The post What Kind of Math is on the GMAT? Breakdown of Quant Concepts by Frequency appeared first on Magoosh GMAT Blog.
]]>The biggest secret to GMAT Quant success is a simple one: identify and study the correct Quant concepts. But what are the GMAT math topics? And which ones are the most important?
To answer all of these questions, I looked at official GMATPrep tests 3 and 4, and the Official Guide for the GMAT Review 2019. Read on to find out what I learned from analyzing the GMAT quantitative topics in these 766 official questions.
Here is a GMAT Quant section breakdown (with category descriptions as needed):
Word problem interpretation is the most frequent concept, showing up on well over half of the questions. Integer properties and arithmetic come second, appearing on nearly a third of all of the questions.
The table below lists GMAT Quant concepts in order of most-to-least frequent. (And again, the most frequent concepts are obviously the most important!) Click the topic names to read on a given topic in more detail. In the case of data interpretation, the link goes to an IR resource that is also relevant to Quant.
You can treat the table and its links as a GMAT Quantitative syllabus of sorts. Follow the links to learn everything about arithmetic, geometry, and proportion, everything about probability, stats, and counting, etc… Just about anything you’d need to know can be seen or accessed in the table!
GMAT Quant concept | Percentage frequency |
---|---|
Word Problems | 58.2% |
Integer properties and arithmetic | 31.1% |
Algebra | 16.3% |
Percents, ratios, and fractions | 13.7% |
Two dimensional geometry | 10.6% |
Statistics | 6.3% |
Powers and roots | 6.3% |
Probability and combinatronics | 5% |
Inequalities | 4.7% |
Sequences | 3.2% |
Coordinate geometry | 2.9% |
Data interpretation | 0.9% |
Three dimensional geometry | 0.8% |
Functions | 0.4% |
Note: Some questions tested multiple concepts and were thus counted more than one time in more than one category. As a result, the percentages in the chart above add up to more than 100%.
As you can see in the table above, not all GMAT Quant concepts are created equal. Certain GMAT Quant topics will definitely appear more often than others.
Clearly, the GMAT loves to test its Quant concepts through word problems. Word problems can overlap with just about topic: statistics, algebra, inequalities, you name it. There can even be coordinate word problems and absolute value word problems! So make sure you build math-related reading comprehension skills as you prepare for the exam.
Several other high-frequency Quant concepts stand out when you look at the table above. Word problems, integer properties, arithmetic, algebra, percents, ratios, fractions, and geometry are the most important. These topics all are clearly vital to GMAT Quant success.
Lower on the chart, you can see some concepts that seem a good deal less important. Sequences, the coordinate plane, three dimensional objects, functions, and data interpretation don’t occur all that often; these topics have minimal importance in GMAT Quant.
Not so fast though. Let’s take a closer look at that last “unimportant” GMAT Quant concept I mentioned.
Although it is clearly not that important in the GMAT Quant section, data interpretation is still a big part of the GMAT as a whole. Remember, the Integrated Reasoning section consists primarily of math data interpretation questions. So be sure to study this concept as part of your overall GMAT prep.
Speaking of other sections of the GMAT, make sure you understand where these Quant concepts fit into the test as a whole. The GMAT maths syllabus should be seen as part of the syllabus for the entire exam.
So be sure to check out my colleague Rachel’s Complete Hassle-Free Guide to the GMAT test. Or for a quicker snapshot of the most common question, GMAT-wide, see Mike’s “Most Common GMAT Topics and Questions.”
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]]>The post GMAT Sample Questions for You to Practice appeared first on Magoosh GMAT Blog.
]]>Problem Solving (PS) questions consist of an algebraic/word prompt and five multiple choice answers. Those who have taken math tests before should already be comfortable with the format of PS questions. See the selected GMAT sample questions below:
A car drives 40 miles on local roads at 20 mph, and 180 miles on the highway at 60 mph, what is the average speed of the entire trip?
(A) 36 mph
(B) 40 mph
(C) 44 mph
(D) 52 mph
(E) 58 mph
Answer: C. See GMAT Distance and Work: Rate Formula for an explanation of the answer.
How many odd factors does 210 have?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 8
Answer: E. See GMAT Math: Factors for an explanation of the answer.
Data Sufficiency (DS) questions are a format unique to the GMAT. Unlike traditional math questions where the goal is to find the exact answer, the goal of DS questions is to determine whether the supplied statements would enable one to find an exact answer. As such, DS questions typically require some particular strategies based upon answer elimination and logical shortcuts (e.g., you should avoid calculating the exact answer when possible).
The five answer choices to DS problems are always the same, so it’s a good idea to memorize the choices below:
(B) Statement 2 alone is sufficient but statement 1 alone is not sufficient to answer the question asked.
(C) Both statements 1 and 2 together are sufficient to answer the question but neither statement is sufficient alone.
(D) Each statement alone is sufficient to answer the question.
(E) Statements 1 and 2 are not sufficient to answer the question asked and additional data is needed to answer the statements.
And now, a couple examples for practice:
ABC is an equilateral triangle, and point D is the midpoint of side BC. A is also a point on circle with radius r = 3. What is the area of the triangle?
Statement #1: The line that passes through A and D also passes through the center of the circle.
Statement #2: Including point A, the triangle intersects the circle at exactly four points.
Answer: C. See our GMAT Data Sufficiency Geometry Practice Questions for an explanation of the answer.
Is |x – 6| > 2?
Statement #1: |x – 4| > 3
Statement #2: |x – 8| > 1
Answer: C. See GMAT Math: Arithmetic with Inequalities for an explanation of the answer.
Reading Comprehension (RC) questions begin with a two- to four-paragraph passage. Then, you’ll see a small set of questions based upon the passage. There are a few different types of RC questions, but they all fundamentally test your ability to apply information obtained from a written source.
Try this free practice problem from Magoosh.
Here’s another free sample problem based upon the same passage as the above.
Critical Reasoning (CR) problems present a paragraph that lays out the foundation of some argument. From there, you will be tasked with strengthening the argument, weakening the argument, drawing an inference, or identifying some other aspect of the argument. To succeed with these questions, you must learn the different elements of an argument.
A minor league baseball franchise experienced a drop in attendance this week after they suffered three losses by margins of ten runs or more last week. Many spectators of those games wrote letters to the editors of the local sporting news, complaining of the poor play of the team in those three losses. Nevertheless, the front office of this baseball franchise maintains that the team’s poor play in those three losses has nothing to do with this week’s decline in attendance.
Which of the following, if true, most strongly supports the position held by the front office of the baseball franchise?
(A) The spectators who wrote letters to the local sporting news were long-standing fans of this minor league baseball team.
(B) Many minor league baseball franchises attribute a drop in attendance to the quality of play of the team only after a string of losses.
(C) Other minor league teams in that region of the state reported a similar drop in attendance this week.
(D) This was not the first time this team suffered multiple lopsided losses in a single week, prompting similar letters to the local sporting news.
(E) This minor league team is over four hours from the closest major league team, so many of the minor league team’s fans do not often attend major league games.
Answer: C. See Introduction to GMAT Critical Reasoning for an explanation of the answer.
State politicians are optimistic that the state’s economic downturn will not be as severe as had been predicted. Their hopes are bolstered by the data released last week: the jobless rate declined two full percentage points in the last six months. But, many economists have pointed out the flight of unemployed residents to the bordering states where jobs are plentiful. Furthermore, many out of work residents have been rehired at minimum wage: virtually all new jobs in the state in the past year have been minimum wage jobs. Economists cast grave doubts on the economic well-being of the state.
In the argument given, the two portions in boldface play which of the following roles?
(A) The first is evidence in support of the conclusion; the second is that conclusion.
(B) The first is evidence opposed to the conclusion; the second is an interpretation of the conclusion.
(C) The first is an interpretation that calls the conclusion into question; the second is that conclusion.
(D) The first is a conclusion the argument calls into question; the second is the evidence that calls it into question.
(E) The first is evidence taken to support a conclusion; the second is a position that opposes that conclusion.
Answer: E. See GMAT Critical Reasoning: Boldface Structure Questions for an explanation of the answer.
Sentence Correction (SC) problems begin with a sentence that includes an underlined portion. Each answer choice contains a variation of the underlined portion. (Note that answer choice A is always the same as the original sentence.) Your task is to select an answer choice that makes the most grammatical and logical sense. The best strategy for tackling SC questions is to employ ‘splits’ among the answer choices.
