Cracking the GRE with 4 Practice Tests, 2016 Edition
Part III. How to Crack the Math Section
Chapter 14. Math Et Cetera
There are a few more math topics that may appear on the GRE that don’t fit nicely into the preceding chapters. This chapter looks at some of these leftover topics, including probability, permutations and combinations, and factorials. The topics in this chapter are not essential to your GRE Math score, because these areas are not tested as frequently as the topics detailed earlier. However, if you feel confident with the previous math topics, and you’re looking to maximize your GRE Math score, this chapter will show you all you need to know to tackle these more obscure GRE problems.
OTHER MATH TOPICS
The bulk of the GRE Math section tests your knowledge of fundamentals, basic algebra, and geometry. However, there are a few other topics that may appear. These “et cetera” concepts usually show up only once or twice per test (although at higher scoring levels they may appear more frequently) and often cause anxiety among test takers. Many test takers worry excessively about probability problems, for example, even though knowledge of more familiar topics such as fractions and percents will be far more important in determining your GRE math score. So tackle these problems only after you’ve mastered the rest. If you find these concepts more difficult, don’t worry—they won’t make or break your GRE score.
These topics show up
rarely on the GRE, but if
you’re going for a very
high score, they are useful
to know.
PROBABILITY
If you flip a coin, what’s the probability that it will land heads up? The probability is equal to one out of two, or . What is the probability that it won’t land heads up? Again, one out of two, or . If you flip a coin nine times, what’s the probability that the coin will land on heads on the tenth flip? Still 1 out of 2, or . Previous flips do not affect the outcome of the current coin flip.
You can think of probability as just another type of fraction. Probabilities express a special relationship, namely the chance of a certain outcome occurring. In a probability fraction, the denominator is the total number of possible outcomes that may occur, while the numerator is the number of outcomes that would satisfy the criteria. For example, if you have 10 shirts and 3 of them are black, the probability of selecting a black shirt from your closet without looking is .
Think of probability in terms of fractions:
· If it is impossible for something to happen—if no outcomes satisfy the criteria—then the numerator of the probability fraction is 0 and the probability is equal to 0.
· If something is certain to happen—if all possible outcomes satisfy the criteria—then the numerator and denominator of the fraction are equal and the probability is equal to 1.
· If it is possible for something to occur, but it will not definitely occur, then the probability of it occurring is between 0 and 1.
Since probability is
expressed as a fraction, it
can also be expressed as
a decimal or a percentage.
A probability of one half is
equivalent to a probability
of 0.5 or 50%.
Let’s see how it works.
At a meeting of 375 members of a neighborhood association, of the participants have lived in the community for less than 5 years and of the attendees have lived in the neighborhood for at least 10 years. If a member of the meeting is selected at random, what is the probability that the person has lived in the neighborhood for at least 5 years but less than 10 years?
Here’s How to Crack It
In order to solve this problem, we need to put together our probability fraction. The denominator of our fraction is going to be 375, the total number of people from which we are selecting. Next we need to figure out how many attendees satisfy the criteria of having lived in the neighborhood for more than 5 years but fewer than 10 years.
What number goes on the
bottom of the probability
fraction?
First, we know that of the participants have lived in the neighborhood for less than 5 years. of 375 is 75 people, so we can take them out of the running. Also, of the attendees have lived in the neighborhood for at least 10 years. of 375 (be careful not to use 300 as the total!) is 250, so we can also remove them from consideration. Thus, if 75 people have lived in the neighborhood for less than 5 years and 250 have lived for at least 10, the remaining people are the ones we want. 250 + 75 is 325, so that leaves us with 50 people who satisfy the criteria. We need to make 50 the numerator of our fraction, which gives us . This reduces to , and answer choice (A) is the best answer.
Two Important Laws of Probability
When you want to find the probability of a series of events in a row, you multiply the probabilities of the individual events. What is the probability of getting two heads in a row if you flip a coin twice? The probability of getting a head on the first flip is . The probability is also that you’ll get a head on the second flip, so the combined probability of two heads is × , which equals . Another way to look at it is that there are four possible outcomes: HH, TT, HT, TH. Only one of those outcomes consists of two heads in a row. Thus, of the outcomes consist of two heads in a row. Sometimes the number of outcomes is small enough that you can list them and calculate the probability that way.
