions on the GRE Quantitative section, the concepts involved in solving this question type are not of themselves very advanced, you just have to be careful in their application. Let’s go through an example problem:

Word problems require you to translate words into math. You will want to learn key phrases that often come up and their math equivalents. With this problem, we will rewrite the phrase “Steve purchased 3 more than half as many pencils as Joanne” into math. To start, Let the variable S mean the number of pencils Steve purchased.

“Steve purchased 3 more than half as many pencils as Joanne”

S = “3 more than half as many pencils as Joanne”

The phrase “more than” means addition.  The value of 3 is added to “half as many pencils as Joanne.”

S = “3 more than half as many pencils as Joanne”

S  = 3 + “half as many pencils as Joanne”

“Half as many” means half, or divided by two. Remember that the number of pencils purchased by Joanne is represented by p

S  = 3 + “half as many pencils as Joanne”

S  = 3 + “half of p

S  = 3 + (p/2)

Now, simplify by combining 3 and (p/2). Remember that to add fractions, you need a common denominator.

The number of pencils Steve purchased, or S, is the same as .  Therefore, the two values in the columns are equal and the answer is C.

Test Masters offers the most comprehensive and successful GRE course available; every Test Masters GRE course, whether it is online or in-class, comes with a 10 point Score Increase Guarantee.