Company policy restricts employees to, at most, three personal days in a month, and even less if the number of Fridays in the month is more than four.
(A) even less if the number of Fridays in the month is more than four
(B) even less if the amount of Fridays in the month is more than four
(C) even fewer if the number of Fridays in the month is greater than four
(D) even fewer if the amount of Fridays in the month is more than four
(E) even less if the number of Fridays in the month is greater than four
Answer: C. See GMAT Sentence Correction: Comparative & Qualitative Idioms for an explanation of the answer.
Although offering a dynamic range simply absent on the harpsichord, the original fortepiano, invented in the early eighteenth century — indeed the name comes from the Italian forte (“loud”) + piano (“soft”) — the fortepiano would now sound dynamically limited compared to our modern grand pianos.
(A) Although offering a dynamic range simply absent on the harpsichord, the original fortepiano, invented in the early eighteenth century
(B) Although the original fortepiano, invented in the early eighteenth century, offered a dynamic range simply absent on the harpsichord
(C) Although it offered a dynamic range simply absent on the harpsichord, the original fortepiano, invented in the early eighteenth century
(D) Invented in the early eighteenth century, the original fortepiano offered a dynamic range simply absent on the harpsichord
(E) The original fortepiano, invented in the early eighteenth century, although it offered a dynamic range simply absent on the harpsichord
Answer: B. See our Run-On Sentences in GMAT Sentence Correction for an explanation of the answer.
How did you do with these GMAT sample questions? Which math or verbal questions did you ace, and which do you need more practice in?
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]]>Many students have questions about the GMATs Integrated Reasoning (IR) section. “Is integrated reasoning part of the GMAT score?” Well…yes and no. Your IR score will be submitted to schools along with your verbal and quantitative score. However, the IR score is totally separate from the “total score,” which consists solely of the Q & V sections. But nonetheless, admission committees have started giving considerable attention to integrated reasoning GMAT scores, so it’s important to perform well on each section. Continue reading below for specifics on precisely how the IR section scored.
Fact: The current version of the GMAT features a Verbal Section, a Quantitative Section, a single AWA essay, and the new Integrated Reasoning (IR) section. The sequence of the new test will be:
1) AWA essay = Analysis of Argument, 30 minutes
2) IR section = 12 questions, 30 minutes
3) optional break, up to 5 minutes
4) Q section = 31 questions, 62 minutes
5) optional break, up to 5 minutes
6) V section = 36 questions, 65 minutes
This the traditional order. As of July 11, 2017, this is one of three possible orders of the sections: students now have some choice about section order.
Fact: the IR section consists of four question types —
a) Graphics Interpretation (GI)
b) Two-Part Analysis (2PA)
c) Table Analysis (TA)
d) Multi-Source Reasoning (MSR)
Fact: all four question types will appear on everyone’s IR sections.
Fact: the breakdown by question type will differ from one person’s IR section to another person’s only because of the experimental questions.
In other words, everyone will have the same breakdown by question type for the questions that actually count toward their score. However, extra experimental questions are added in to this baseline, resulting in different IR section breakdowns for different people.
GMAC has revealed neither what that fundamental breakdown is nor how many of the 12 questions will be experimental. Let’s examine a hypothetical scenario just to understand:
Let’s say the graded IR questions consist of 2 GIs, 2 2PAs, 2 TAs, and 2 MSRs, for a total of eight (these are my made-up numbers). For everyone taking the test, let’s say those are the eight questions that are graded. The other four questions would be experimental questions, and will be different for different users. Thus, Abe might get an IR section with 3 GIs, 3 2PAs, 3 TAs, and 3 MSRs. Betsy might get an IR section with 2 GIs, 3 2PAs, 3 TAs, and 4 MSRs. Cathy might get an IR section with 2 GIs, 6 2PAs, 2 TAs, and 2 MSRs.
In each case, only the baseline eight questions count toward the score, and the others are experiments. (The numbers in this example are purely speculative: we have no idea what GMAC has up their sleeve.)
Here’s the kicker, though. As our hypothetical friend Cathy is working through her IR section, she may start to think: “Gee, I’m seeing a lot of 2PA questions! Some of them must be experimental!” Quite true. But the catch is, among those six 2 PA questions, the two that actually count could be the first two, or the last two, or any combination. Those comfortable with combinations will see that there are actually 6C2 = 15 different ways that the two that count could be scrambled among the four experimental questions.
As the test taker, even if you do have strong suspicions about which question types the experimental questions were, you will have no way of knowing, as you are working on a particular question, whether it counts or is experimental. Therefore, you have to treat every single question as if it counts, same as on the Q & V sections.
Fact: the IR section is not computer adaptive. You are randomly assigned 12 questions as a group, and move through that sequence regardless of whether you are getting questions right or wrong.
Fact: The GMAT score report will consist of (a) V score, (b) Q score, (c) Total Score (combination of your V & Q scores), (d) AWA score, and (e) IR score.
Fact: the IR score will be an integer from 1 to 8. There is no partial credit on the IR section. For example, in a TA question in which there are three dichotomous prompts (e.g. true/false), you must get all three right to get credit for that one question. If you get at least one of the three parts wrong, the whole question is marked wrong.
Fact: The number of IR questions you get right will constitute a raw score. The GMAC, using some arcane alchemy known only to them, will convert that raw score into a scaled score (1 – 8), which will be accompanied by percentiles.
Notice: Because of the statistical magic GMAC uses in converting raw scores to scaled scores (on IR, Q, & V sections), what may seem to your advantage or disadvantage may not work out that way. For example, the fact that there’s no partial credit is challenging: it makes it harder to earn points on individual questions. BUT, harder for everyone means that lower raw scores are needed to get a higher percentile grade. By contrast, if all the questions are very easy, that means most people will get them right, which means it will be “crowded” at the top, and much harder to place in a high percentile. Therefore, what matters is not how inherently easy or hard the test is—what matters is how well you perform, compared to other test takers.
Given your inherent talents, what will maximize your GMAT skills with respect to others taking the GMAT? Sign up for Magoosh, and you will learn all the content and strategy you will need.
Editor’s Note: This post was originally published in May 2012 and has been updated for freshness, accuracy, and comprehensiveness.
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]]>For decades, the GMAT had a very fixed format: first, the AWA; then the IR; then, after an optional break, the Quant section; finally, after a second optional break, the Verbal Section. That was it, no choice.
This will change on July 11, 2017. Starting on that day, all GMAT test takers worldwide will have a choice of one of three possible orders. Yes, theoretically there would be a possible 4! = 24 possible permutations, but right now, the GMAT is just allowing three of those. Here are the three orders:
Order 1: AWA, then IR, first break, Quant, second break, Verbal
Order 2: Verbal, first break, Quant, second break, IR, then AWA last
Order 3: Quant, first break, Verbal, second break, IR, then AWA last
Notice that Order #1 is the traditional order, the order fixed on all GMATs before July 11 of this year. Starting on July 11, 2017, you will get to choose one of these three. When you sit down in the Pearson VUE center to take your GMAT, after the introductory screens, the section order selection screen will be the very last screen you see before your official GMAT begins.
What about GMAT Prep? The folks at GMAC are working on that. Starting on July 31, 2017, everything associated with GMAT prep, including the purchasable exam packs and question packs, will be updated to reflect the section order selection. If you have GMAT prep already on your computer, then after that date, you simply can install the new version as a software update, and it automatically will update all the GMAT prep products you already own. Obviously, if you download GMAT prep or purchase any of the additional products after that date, everything already will have the section order selection built in.
So, if you are about to take the GMAT, and you will be faced with this choice, what does this mean? Is there one choice that’s ideal for everyone? Of course not! It very much depends on your preferences, your relative strengths, and your needs.
For example, I can do math even when I’m tired and tapped out, but I need to be sharper to do verbal, so I would probably choose order #2. That’s me, but my friend Chris might save Verbal for last, the traditional order, because Verbal is his strong suit. My general advice is that if one section is a huge challenge for you, maybe you should get that section out of the way first: with this in mind, order #2 might be better for many non-native English speakers, while order #3 might be better for American math-phobes. On the other hand, if you feel as if a difficult section might fluster you and prevent you from doing your best on other sections, perhaps you want to take this difficult section later. As always, as Socrates found, it’s very hard to improve on the Delphic maxim of “know thyself.” In your practice, experiment with some orders, and see what feels good.