Probability of A and B
= Probability of A
× Probability of B
Occasionally, instead of finding the probability of one event AND another event happening, you’ll be asked to find the probability of either one event OR another event happening. In this situation, instead of multiplying the probabilities, you add them. Let’s say you have a normal deck of 52 cards. If you select a card at random, what’s the probability that you select a 7 or a 4? The probability of selecting a 7 is , which reduces to . The probability of selecting a 4 is the same; . Therefore the probability of selecting a 7 or a 4 is .
Probability of A or B
= Probability of A
+ Probability of B
Let’s look at a problem:
When a pair of six-sided dice, each with faces numbered 1 to 6, is rolled once, what is the probability that the result is either a 3 and a 4 or a 5 and a prime number?
Give your answer as a fraction.
Here’s How to Crack It
Probability is fundamentally about counting. You need to be able to count all the things that can happen and count all the situations that meet the conditions of the problem. Sometimes, the easiest way to count both everything that can happen and the situations that meet the condition is to write everything out. In this case, let’s use a table:
Each cell of this table represents a result when the dice are rolled. For example, the cell at the intersection of the row shown as 1 and the column shown as 1 would represent that 1 was showing on each of the two die. This cell has been marked with an because it does not meet either condition of the problem.
The cells marked with a are the only dice rolls that meet one of the conditions of the problem. To finish, just count the marks—there are 7. (Remember that 1 is not prime. That’s why combinations such as 5 and 1 are not checked.) Next, count the total possibilities—there are 36. So, the probability of rolling either a 3 and a 4 or a 5 and prime number is .
One last important thing you should know about probabilities is that the probability of an event happening and the probability of an event not happening must add up to 1. For example, if the probability of snow falling on one night is , then the probability of no snow falling must be . If the probability that it will rain is 80%, then the probability that it won’t rain must be 20%. The reason this is useful is that, on some GRE probability problems, it will be easier to find the probability that an event doesn’t occur; once you have that, just subtract from 1 to find the answer.
Let’s look at the following example.
Dipak has a 25% chance of winning each hand of blackjack he plays. If he has $150 and bets $50 a hand, what is the probability that he will still have money after the third hand?
Since probabilities are just
fractions, they can also be
expressed as percents.
Here’s How to Crack It
If Dipak still has money after the third hand, then he must have won at least one of the hands, and possibly more than one. However, directly calculating the probability that he wins at least one hand is tricky because there are so many ways it could happen (for example, he could lose-lose-win, or W-W-L or W-L-W or L-W-L, and so on). So think about it this way: The question asks for the probability that he will win at least one hand. What if he doesn’t? That would mean that he doesn’t win any hands at all. If we calculate the probability that he loses every hand, we can then subtract that from 1 and find the corresponding probability that he wins at least one hand. Since Dipak has a 25% chance of winning each hand, this means that he has a 75% chance of losing it, or (the answers are in fractions, so it’s best to work with fractions). To find the probability that he loses all three hands, simply multiply the probabilities of his losing each individual hand. so there is a probability that he will lose all three hands. Subtracting this from 1 gives you the answer you’re looking for. . The answer is (D).
Given events A and B, the probability of
· A and B = (Probability of A) × (Probability of B)
· A or B = Probability of A + Probability of B - Probability of A and B
Given event A
· Probability of A + Probability of Not A = 1
FACTORIALS
The factorial of a number is equal to that number times every positive whole number smaller than that number, down to 1. For example, the factorial of 6 is equal to 6 × 5 × 4 × 3 × 2 × 1, which equals 720. The symbol for a factorial is ! so 4! doesn’t mean we’re really excited about the number 4, it means 4 × 3 × 2 × 1, which is equal to 24. (0! is equal to 1, by the way.) When factorials show up in GRE problems, always look for a shortcut like canceling or factoring. The point of a factorial problem is not to make you do a lot of multiplication. Let’s try one.
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Here’s How to Crack It
Let’s tackle Quantity A. We definitely don’t want to multiply out the factorials since that would be pretty time-consuming: 12! and 11! are both huge numbers. Instead let’s look at what they have in common. What we’re really talking about here is . Now it’s clear that both factorials share everything from 11 on down to 1. The entire bottom of the fraction will cancel and the only thing left on top will be 12, so the value of Quantity A is 12. For Quantity B, we can also write out the factorials and get . The 2 and the 1 in the bottom cancel, and the only thing left on top will be 4 × 3, which is equal to 12. The two quantities are equal and the answer is (C).
PERMUTATIONS AND COMBINATIONS
The basic definition of a permutation is an arrangement of things in a particular order. Suppose you were asked to figure out how many different ways you could arrange five statues on a shelf. All you have to do is multiply 5 × 4 × 3 × 2 × 1, or 120. (Yes, this is another application of factorials.) You have five possible statues that could fill the first slot on the shelf, then, once the first slot is filled, there are four remaining statues that could fill the second slot, three that could fill the third slot, and so on, down to one.
Permutation problems
often ask for
arrangements, orders,
schedules, or lists.
Now suppose that there are five people running in a race. The winner of the race will get a gold medal, the person who comes in second will get a silver medal, and the person who comes in third will get a bronze medal. You’re asked to figure out how many different orders of gold-silver-bronze winners there can be. (Notice that this is a permutation because the order definitely matters.)
First, ask yourself how many of these runners can come in first? Five. Once one of them comes in first, she’s out of the picture, so how many can then come in second? Four. Once one of them comes in second, she’s out of the picture, so how many of them can come in third? Three. And now you’re done because all three slots have been filled. The answer is 5 × 4 × 3, which is 60.
To solve a permutation
· Figure out how many slots you have.
· Write down the number of options for each slot.
· Multiply them.
The difference between a permutation and a combination is that in a combination, the order is irrelevant. A combination is just a group, and the order of elements within the group doesn’t matter. For example, suppose you were asked to go to the store and bring home three different types of ice cream. Now suppose that when you got to the store, there were five flavors in the freezer—chocolate, vanilla, strawberry, butter pecan, and mocha. How many combinations of three ice cream flavors could you bring home? Notice that the order doesn’t matter, because bringing home chocolate, strawberry, and vanilla is the same thing as bringing home strawberry, vanilla, and chocolate. One way to solve this is the brute force method; in other words, write out every combination.
Combination problems
usually ask for groups,
teams, or committees.
VCS VCB VCM VSB VSM VBM CSB CSM CBM SBM
That’s 10 combinations, but there’s a quicker way to do it. Start by filling in the three slots as you would with a permutation (there are three slots because you’re supposed to bring home three different types of ice cream). Five flavors could be in the first slot, four could be in the second, and three could be in the third. So far, that’s 5 × 4× 3. But remember, this takes into account all the different orders that three flavors can be arranged in. We don’t want that, because the order doesn’t matter in a combination. So we have to divide 5 × 4 × 3 by the number of ways of arranging three things. In how many ways can three things be arranged? That’s 3!, 3 × 2 × 1, which is 6. Thus we end up with , which is equal to , or 10. Bingo.
Does the order matter?
To solve a combination
· Figure out how many slots you have.
· Fill in the slots as you would a permutation.
· Divide by the factorial of the number of slots.
The denominator of the fraction will always cancel out completely, so you can cancel first before you multiply.
Here’s an example:
Brooke wants to hang three paintings in a row on her wall. She has six paintings to choose from. How many arrangements of paintings on the wall can she create?
6
30
90
120
720
Always cross off wrong
answer choices on your
scratch paper.