One thing I definitely would recommend: get all your thinking or experimenting done long long before you walk into Pearson to take your GMAT. Your preferred GMAT section order is definitely a decision that should be done and set well before the time you sit down to take the GMAT.
The choices of section order should be made long before you walk into your GMAT. Preparing for the GMAT also should involve learning all the content and strategies for all the sections on the test. That’s precisely where Magoosh can help you! Magoosh can help you attain mastery on all four sections of the GMAT!
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]]>First, three practice GMAT grammar subordinate conjunctions questions, each somehow involving buses!
1) While spark ignitions start the combustion in gasoline engines, typical in automobiles, high compression of gases, with high temperatures, are igniting the combustion in diesel engines.
(A) high compression of gases, with high temperatures, are igniting the combustion in diesel engines
(B) the high temperatures made by high compression of gases, igniting the fuel in diesel engines
(C) diesel engines highly compress the gasses, and this high compression, which causes high temperatures, ignites the combustion of the fuel in the engine
(D) high compression of gases, producing high temperatures, ignite the combustion in diesel engines
(E) it is the high compression of gases, causing high temperatures, that ignites the fuel in diesel engines
2) Even though the original AEC Routemaster has been retired, still this red double-decker bus is familiar, and it has been an icon of culture in Britain.
(A) still this red double-decker bus is familiar, and it has been an icon of culture in Britain
(B) the familiarity of this red double-decker bus still remains, as does its role as a British cultural icon
(C) this familiar red double-decker bus remaining a cultural icon in Britain
(D) this familiar red double-decker bus remains a British cultural icon
(E) the British are familiar with this red double-decker bus and still consider it to be one of their cultural icons
3) The end of the Montgomery Bus Boycott was the Browder v. Gayle (1956) ruling, an apparent victory for civil rights, with most blacks in Montgomery, with the experience of massive discrimination in all sectors of life, resigning themselves to the back of the bus by the early 1960s.
(A) The end of the Montgomery Bus Boycott was the Browder v. Gayle (1956) ruling, which should have been a victory for civil rights, with most blacks in Montgomery, with the experience of massive discrimination in all sectors of life, resigning themselves to the back of the bus by the early 1960s
(B) While the Montgomery Bus Boycott ended with an apparent victory for civil rights in the Browder v. Gayle (1956) ruling, most blacks in Montgomery, in the face of massive discrimination in all sectors of life, resigned themselves to the back of the bus by the early 1960s
(C) The Browder v. Gayle (1956) ruling was apparently a victory for civil rights when it ended the Montgomery Bus Boycott, and because of this, most blacks in Montgomery by the early 1960s resigned themselves to the back of the bus, with the experiencing of massive discrimination in all sectors of life
(D) Because of the Montgomery Bus Boycott, the civil rights movement incorrectly believed it had a victory to the Browder v. Gayle (1956) ruling, and by the early 1960s, the blacks with resignation moved to the back of the bus, because they experienced massive discrimination in all sectors of life
(E) Despite a victory that was not a victory for civil rights in the Browder v. Gayle (1956) ruling, the ending of the Montgomery Bus Boycott, in the early 1960s, most blacks in Montgomery facing massive discrimination in all sectors of life and therefore resigning themselves to the back of the bus
Explanations will come at the end of this blog.
In the big world of grammar, there are two kinds of clauses: those that can stand alone on their own two feet and those that cannot. The first kind, independent clauses, can stand alone as complete sentences. The latter kind, dependent clauses, or subordinate clauses, do not work on their own as a complete sentence: they have to be part of a larger sentence, a sentence that is anchored by at least one independent clause. Here are some examples of independent clauses as stand-alone sentences:
By contrast, here are three subordinate clauses utterly failing to constitute complete sentences when they stand alone:
Notice that all three of those give the impression that something is missing, as if more of the sentence is about to occur. Subordinate clauses don’t work on their own; they are set up to play a supporting role to the independent clause.
There are four basic categories of subordinate clauses:
(i) Substantive Clauses: in a sentence, these clauses take the role of a noun, such as subject, direct object, or object of a preposition. These are also called “noun clauses” or “nominal relative clauses.”
(ii) Relative Clauses: these begin with a relative pronoun or adverb, and will act, respectively, as a noun-modifier or verb-modifier.
(iii) Clauses that begin with subordinate conjunctions: discussed below.
(iv) Comparative Clauses: these typically begin with “than” and complete a comparison.
All four of these share two features: (1) they cannot stand alone as separate sentences and (2) their entire purpose, their raison d’être, is to support the larger sentence.
Subordinate conjunctions are words that begin a common category of subordinate clauses. All of these clauses function as adverbial clauses, that is, as verb modifiers. One handy mnemonic for the subordinate conjunctions is “on a white bus”:
O = only if, once
N = now that
A = although, after, as
WH = while, when, whereas, whenever, wherever, whether
H = how
I = if, in case, in order that
T = though
E = even though, even if
B = because, before
U = until, unless
S = since, so, so that
You certainly don’t need to memorize this list. It’s helpful, especially for non-native speakers, to recognize these words and be familiar with them. Any of these begins a clause that modifies the independent clause and could not stand on its own as a free-standing separate sentence.
By definition, any clause has a [noun] + [verb] unit at its core. Every clause, whether independent or subordinate, needs to have a full verb. Nevertheless, the following sentence is 100% correct:
10) Though polite and refined in person, Boris Karloff was known for playing monsters on screen.
It appears that there’s a problem in the “though” clause: the subordinate conjunction “though” is followed by only adjectives. There’s neither a noun or a verb, it would seem. How can this be correct? In fact, it’s perfectly fine if simple words [pronoun] + [“to be” verb] are omitted. For example, with the omitted words, this sentence would be:
10a) Though [he was] polite and refined in person, Boris Karloff was known for playing monsters on screen.
When the omitted words are included, we see that the “though” clause was a full bonafide clause all along.
While the GMAT is not going to expect you to know these grammar terms, it’s important to have good instincts about clauses, both independent and dependent. Pay attention when you read to how these clauses behave—seeing what these clauses do in sentence after sentence will help you understand them more deeply.
1) Traditionally, most buses and truck had diesel engines, although many buses today run on alternative fuels. We have to consider all five answer separately.
(A) the progressive “are igniting” is wrong. This is incorrect.
(B) the famous missing verb mistake! This is incorrect.
(C) wordy, repetitive, awkward. Far from ideal.
(D) SVA mistake: “high compression (singular) . . . ignite (plural).” This is incorrect.
(E) Correct and elegant. This uses the sophisticated idiom “it is A that does X.”
(E) is much better than (C) and is the best answer.
2) The famous red AEC Routemaster double-decker buses!
(A) this is not grammatically wrong, but it is wordy and awkward; we dearly hope to find a better answer than this so we don’t have to settle for this pathetic loser!
(B) “still remains” is redundant. The GMAT has zero toleration for redundancy. This is incorrect.
(C) the famous missing verb mistake! This is incorrect.
(D) elegant, direct, flawless.
(E) “consider X to be” is a idiom mistake. Also, this is wordy and awkward. This is incorrect.
The best answer by far is (D)
3) This question is about a sad chapter in American history. The historic Montgomery Bus Boycott was not followed immediately by more progress; instead, there was the “three steps forward, two steps back” pattern has been characteristic of civil rights progress up to this day.
The answers are long and have to be considered separately.
(A) This choice uses the “with” + [noun] + [participle] structure in a construction that incorrectly replaces a full clause. This is incorrect.
(B) No obvious flaws. This is promising.
(C) The “because of this” in this choice is not an appropriate way to express the logical contrast in the prompt, so this changes the meaning. Also, “the experiencing” is very awkward. This is incorrect.
(D) This also changes the meaning: it wasn’t “because of the Montgomery Bus Boycott” that something was incorrectly believed. This also lacks the strong logical contrast in the prompt. This is incorrect.
(E) the famous missing verb mistake! This is incorrect.
The only possible answer is (B).
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]]>Since the GMAT is typically used for admissions to American universities, it is conducted only in English. It is therefore more challenging for non-native English speakers to take the GMAT than it is for native English speakers. Non-native speakers must navigate the same challenges that English speakers encounter, but in a second language.