Here’s How to Crack It
The first thing you need to do is determine whether the order matters. In this case it does, because we’re arranging the paintings on the wall. Putting the Monet on the left and the Van Gogh in the middle isn’t the same arrangement as putting the Van Gogh on the left and the Monet in the middle. This is a permutation question. We have three slots to fill because we’re arranging three paintings. There are 6 paintings that could fill the first slot, 5 paintings that could fill the second slot, and 4 paintings that could fill the third slot. So we have 6 × 5 × 4, which equals 120. Thus, the correct answer is (D).
Here’s another example:
A pizza may be ordered with any of eight possible toppings.
Quantity A |
Quantity B |
The number of different ways to order a pizza with three different toppings |
The number of different ways to order a pizza with five different toppings |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Here’s How to Crack It
First, note that for both quantities we’re dealing with a combination, because the order of toppings doesn’t matter. A pizza with mushrooms and pepperoni is the same thing as a pizza with pepperoni and mushrooms. Let’s figure out Quantity A first.
We have eight toppings and we’re picking three of them. That means we have three slots to fill. There are 8 toppings that could fill the first slot, 7 that could fill the second slot, and 6 that could fill the third, so we have 8 × 7 × 6. Since this is a combination, we have to divide by the factorial of the number of slots. In this case we have three slots, so we have to divide by 3!, or 3 × 2 × 1. So our problem looks like this: . To make the multiplication easier, let’s cancel first. The 6 on top will cancel with the 3 × 2 on the bottom, leaving us with , which is 56. Thus, there are 56 ways to order a three-topping pizza with eight toppings to choose from. Now let’s look at Quantity B.
We still have eight toppings, but this time we’re picking five of them so we have five slots to fill. There are 8 toppings that could fill the first slot, 7 that could fill the second slot, 6 that could fill the third, 5 that could fill the fourth, and 4 that could fill the fifth. That’s 8 × 7 × 6 × 5 × 4, but we still have to divide by the factorial of the number of slots. We have five slots, so that means we need to divide by 5!, or 5 × 4 × 3 × 2 × 1. Thus we have . We definitely want to cancel first here, rather than doing all that multiplication. The 5 on top will cancel with the 5 on the bottom. Likewise, the 4 on top will cancel with the 4 on the bottom. The 6 on top will cancel with the 3 × 2 on the bottom, leaving us again with , which is 56. Therefore, there are also 56 ways to order a five-topping pizza with eight toppings to choose from. The two quantities are equal, and the answer is (C).
Let’s try one more:
Nicole needs to form a committee of 3 from a group of 8 research attorneys to study possible changes to the Superior Court. If two of the attorneys are too inexperienced to serve together on the committee, how many different arrangements of committees can Nicole form?
20
30
50
56
336
Here’s How to Crack It
This problem is a little more complicated than an ordinary combination problem, because an extra condition has been placed on the committee. Without that condition, this would be a fairly ordinary combination problem, and we’d simply calculate how many groups of three can be created with eight people to choose from.
There’s more than one way to approach this problem. First, you should realize that there are two ways that we could form this committee. We could have three experienced attorneys, or we could have two experienced attorneys and one inexperienced attorney. If we find the number of ways to create each of those two possibilities, we can add them together and have our answer. It’s fairly straightforward to calculate the number of ways to have three experienced attorneys on a committee: There are three slots to fill, and we have 6 options for the first slot, 5 for the second, and 4 for the third. Here the order doesn’t matter, so we divide by 3! to get = 20. Thus there are 20 ways to create the committee using three experienced attorneys. What about creating a committee that has two experienced attorneys and one inexperienced attorney? We have 6 options for the first experienced attorney and 5 options for the second. Order doesn’t matter so we divide by 2!. So far we have . Next we have 2 options for the inexperienced attorney, so now we have to multiply by 2, and our calculation is = 30. As you can see, there are 30 ways to create the committee using two experienced attorneys and one inexperienced attorney. Adding 20 and 30 gives us 50 total committees, and the answer is (C).