There’s no getting around this: if your English is weak, everything you do in English will suffer. So there’s just no “easy” route to scoring well on the GMAT without a certain level of English fluency. Furthermore, your end goal is, presumably, to study at a graduate school in an English-speaking country, most likely the United States. If your English isn’t strong enough for the GMAT, how are you going to succeed in your graduate studies?! It won’t be easy!
That said, there are things you can do to improve your English in a measurable way, so don’t get discouraged!
If immersing yourself in English on a daily basis isn’t a viable option, then you might want to take ESL courses. Again, it just isn’t possible to perform well on the GMAT if your English isn’t up to the task.
If you’re having trouble improving your English, you might consider first focusing solely on improving your English before you even start studying for the GMAT. Depending on your financial means, there are many ESL course options, including:
Always carry a dictionary! One of the key components of fluency that students often lack is a rich English vocabulary. If you find yourself using only “simple” words, then you probably need to enrich your vocabulary.
You can do this in a variety of ways. When you see a word you don’t understand, no matter where you see it, look it up in a dictionary right away. Make a flashcard with the definition. Memorize it. But also use it. Use the new word in speech and writing. Look for examples of how it’s used. Using these new words “in context” is a much better way of truly understanding the definition of a word. It’s better than just memorizing a definition on a flashcard. Sometimes, a single definition just doesn’t capture the nuance of a word’s meaning or when the word should be employed.
If you’re applying to a quant-heavy program and are thinking about lightly studying or skipping the verbal section entirely, then think again. Even if you’re applying to quant-heavy programs (engineering, math, etc.), the verbal section of the GMAT is almost always still important to admissions.
But good verbal skills are also important for the quant section. You’ll have to decipher complex sets of instructions, and it’s easy to miss a small detail if your verbal skills aren’t up to par. This is especially true when reading charts and graphs. Oftentimes, information in the “fine print” will clarify what a graph means. If you don’t fully understand that fine print, you’ll miss important info.
In short, even if you’re applying to engineering or math or other STEM programs, you’re still going to need strong verbal skills to succeed on the quant portions of the GMAT.
In some ways, people for whom English is a second language actually have an advantage on the GMAT. Many native English speakers have not thoroughly memorized grammar rules because they learned English by ear, in childhood. Native speakers of a language usually don’t learn that language by studying and practicing grammar rules.
Non-native speakers, by contrast, typically do learn English by rehearsing grammar rules, and thus tend to be very well versed in grammar. The GMAT thus provides ample opportunities for non-native English speakers to use a skill they have honed more sharply than many lifelong Anglophones. Unlike (surprisingly many!) native speakers, you know grammar rules and can identify and fix erroneous sentences. Use this to your advantage!
You’re not alone: plenty of students for whom English is a second language take the GMAT. Only about one third of all GMAT test contenders come from the United States, so it’s quite possible that you’re in the majority if you’re taking the GMAT as a non-native speaker. While taking the GMAT as a non-native speakers requires solid English skills, if your English is already strong, you shouldn’t have to study any differently than a native speaker. Remember to keep your advantages over native speakers in mind. On grammar issues, for example, you might know more than a native speaker!
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]]>First, consider these three practice questions.
1. In the equation above, x =
2. Triangle ABC is an equilateral triangle with an altitude of 6. What is its area?
3. In the equation above, x =
The second one throws in a little geometry. You may want to review the properties of the 30-60-90 Triangle and the Equilateral Triangle if those are unfamiliar. The first one is just straightforward arithmetic. The third is quite hard. For any of these, it may well be that, even if you did all your multiplication and division correctly, you wound up with an answers of the form —something divided by the square root of something—and you are left wondering: why doesn’t this answer even appear among the answer choices? If this has you befuddled, you have found exactly the right post.
When we first met fractions, in our tender prepubescence, both the numerators and denominators were nice easy positive integers. As we now understand, any kind of real number, any number on the entire number line, can appear in the numerator or denominator of a fraction. Among other things, radicals—that is, square-root expressions—can appear in either the numerator or denominator. There’s no particular issue if we have the square-root in a numerator. For example,
is a perfectly good fraction. In fact, those of you who ever took trigonometry might even recognize this special fraction. Suppose, though, we have a square root in the denominator: what then? Let’s take the reciprocal of this fraction.
This is no longer a perfectly good fraction. Mathematically, this is a fraction “in poor taste”, because we are dividing by a square root. This fraction is crying out for some kind of simplification. How do we simplify this?
By standard mathematical convention, a convention the GMAT follows, we don’t leave square-roots in the denominator of a fraction. If a square-root appears in the denominator of a fraction, we follow a procedure called rationalizing the denominator.
We know that any square root times itself equals a positive integer. Thus, if we multiplied a denominator of the square root of 3 by itself, it would be 3, no longer a radical. The trouble is—we can’t go around multiplying the denominator of fractions by something, leaving the numerator alone, and expect the fraction to maintain its value. BUT, remember the time-honored fraction trick—we can always multiply a fraction by A/A, by something over itself, because the new fraction would equal 1, and multiplying by 1 does not change the value of anything.
Thus, to simplify a fraction with the square root of 3 in the denominator, we multiply by the square root of 3 over the square root of 3!
That last expression is numerically equal to the first expression, but unlike the first, it is now in mathematical “good taste”, because there’s no square root in the denominator. The denominator has been rationalized (that is to say, the fraction is now a rational number).
Sometimes, some canceling occurs between the number in the original numerator and the whole number that results from rationalizing the denominator. Consider the following example:
That pattern of canceling in the simplification process may give you some insight into practice problem #1 above.
This is the next level of complexity when it comes to dividing by square roots. Suppose we are dividing a number by an expression that involves adding or subtracting a square root. For example, consider this fraction:
This is a fraction in need of rationalization. BUT, if we just multiply the denominator by itself, that WILL NOT eliminate the square root — rather, it will simply create a more complicated expression involving a square root. Instead, we use the difference of two squares formula, = (a + b)(a – b). Factors of the form (a + b) and (a – b) are called conjugates of one another. When we have (number + square root) in the denominator, we create the conjugate of the denominator by changing the addition sign to a subtraction sign, and then multiply both the numerator and the denominator by the conjugate of the denominator. In the example above, the denominator is three minus the square root of two. The conjugate of the denominator would be three plus the square root of two. In order to rationalize the denominator, we multiply both the numerator and denominator by this conjugate.
Notice that the multiplication in the denominator resulted in a “differences of two squares” simplification that cleared the square roots from the denominator. That final term is a fully rationalized and fully simplified version of the original.
Having read these posts about dividing by square roots, you may want to give the three practice questions at the top of this article another try, before reading the explanations below. If you have any questions on dividing by square roots or the explanations below, please ask them in the comments sections! And good luck conquering these during your GMAT!
1) To solve for x, we will begin by cross-multiplying. Notice that
because, in general, we can multiply and divide through radicals.
Cross-multiplying, we get
You may well have found this and wondered why it’s not listed as an answer. This is numerically equal to the correct answer, but of course, as this post explains, this form is not rationalized. We need to rationalize the denominator.
Answer = (D)
2) We know the height of ABC and we need to find the base. Well, altitude BD divides triangle ABC into two 30-60-90 triangles. From the proportions in a 30-60-90 triangle, we know:
Now, my predilection would be to rationalize the denominator right away.
Now, AB is simplified. We know AB = AC, because the ABC is equilateral, so we have our base.
Answer = (C)
3) We start by dividing by the expression in parentheses to isolate x.
Of course, this form does not appear among the answer choices. Again, we need to rationalize the denominator, and this case is a little trickier because we have addition in the denominator along with the square root. Here we need to find the conjugate of the denominator—changing the plus sign to a minus sign—and then multiply the numerator and denominator by this conjugate. This will result in:
Answer = (A)
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]]>The post Challenging GMAT Math Practice Questions appeared first on Magoosh GMAT Blog.
]]>
1) Let abcd be a general four-digit number and all the digits are non-zero. How many four-digits numbers abcd exist such that the four digits are all distinct and such that a + b + c = d?
(A) 6
(B) 7
(C) 24
(D) 36
(E) 42
2) Let abcd be a general four-digit number. How many odd four-digits numbers abcd exist such that the four digits are all distinct, no digit is zero, and the product of a and b is the two digit number cd?
(A) 4
(B) 6
(C) 12
(D) 24
(E) 36
3) There are 500 cars on a sales lot, all of which have either two doors or four doors. There are 165 two-door cars on the lot. There are 120 four-door cars that have a back-up camera. Eighteen percent of all the cars with back-up cameras have standard transmission. If 40% of all the cars with both back-up cameras and standard transmission are two-door cars, how many four-door cars have both back-up cameras and standard transmission?
(A) 18
(B) 27
(C) 36
(D) 45
(E) 54
4) At Mnemosyne Middle School, there are 700 students: all the students are boys or girls in the 4^{th} or 5^{th} grade. There are 320 students in the 4^{th} grade, and there are 210 girls in the 5^{th} grade. Fifty percent of the 5^{th} graders and 40% of the 4^{th} graders take Mandarin Chinese. Ninety 5^{th} grade boys do not take Mandarin Chinese. The number of 4^{th} grade girls taking Mandarin Chinese is less than half of the number of 5^{th} grade girls taking Mandarin Chinese. Which of the following could be the number of 4^{th} grade boys in Mandarin Chinese?
(A) 10
(B) 40
(C) 70
(D) 100
(E) 130
5) A hundred identical cubic boxes are currently arranged in four cubes: a single cubic box, a 2 x 2 x 2 cube, a 3 x 3 x 3 cube, and a 4 x 4 x 4 cube. These four are not touching each other. All outward faces are painted and all inward faces are not painted. These four cubes are going to be dismantled and reassembled as a flat 10 x 10 square. The top and all the edges of this 10 x 10 square must be painted, but there is no requirement for paint on the bottom. How many individual faces will have to be painted to accommodate the requirements of this new design?
(A) 0
(B) 5
(C) 9
(D) 16
(E) 27
6) Twelve points are spaced evenly around a circle, lettered from A to L. Let N be the total number of isosceles triangles, including equilateral triangles, that can be constructed from three of these points. A different orientation of the same lengths counts as a different triangle, because a different combination of points form the vertices. What is the value of N?
(A) 48
(B) 52
(C) 60
(D) 72
(E) 120
7) Theresa is a basketball player practicing her free throws. On her first free throw, she has a 60% chance of making the basket. If she has just made a basket on her previous throw, she has a 80% of making the next basket. If she has just failed to make a basket on her previous throw, she has a 40% of making the next basket. What is the probability that, in five throws, she will make at least four baskets?
8) Suppose a “Secret Pair” number is a four-digit number in which two adjacent digits are equal and the other two digits are not equal to either one of that pair or each other. For example, 2209 and 1600 are “Secret Pair” numbers, but 1333 or 2552 are not. How many “Secret Pair” numbers are there?
(A) 720
(B) 1440
(C) 1800
(D) 1944
(E) 2160
9) In the coordinate plane, a circle with its center on the negative x-axis has a radius of 12 units, and passes through (0, 6) and (0, – 6). What is the area of the part of this circle in the first quadrant?
10) In the coordinate plane, line L passes above the points (50, 70) and (100, 89) but below the point (80, 84). Which of the following could be the slope of line L?
(A) 0
(B) 1/2
(C) 1/4
(D) 2/5
(E) 6/7
11) At the beginning of the year, an item had a price of A. At the end of January, the price was increased by 60%. At the end of February, the new price was decreased by 60%. At the end of March, the new price was increased by 60%. At the end of April, the new price was decreased by 60%. On May 1^{st}, the final price was approximately what percent of A?
(A) 41%
(B) 64%
(C) 100%
(D) 136%
(E) 159%
12) Suppose that, at current exchange rates, $1 (US) is equivalent to Q euros, and 1 euro is equivalent to 7Q Chinese Yuan. Suppose that K kilograms of Chinese steel, worth F Chinese Yuan per kilogram, sold to a German company that paid in euros, can be fashioned into N metal frames for chairs. These then are sold to an American company, where plastic seats & backs will be affixed to these frames. If the German company made a total net profit of P euros on this entire transaction, how much did the US company pay in dollars for each frame?
13) At the Zamenhof Language School, at least 70% of the students take English each year, at least 40% take German each year, and between 30% and 60% take Italian each year. Every student must take at least one of these three languages, and no student is allowed to take more than two languages in the same year. What is the possible percentage range for students taking both English and German in the same year?
(A) 0% to 70%
(B) 0% to 100%
(C) 10% to 70%
(D) 10% to 100%
(E) 40% to 70%
14) On any given day, the probability that Bob will have breakfast is more than 0.6. The probability that Bob will have breakfast and will have a sandwich for lunch is less than 0.5. The probability that Bob will have breakfast or will have a sandwich for lunch equals 0.7. Let P = the probability that, on any given day, Bob will have a sandwich for lunch. If all the statements are true, what possible range can be established for P?
(A) 0 < P < 0.6
(B) 0 ≤ P < 0.6
(C) 0 ≤ P ≤ 0.6
(D) 0 < P < 0.7
(E) 0 ≤ P < 0.7
(A) – 64
(B) – 7
(C) 38
(D) 88
(E) 128
Explanations for this problem are at the end of this article.
Here are twenty-eight other articles on this blog with free GMAT Quant practice questions. Some have easy questions, some have medium, and few have quite challenging questions.
1) GMAT Geometry: Is It a Square?
2) GMAT Shortcut: Adding to the Numerator and Denominator
3) GMAT Quant: Difficult Units Digits Questions
4) GMAT Quant: Coordinate Geometry Practice Questions
5) GMAT Data Sufficiency Practice Questions on Probability
6) GMAT Quant: Practice Problems with Percents
7) GMAT Quant: Arithmetic with Inequalities
8) Difficult GMAT Counting Problems
9) Difficult Numerical Reasoning Questions
10) Challenging Coordinate Geometry Practice Questions
11) GMAT Geometry Practice Problems
12) GMAT Practice Questions with Fractions and Decimals
13) Practice Problems on Powers and Roots
14) GMAT Practice Word Problems
15) GMAT Practice Problems: Sets
16) GMAT Practice Problems: Sequences
17) GMAT Practice Problems on Motion
18) Challenging GMAT Problems with Exponents and Roots
19) GMAT Practice Problems on Coordinate Geometry
20) GMAT Practice Problems: Similar Geometry Figures
20) GMAT Practice Problems: Variables in the Answer Choices
21) Counting Practice Problems for the GMAT
22) GMAT Math: Weighted Averages
23) GMAT Data Sufficiency: More Practice Questions
24) Intro to GMAT Word Problems, Part I
25) GMAT Data Sufficiency Geometry Practice Questions
26) GMAT Data Sufficiency Logic: Tautological Questions
27) GMAT Quant: Rates and Ratios
28) Absolute Value Inequalities
These are hard problems. When you read the solutions, don’t merely read them passively. Study the strategies used, and do what you can to retain them. Learn from your mistakes!
1) We need sets of three distinct integers {a, b, c} that have a sum of one-digit number d. There are seven possibilities:
For each set, the sum-digit has to be in the one’s place, but the other three digits can be permutated in 3! = 6 ways in the other three digits. Thus, for each item on that list, there are six different possible four-digit numbers. The total number of possible four-digit numbers would be 7*6 = 42. Answer = (E)
2) The fact that abcd is odd means that cd must be an odd number and that a & b both must be odd. That limits the choices significantly. We know that neither a nor b can equal 1, because any single digit number times 1 is another single digit number, and we need a two-digit product—there are no zeros in abcd. We also know that neither a nor b can equal 5, because any odd multiple of 5 ends in 5, and we would have a repeated digit: the requirement is that all four digits be distinct.
Therefore, for possible values for a & b, we are limited to three odd digits {3, 7, 9}. We can take three different pairs, and in each pair, we can swap the order of a & b. Possibilities:
Those six are the only possibilities for abcd.
Answer = (B)
3) Total number of cars = 500
2D cars total = 165, so
4D cars total = 335
120 4D cars have BUC
“Eighteen percent of all the cars with back-up cameras have standard transmission.”
18% = 18/100 = 9/50
This means that the number of cars with BUC must be a multiple of 50.