Here’s another way that you could solve the problem. If there were no conditions placed on the committee, we could just calculate , which would give us 56 committees. But we know some of those committees are not allowed; any committee that has the two inexperienced attorneys on it isn’t allowed. How many of these types of committees are there? Let’s call the inexperienced attorneys A and B. An unacceptable committee would be A B __, in which the last slot could be filled by any of the experienced attorneys. Since there are 6 experienced attorneys, there are 6 unacceptable committees. Subtracting them from 56 gives us 50 acceptable committees. Hey, the answer’s still (C)!
FUNCTIONS AND THE GRE
F(x) Notation
ETS often employs the use of function notation to create difficult problems. Generally speaking, the function notation is a style of math problem that causes test takers to be nervous. The function notation, f(x), is unfamiliar to look at, seems difficult and involved, and evokes memories of graphs and charting lines that you may have learned in high school geometry.
The good news is that pure function problems on the GRE are much more straightforward than that and become very manageable if you utilize Plugging In strategies.
The easiest way to think about a function question is to look at an example. Take f(x) = x + 2, for instance. All this problem is stating is that for any value of x, the function f(x) is that value plus 2. Let’s say that x = 3; therefore, to solve this problem, take the value of x and plug it into the given equation. So if x = 3, the equation now reads f(3) = 3 + 2, or f(3) = 5. To solve function notation problems, all you need to do is read the instructions carefully and fill in the values for the variables where appropriate. If you used the same equation, but the value of x is 10, then the function is now f(10) = 10 + 2, so f(10) = 12.
Sometimes a function problem gives a restriction such as x ≠ 0. If this is the case, you know that x could be equal to any value but 0, and this is generally for a good reason. If , then x cannot equal 0 because a number cannot be divided by 0.
Try this example of a function question on the GRE.
If − 3 ≤ g ≤2 and f(g) = −2g and g is an integer, then which of the following integers could be a value of f(g) ?
Indicate all such integers.
−6
−5
−2
0
2
4
6
Here’s How to Crack It
This is a function problem with restrictions, so find all the different values that can be plugged in for g. Since g is an integer that is equal to or between −3 and 2 then there is a range for its values. Therefore, f(g) (which equals −2g) is all the integer values between the high and low end of the range of g multiplied by 2. In other words, plug in 2 and −3 for g in the function and figure out what the range is. If g = −3, then f(g) is f(−3)= −2(−3) = 6. And if g = 2, then f(g) is f(2) = −2(2) = −4. So the range of f(g) is −4 ≤ f(g) ≤ 6. Choices (A) and (B) are less than −4 and fall out of the range. The rest of the integers fall in the range, so are possible values of f(g). Therefore the correct answer is (C), (D), (E), (F), and (G).
Evaluating functions is all about following the directions. Just plug in the values for the variable and solve.
Functions With Uncommon Symbols:
The GRE also tries to scare students using functions in another way: picking strange symbols and putting them in a problem. When you see a funny symbol that you have never seen before, don’t stress out! It’s just a function problem in disguise. Follow the directions to find the correct answer.
A problem with funny symbols may look something like this:
If the operation is defined for all integers x and y as x y = x ^{2} + y − 2, then which of the following is equal to 4 −3 ?
21
17
15
11
10
Here’s How to Crack It
Remember, this is a function problem, so just follow the directions. The problem wants to know the value of 4 −3, and it states that x y = x^{2} + y − 2. To solve this problem, plug in x = 4 and y = −3. So 4 −3 = 42 + (−3) − 2. Now, solve: 4 −3 = 16 − 3 − 2 = 11. The correct answer is (D).
You may get a different symbol when you get another problem like this, but the process is still the same. Just plug in the values given for the variables and solve the problem. If you have worked your way through this book and mastered the content, then there won’t be any actual mathematical symbols on the GRE that are unfamiliar to you. If you see a symbol like that, it’s a function problem!
Let’s look at one more example.
For any non-negative integer x, let x* = x − 1
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Here’s How to Crack It
Just follow the directions—15* = 15 – 1, or 14, and 3* = 3 – 1, or 2. So Quantity A is , or 7. Don’t forget PEMDAS for Quantity B. First, is 5. Then, 5* = 5 – 1, or 4. So because Quantity A is 7 and Quantity B is 4, the correct answer is (A).