How many 2D cars can we add to 120 4D cars to get a multiple of 50? We could add 30, or 80, or 130, but after that, we would run out of 2D cars. These leaves three possibilities for the total number with BUC:
If a total of 150 have BUC, then 18% or 27 of them also have ST.
If a total of 200 have BUC, then 18% or 36 of them also have ST.
If a total of 250 have BUC, then 18% or 45 of them also have ST.
Then we are told: “40% of all the cars with both back-up cameras and standard transmission are two-door car.”
40% = 40/100 = 2/5
This means that number of cars with both back-up cameras and standard transmission must be divisible by 5. Of the three possibilities we have, only the third words.
Total cars with BUC cams = 250 (120 with 4D and 130 with 2D)
18% or 45 of these also have ST.
40% of that is 18, the number of 2D cars with both BUC and ST.
Thus, the number of 4D cars with both BUC and ST would be
45 – 18 = 27
Answer = (B)
4) 700 student total
4G = total number of fourth graders
5G = total number of fifth graders
We are told 4G = 320, so 5G = 700 – 320 = 380
5GM, 5GF = fifth grade boys and girls, respectively
We are told 5GF = 210, so 5GM = 380 – 210 = 170
4GC, 5GC = total number of 4^{th} or 5^{th} graders, respectively taking Chinese
We are told
5GC = 0.5(5G) = 0.5(380) = 190
4GC = 0.4(4G) = 0.4(320) = 128
4GFM, 4GMC, 5GFC, 5GMC = 4^{th}/5^{th} grade boys & girls taking Chinese
We are told that, of the 170 fifth grade boys, 90 do not take Chinese, so 170 = 90 = 80 do. Thus 5GMC = 80.
5GMC + 5GFC = 5GC
80 + 5GFC = 190
5GFC = 110
We are told:
4GFM < (0.5)(5GFC)
4GFM < (0.5)(100)
4GFM < 55
Thus, 4GFM could be as low as zero or as high as 54.
4GMC = 4GC – 4GFM
If 4GFM = 0, then 4GMC = 128 – 0 = 128
If 4GFM = 54, then 4GMC = 128 – 54 = 74
Thus, fourth grade boys taking Mandarin Chinese could take on any value N, such that 74 ≤ N ≤ 128. Of the answer choices listed, the only one that works is 100.
Answer = (D)
5) The single cube has paint on all six sides. Each of the eight boxes in the 2 x 2 x 2 cube has paint on three sides (8 corner pieces). In the 3 x 3 x 3 cube, there are 8 corner pieces, 12 edge pieces (paint on two sides), 6 face pieces (paint on one side), and one interior piece (no paint). In the 4 x 4 x 4 cube, there are 8 corner pieces, 24 edge pieces, 24 face pieces, and 8 interior pieces. This chart summarizes what we have:
For the 10 x 10 flat square, we will need 4 corner pieces that have paint on three sides, 32 edge pieces that have paint on two sides (top & side), and 64 middle pieces that have paint on one side (the top).
We could use either the single total box or any of the 24 corner boxes for the four corners of the square. That leaves 21 of these, and 36 edge boxes, more than enough to cover the 32 edges of the square. The remaining ones, as well as all 30 face boxes, can be turned paint-side-up to fill in the center. The only boxes that will need to be painted, one side each, are the 9 interior boxes. Thus, we have 9 sides to paint.
Answer = (C)
6) Here’s a diagram.
First, let’s count the equilateral triangles. They are {AEI, BFJ, CGK, DHL}. There are only four of them.
Now, consider all possible isosceles triangles, excluding equilateral triangles, with point A as the vertex. We could have BAL, CAK, DAJ, and FAH. All four of those have a line of symmetry that is vertical (through A and G). Thus, we could make those same four triangles with any other point as the vertex, and we would never repeat the same triangle in the same orientation. That’s 4*12 = 48 of these triangles, plus the 4 equilaterals, is 52 total triangles.
Answer = (B)
7) There are five basic scenarios for this:
Case I: (make)(make)(make)(make)(any)
If she makes the first four, then it doesn’t matter if she makes or misses the fifth!
Case II: (miss)(make)(make)(make)(make)
Case III: (make)(miss)(make)(make)(make)
Case IV: (make)(make)(miss)(make)(make)
Case V: (make)(make)(make)(miss)(make)
Put in the probabilities:
Case I: (0.6)(0.8)(0.8)(0.8)
Case II: (0.4)(0.4)(0.8)(0.8)(0.8)
Case III: (0.6)(0.2)(0.4)(0.8)(0.8)
Case IV: (0.6)(0.8)(0.2)(0.4)(0.8)
Case V: (0.6)(0.8)(0.8)(0.2)(0.4)
Since all the answers are fractions, change all of those to fractions. Multiply the first by (5/5) so it has the same denominator as the other products.
Case I: (3/5)(4/5)(4/5)(4/5)(5/5) = 960/5^5
Case II: (2/5)(2/5)(4/5)(4/5)(4/5) = 256/5^5
Case III: (3/5)(1/5)(2/5)(4/5)(4/5) = 96/5^5
Case IV: (3/5)(4/5)(1/5)(2/5)(4/5) = 96/5^5
Case V: (3/5)(4/5)(4/5)(1/5)(2/5) = 96/5^5
Add the numerators. Since 96 = 100 – 4, 3*96 = 3(100 – 4) = 300 – 12 = 288.
288 + 256 + 960 = 1504
P = 1504/5^5
Answer = (E)
8) There are three cases: AABC, ABBC, and ABCC.
In case I, AABC, there are nine choices for A (because A can’t be zero), then 9 for B, then 8 for C. 9*9*8 = 81*8 = 648.
In case II, ABBC, there are 9 choices for A, 9 for B, and 8 for C. Again, 648.
In case III, ABCC, there are 9 choices for A, 9 for B, and 8 for C. Again, 648.
48*3 = (50 – 2)*3 = 150 – 6 = 144
3*648 = 3(600 + 48) = 1800 + 144 = 1948
Answer = (D)
9)
We know that the distance from A (0,6) to B (0, – 6) is 12, so triangle ABO is equilateral. This means that angle AOB is 60°. The entire circle has an area of
A 60° angle is 1/6 of the circle, so the area of sector AOB (the “slice of pizza” shape) is
The area of an equilateral triangle with side s is
Equilateral triangle AOB has s = 12, so the area is
If we subtract the equilateral triangle from the sector, we get everything to the right of the x-axis.
Again, that’s everything to the right of the x-axis, the parts of the circle that lie in Quadrants I & IV. We just want the part in Quadrant I, which would be exactly half of this.
Answer = (C)
10) One point is (50, 70) and one is (100, 89): the line has to pass above both of those. Well, round the second up to (100, 90)—if the line goes above (100, 90), then it definitely goes about (100, 89)!
What is the slope from (50, 70) to (100, 90)? Well, the rise is 90 – 70 = 20, and the run is 100 – 50 = 50, so the slope is rise/run = 20/50 = 2/5. A line with a slope of 2/5 could pass just above these points.
Now, what about the third point? For the sake of argument, let’s say that the line has a slope of 2/5 and goes through the point (50, 71), so it will pass above both of the first two points. Now, move over 5, up 2: it would go through (55, 73), then (60, 75), then (65, 77), then (70, 79), then (75, 81), then (80, 83). This means it would pass under the third point, (80, 84). A slope of 2/5 works for all three points.
We don’t have to do all the calculations, but none of the other slope values works.
Answer = (D)
11) The trap answer is 100%: a percent increase and percent decrease by the same percent do not cancel out.
Let’s say that the A = $100 at the beginning of the year.
End of January, 60% increase. New price = $160
End of February, 60% decrease: that’s a decrease of 60% of $160, so that only 40% of $160 is left.
10% of $160 = $16
40% of $160 = 4(16) = $64
That’s the price at the end of February.
End of March, a 60% increase: that’s a increase of 60% of $64.
10% of $64 = $6.40
60% of $64 = 6(6 + .40) = 36 + 2.4 = $38.40
Add that to the starting amount, $64:
New price = $64 + $38.40 = $102.40
End of April, 60% decrease: that’s a decrease of 60% of $102.40, so that only 40% of $102.40 is left.
At this point, we are going to approximate a bit. Approximate $102.40 as $100, so 40% of that would be $40. The final price will be slightly more than $40.
Well, what is slightly more than $40, as a percent of the beginning of the year price of $100? That would be slightly more than 40%.
Answer = (A)
12) The K kilograms, worth F Chinese Yuan per kilogram, are worth a total of KF Chinese Yuan. The German company must pay this amount.
Since 1 euro = (7Q) Chinese Yuan, then (1/(7Q)) euro = 1 Chinese Yuan, and (KF/7Q) euros = KF Chinese Yuan. That’s the amount that the Germans pay to the Chinese.
That is the German company’s outlay, in euros. Now, they make N metal chairs, and sell them, making a gross profit of P euros.
That must be the total revenue of the German company, in euros. This comes from the sale to the American company. Since $1 = Q euros, $(1/Q) = 1 euro, so we change that entire revenue expression to euros to dollars, we divide all terms by Q.
That must be the total dollar amount that leaves the American company and goes to the German company. This comes from the sale of N metal frames for chairs, so each one must have been 1/N of that amount.
Answer = (A)
13) First, we will focus on the least, the lowest value. Suppose the minimum of 70% take English, and the minimum of 40% take German. Even if all 30% of the people not taking English take German, that still leaves another 10% of people taking German who also have to be taking English. Thus, 10% is the minimum of this region.
Now, the maximum. Both the German and English percents are “at least” percents, so either could be cranked up to 100%. The trouble is, though, that both can’t be 100%, because some folks have to take Italian, and nobody can take three languages at once. The minimum taking Italian is 30%. Let’s assume all 100% take German, and that everyone not taking Italian is taking English: that’s 70% taking English, all of whom also would be taking German. Thus, 70% is the maximum of this region.
Answer = (C)
14) Let A = Bob eats breakfast, and B = Bob has a sandwich for lunch. The problem tells us that:
P(A) > 0.6
P(A and B) < 0.5
P(A or B) = 0.7
First, let’s establish the minimum value. If Bob never has a sandwich for lunch, P(B) = 0, then it could be that P(A and B) = 0, which is less than 0.5, and it could be that P(A) = 0.7, which is more than 0.6, so that P(A or B) = 0.7. All the requirements can be satisfied if P(B) = 0, so it’s possible to equal that minimum value.
Now, the maximum value. Since P(A or B) = 0.7, both P(A) and P(B) must be contained in this region. See the conceptual diagram.
The top line, 1, is the entire probability space. The second line, P(A or B) = 0.7, fixes the boundaries for A and B. P(A) is the purple arrow, extending from the right. P(B) is the green arrow extending from the left. The bottom line, P(A and B) < 0.5, is the constraint on their possible overlap.
Let’s say that P(A) is just slightly more than 0.6. That means the region outside of P(A), but inside of P(A or B) is slightly less than 1. That’s the part of P(B) that doesn’t overlap with P(A). Then, the overlap has to be less than 0.5. If we add something less than 1 to something less than 5, we get something less than 6. P(B) can’t equal 0.6, but it can any value arbitrarily close to 0.6.
Thus, 0 ≤ P(B) < 0.6.
Answer = (B)
15)
Answer = (E)
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]]>The post Matemática GMAT: Decimais Exatos e Periódicos appeared first on Magoosh GMAT Blog.
]]>Inteiros são números positivos e negativos reais, incluindo o zero. Aqui estão alguns inteiros:
{ … -3, -2, -1, 0, 1, 2, 3, …}
Quando fazemos uma razão entre dois inteiros, nós temos um número racional. Um número racional é qualquer número no formato a/b, onde a e b são inteiros e b ≠ 0. Números racionais são o conjunto de todas frações feitas com fatores inteiros. Perceba que todos os inteiros estão incluídos no conjunto de números racionais, pois, por exemplo, 3/1 = 3.
Quando fazemos um decimal de uma fração, uma das duas coisas acontece. Ou termina (decimal exato) ou repete (continua para sempre em um padrão, chamados de decimais periódicos). Números racionais exatos incluem:
1/2 = 0.5
1/8 = 0.125
3/20 = 0.15
9/160 = 0.05625
Números racionais periódicos incluem:
1/3 = 0.333333333333333333333333333333333333…
1/7 = 0.142857142857142857142857142857142857…
1/11 = 0.090909090909090909090909090909090909…
1/15 = 0.066666666666666666666666666666666666…
O GMAT não dará uma fração complicada como 9/160 e esperará que você descubra qual sua expressão decimal. MAS, o GMAT pode fornecer uma fração como 9/160 e perguntar se é exato ou não. Mas como saber?
Bem, primeiro de tudo, qualquer terminação decimal (como 0.0376) é, essencialmente, uma fração com uma potência de dez no denominador. Por exemplo, 0.0376 = 376/10000 = 47/1250. Note que simplificamos esta fração, dividindo o numerador por 8. O dez é múltiplo de 2 e 5, então qualquer potência de dez irá ser potência de 2 e de 5, e algumas podem ser canceladas por fatores no numerador, mas nenhum outro fator será introduzido no denominador. Então, se a fatoração primária do denominador de uma fração possui apenas múltiplos de 2 e múltiplos de 5, então pode ser escrita como algo com potência de 10, o que significa que sua expressão decimal será exata.
Se a fatoração primária do denominador de uma fração possui apenas múltiplos de 2 e de 5, as expressões decimais são exatas. Se há algum fator primário no denominador que não seja 2 ou 5, então a expressão decimal é periódica. Deste modo,
1/24 é periódica (há um múltiplo de 3)
1/25 é exata (apenas múltiplos de 5)
1/28 é periódica (há um múltiplo de 7)
1/32 é exata (apenas múltiplos de 2)
1/40 é exata (apenas múltiplos de 2 e 5)
Note que, contanto que a fração esteja nos seus menores termos, o numerador não importa. Já que 1/40 é exata, então 7/40, 13/40 ou qualquer outro inteiro sobre 40 também é.
Já que 1/28 é periódica, então 5/28 e 15/28 e 25/28 também são. Note, entretanto, que 7/28 não é periódica por causa do cancelamento: 7/28 = 1/4 = 0.25.
Há alguns decimais que são úteis como atalhos, tanto para conversões de fração-para-decimal quanto para conversões de fração-para-porcentagem. Esses são
1/2 = 0.5
1/3 = 0.33333333333333333333333333…
2/3 = 0.66666666666666666666666666…
1/4 = 0.25
3/4 = 0.75
1/5 = 0.2 (e vezes 2, 3 e 4 para decimais fáceis)
1/6 = 0.166666666666666666666666666….
5/6 = 0.833333333333333333333333333…
1/8 = 0.125
1/9 = 0.111111111111111111111111111… (e vezes outros dígitos para outros decimais fáceis)
1/11 = 0.09090909090909090909090909… (e vezes outros dígitos para outros decimais fáceis)
Há outra categoria de decimais periódicos (que continuam para sempre) e eles não têm padrões de repetição. Esses números, os decimais periódicos não repetitivos, são chamados de números irracionais. É impossível escrever qualquer forma deles como razão de dois inteiros. O Sr. Pitágoras (c. 570 – c. 495 aC) foi o primeiro a provar um número irracional: ele provou que a raiz quadrada de 2 — — é irracional. Nós todos sabemos: toda raíz quadrada de um inteiro cuja solução não é outro número inteiro é irracional Outro número racional famoso é o , ou pi, a razão da circunferência de um círculo pelo seu diâmetro. Por exemplo,
= 3.1415926535897932384626433832795028841971693993751058209749445923078164
0628620899862803482534211706798214808651328230664709384460955058223172535940812848111745
0284102701938521105559644622948954930381964428810975665933446128475648233786783165271201
909145648566923460348610454326648213393072602491412737…
Esses são os primeiros trezentos dígitos do pi, e eles nunca se repetem: continuam para sempre sem padrões repetitivos. Há uma infinidade de outros números irracionais: na verdade, a infinidade de números irracionais é infinitamente maior que a infinidade de números racionais, mas isto leva a uma matemática (http://en.wikipedia.org/wiki/Aleph_number) que é muito mais avançada que a do GMAT.
1)
(A) 2/27
(B) 3/2
(C) 3/4
(D) 3/8
(E) 9/16
1) A partir dos nossos atalhos, nós sabemos que 0.166666666666… = 1/6, e 0.444444444444… = 4/9. Portanto (1/6)*(9/4) = 3/8. Resposta = D
Esta postagem apareceu originalmente em inglês no Magoosh blog e foi traduzida por Jonas Lomonaco.