GROUPS
Group problems, although not too common on the GRE, can be troublesome if you don’t know how to set them up. When confronted by a group problem, use the group equation
You might see one group
problem on the GRE.
T = G_{1} + G_{2} – B + N
In the equation, T represents the Total, G_{1} is one group, G_{2} is the second group, B is for the members in both groups and N is for the members in neither group. Here’s an example of a typical group problem.
A biologist studying breeding groups noted that of 225 birds tagged for the study, 85 birds made nests in pine trees, 175 made nests in oak trees, and 40 birds did not build nests in either type of tree. How many birds built nests in both types of trees?
45
60
75
80
125
Here’s How to Crack It
Let’s use the group equation. The total is 225, one group consists of 85 birds, the other group has 175 birds in it, and we know that 40 birds built nests in neither type of tree. Our equation would look like this:
225 = 85 + 175 – B + 40
All we have to do is solve for B. Simplifying the equation gives us 225 = 300 – B, so B must equal 75. Choice (C) is our answer.
Wrapping Up Math
You’re almost done with
the Math section. Tackle
the Math Drills on the
following pages, then give
yourself a break before
you dive into the
Analytical Writing
Section. Take a walk, eat
a snack, or meet up with a
pal and give youself some
downtime before you dive
into Part IV.
Et Cetera Drill
Here are some math questions to practice on. Remember to check your answers when you finish. You can find the answers in Part V.
1 of 10
A bowl contains 15 marbles, all of which are either red or blue. If the number of red marbles is one more than the number of blue marbles, what is the probability that a marble selected at random is blue?
2 of 10
If ¥(x) = 10x – 1, what is ¥(5) – ¥(3) ?
15
18
19
20
46
3 of 10
Quantity A |
Quantity B |
The greatest odd factor of 78 |
The greatest prime factor of 78 |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
4 of 10
At a recent dog show, there were 5 finalists. One of the finalists was awarded “Best in Show” and another finalist was awarded “Honorable Mention.” In how many ways could the two awards be given out?
5 of 10
Company X budgets $90,000 total on advertising for all of its products per year. Company X budgets $40,000 for all advertising for product A and $30,000 for all advertising for product B. From the budgets for products A and B, $15,000 is budgeted for advertisements that feature both products used as a system.
Quantity A |
Quantity B |
The total amount Company X budgets for advertising products other than products A and B. |
$20,000 |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
6 of 10
Lee randomly selects a 2-digit prime number less than 50. What is the probability that the tens digit is greater than the units digit?
7 of 10
An elected official wants to take five members of his staff to an undisclosed secure location. How many staff members must the elected official employ in order to have a minimum of 20 different groups from which to choose?
7
8
9
10
11
8 of 10
For all real numbers x and y, if x # y = x(x – y), then x # (x # y) =
x^{2} – xy
x^{2} – 2xy
x^{3} – x^{2} – xy
x^{3} – (xy)^{2}
x^{2} – x^{3} + x^{2}y
9 of 10
A jar contains 12 marbles. Each is either yellow or green and there are twice as many yellow marbles as green marbles. If two marbles are to be selected from the jar at random, what is the probability that exactly one of each color is selected?
10 of 10
A set of 10 points lies in a plane such that no three points are collinear.
Quantity A |
Quantity B |
The number of distinct triangles that can be created from the set |
The number of distinct quadrilaterals that can be created from the set |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Comprehensive Math Drill
Let’s do a drill involving all of the math topics we have covered throughout the book. Remember to check your answers when you finish. You can find the answers in Part V.
1 of 20
Line AB is tangent to the circle C at point A. The radius of the circle with center C is 5 and BC = .
Quantity A |
Quantity B |
The length of line segment AB |
The length of line segment AC |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
2 of 20
x ≠ 0
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
3 of 20
The test scores for a class have a normal distribution, a mean of 50, and a standard deviation of 4.
Quantity A |
Quantity B |
Percentage of scores at or above 58 |
Percentage of scores at or below 42 |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
4 of 20
The line y = − x + 1 is graphed on the rectangular coordinate axes.
Quantity A |
Quantity B |
OQ |
OP |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
5 of 20
At a dog show, there are 20 judges and 10 dogs in the final round.
Quantity A |
Quantity B |
The number of distinct pairs of judges |
The number of possible rankings of dogs from first to third place |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
6 of 20
k > 0
l > 1
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
7 of 20
A jeweler made a profit of $800 on the discounted price of a diamond that cost the jeweler $5,400 to purchase. The profit from the discounted price is 80% less than the profit the jeweler would have made if he sold the diamond for its regular, undiscounted price.
Quantity A |
Quantity B |
The profit the jeweler would have made from the regular, undiscounted price of the diamond expressed as a percentage of the price for which the jeweler bought the diamond. |
The difference between the profit on the discounted price and the profit on the regular, undiscounted price expressed as a percentage of the original price. |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
8 of 20
Joe has $200. If he buys a CD player for $150, what is the greatest number of CDs he can buy with the remaining money if CDs cost $12 each?
9 of 20
What is the area of triangle ABC in the figure above?
2
4
4
7
8
10 of 20
By which of the following could 10(9^{6}) be divided by to produce an integer result?
Indicate all such values.
90
100
330
540
720
11 of 20
Roberta drove 50 miles in 2 hours. Her rate in miles per hour is equivalent to which of the following proportions?
Indicate all such proportions.
5 to 20
100 to 4
400 to 16
20 to 500
520 to 20
Questions 12 through 14 refer to the following graph.
12 of 20
For how many of the cities shown was the highest temperature in Year Y greater than or equal to the highest temperature in Year X ?
4
5
7
8
12
13 of 20
What is the approximate percent increase from the lowest average (arithmetic mean) temperature for Years X and Y to the highest average temperature?
60%
82%
140%
188%
213%
14 of 20
The average (arithmetic mean) temperature for any city in Years X and Y is the average of the high and low temperatures for those years. What is the average low temperatures for Baltimore for Years X and Y ?
–9° F
11° F
20° F
44° F
It cannot be determined from the information given.
15 of 20
If |2x – 3| + 2 > 7, which of the following could be the value of x ?
Indicate all such values.
–4
–3
–2
–1
0
1
2
3
16 of 20
If x, y, and z are consecutive odd integers where x < y < z and x + y + z < z, then which of the following could be the value of x ?
Indicate all such values.
–3
–1
0
1
3
17 of 20
If 4^{x} = 1,024, then (4^{x} ^{+ 1}) (5^{x} ^{– 1}) =
10^{6}
(5^{4}) (10^{5})
(4^{4}) (10^{5})
(5^{4}) (10^{4})
(4^{4}) (10^{4})
18 of 20
What is the greatest distance between two vertices of a rectangular solid with a height of 5, a length of 12, and a volume of 780 ?
12
12
13
13
13
19 of 20
If three boys and three girls sit in a row on a park bench, and no boy can sit on either end of the bench, how many arrangements of the children on the bench are possible?
46,656
38,880
1,256
144
38
20 of 20
If 16 is the average (arithmetic mean) of p, 24, and q, what is 16(p + q) ?
180
192
384
524
768
Summary
· Topics such as probability, permutations and combinations, factorials, and functions represent only a small percentage of the math topics tested on the GRE. Make sure you’ve mastered all the more important topics before attempting these.
· Probability is expressed as a fraction. The denominator of the fraction represents the total number of possible outcomes, while the numerator stands for the desired outcomes.
· If a probability question asks for the chance of event A or event B, find the probability of each event and add them together. If the question asks for the probability of event A and event B, multiply the individual probabilities.
· The key to factorial problems is to look for ways to cancel or factor out terms.
· Permutations and combinations are related concepts. A permutation tells you how many arrangements or orderings of things are possible. A combination tells you how many groupings of things are possible.
· Function problems use funny looking symbols as shorthand for the operations to perform on a certain number.
· The group equation is Total = Group_{1} + Group_{2} – Members of Both Groups + Members of Neither Group.