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]]>2) Suponha que você jogue uma moeda não-viciada seis vezes. Qual é a probabilidade de, em seis jogadas, sair pelo menos uma cara?
3) Em um certo jogo, você escolhe uma carta de um baralho padrão de 52 cartas. Se a carta é de copas, você ganha. Se não for de copas, a pessoa a repõe no bolo de cartas, embaralha e puxa novamente. Esse processo é repetido até sair uma de copas, e a questão é calcular: quantas vezes a pessoa precisa puxar antes que consiga uma carta de copas e ganhe? Qual é a probabilidade de puxar pelo menos duas cartas que não sejam de copas nas primeiras duas jogadas, e só pegar a primeira carta de copas pelo menos na terceira jogada?
Há uma regra muito simples e importante relacionando P(A) e P(A’), conectando a probabilidade de qualquer evento acontecer com a probabilidade daquele mesmo evento não acontecer. Para qualquer evento bem definido, é 100% verdade que este pode ou não acontecer. O GMAT não irá fazer perguntas de probabilidade sobre eventos bizarros dos quais, por exemplo, você não saberia dizer se aconteceu ou não, ou eventos complexos que podem, de alguma forma, tanto acontecer quanto não acontecer. Para qualquer evento A em uma questão de probabilidade no GMAT, os dois cenários “A acontece” e “A não acontece” esgotam as possibilidades que podem acontecer. Com certeza, nós podemos falar: um dos dois irá ocorrer. Em outras palavras
P(A OU A’) = 1
Ter a probabilidade de 1 significa certeza garantida. Obviamente, por uma variedade de razões lógicas profundas, os eventos “A” e “não A (ou A’)” são desconexos e não têm sobreposição. A regra do OU, discutida no último post, implica:
P(A) + P(A’) = 1
Subtraia o primeiro termo para isolar P(A’).
P(A’) = 1 – P(A)
Isto é conhecido em probabilidade como regra de um evento complementar, porque a região probabilística na qual um evento não acontece complementa a região na qual ele ocorre. Esta é a ideia crucial no geral, para todas as questões de probabilidade do GMAT, que será muito importante para resolver questões de “pelo menos” em particular.
Suponha que um evento A é uma afirmação envolvendo palavras de “pelo menos” – o que afirmaria os constituintes de “não A”? Em outras palavras, como negamos uma afirmação “pelo menos”? Vamos ser concretos. Suponha que há alguns eventos que envolvam apenas dois resultados: sucesso e fracasso. O evento pode ser, por exemplo, fazer um arremesso de basquete ou jogar uma moeda e tirar cara. Agora, suponha que temos uma “disputa” envolvendo dez desses eventos consecutivos, e nós estamos contando o número de sucessos dessas dez tentativas. O evento A será definido como: A = “há pelo menos 4 sucessos nessas dez tentativas.” Que resultados iriam constituir “não A”? Bem, vamos pensar sobre isso. Em dez tentativas, uma pode ser de nenhum sucesso, exatamente um sucesso, exatamente dois sucessos, até dez sucessos. Há onze resultados possíveis, os números de 0 – 10, para o número de sucessos que pode ocorrer em 10 tentativas. Considere o seguinte diagrama de números de sucessos possíveis em dez tentativas.
Os números em roxo são membros de A, “pelo menos 4 sucessos” em dez tentativas. Portanto, os números verdes são os espaços complementares, a região de “não A”. Em palavras, como iríamos descrever as condições que lhe colocariam na região verde? Nós podemos dizer: “não A” = “três sucessos ou menos” em dez tentativas. A negação, o oposto, de “pelo menos quatro” é “três ou menos”.
Abstraindo disso, a negação ou oposto de “pelo menos n” é a condição “(n – 1) ou menos”. Um caso particularmente interessante é n = 1: a negação ou o oposto de “pelo menos um” é “nenhum.” Esta última afirmação é uma ideia extremamente importante, indiscutivelmente a chave para resolver a maior parte das questões de “pelo menos” que você encontrará no GMAT.
A maior ideia das questões de “pelo menos” no GMAT é: sempre é mais fácil descobrir a probabilidade do complementar. Por exemplo, no cenário acima das dez tentativas de alguma coisa, calcular “pelo menos 4” diretamente iria envolver sete cálculos diferentes (para os casos de 4 a 10), enquanto calcular “três ou menos” iria envolver apenas quatro cálculos separados (para os casos de 0 a 3). No extremo – e extremamente comum – caso de “ao menos um”, a abordagem direta iria envolver o cálculo de um quase caso, mas o cálculo do complementar envolve simplesmente calcular a probabilidade para o caso de “nenhum” e então subtraí-lo de um.
P(A’) = 1 – P(A)
P(pelo menos um sucesso) = 1 – P(nenhum sucesso)
Este é um dos atalhos mais poderosos e que economizam o seu tempo em todo o GMAT.
Considere a simples questão a seguir.
4) Dois dados são arremessados. Qual é a probabilidade de se tirar um 6 em pelo menos um deles?
Acontece que calcular isso diretamente iria envolver um cálculo relativamente longo – a probabilidade de tirar exatamente um 6, em qualquer dado, e a rara probabilidade de ambos darem 6. Este cálculo poderia facilmente levar muitos minutos para ser concluído.
Em vez disso, nós iremos usar o atalho definido acima:
P(A’) = 1 – P(A)
P(pelo menos um 6) = 1 – P(nenhum 6)
Qual a probabilidade de ambos os dados darem nenhum 6? Bem, primeiro, vamos considerar apenas um dado. A probabilidade de arremessar um 6 é de 1/6, então a probabilidade de arremessar algo diferente de 6 (não 6) é 5/6.
P(dois dados, nenhum 6) = P(“não 6” no dado nº 1 E “não 6” no dado nº 2)
Como vimos no último post, a palavra E significa multiplicação. (Claramente, o resultado de cada dado é independente do outro). Então:
P(dois dados, nenhum 6) =(5/6)*(5/6) = 25/36
P(pelo menos um 6) = 1 – P(nenhum 6) = 1 – 25/36 = 11/36
O que poderia ser um cálculo bem longo tornou-seincrivelmente simples com esse atalho. Isto pode ser um enorme quebra-galho para economizar o seu tempo no GMAT!
Após ler este post, tente resolver novamente as três questões práticas acima antes de ler suas respostas abaixo. E mais, aqui está uma questão grátis, com vídeo explicativo, sobre esse mesmo tema:
5) http://gmat.magoosh.com/questions/839
O próximo artigo da série irá explorar as questões de probabilidade que envolvem técnicas de contagem.
1) P(pelo menos uma vogal) = 1 – P(nenhuma vogal)
A probabilidade de pegar uma letra que não seja uma vogal no primeiro conjunto é de 3/5. A probabilidade de não pegar nenhuma vogal no segundo conjunto é 5/6. Para não pegar vogal alguma, não podemos pegar nenhuma no primeiro conjunto E nenhuma no segundo conjunto. De acordo com a regra E, nós multiplicamos as probabilidades.
P(nenhuma vogal) = (3/5)*(5/6) = 1/2
P(pelo menos uma vogal) = 1 – P(nenhuma vogal) = 1 – 1/2 = 1/2
Resposta = C
2) P(pelo menos um H) = 1 – P(nenhum H)
Em uma jogada, P(“não H”) = P(T) = 1/2. Nós precisaríamos que isso acontecesse seis vezes – isto é, seis eventos independentes unidos pela palavra E, o qual significa que são multiplicados entre si.
Resposta = E
3) Um baralho completo de 52 cartas contém 13 cartas de cada um dos quatro naipes. A probabilidade de puxar uma carta de copas em um baralho completo é 1/4. Portanto, a probabilidade de tirar uma carta que “não seja copas” é 3/4.
P(pelo menos três jogadas para ganhar) = 1 – P(ganhar em duas ou menos jogadas)
Além disso,
P(ganhar em duas ou menos jogadas) = P(ganhar em uma jogada OU ganhar em duas jogadas)
= P(ganhar em uma jogada) + P(ganhar em duas jogadas)
Ganhar em uma jogada significa: eu seleciono uma carta do baralho e acontece de ser de copas. Acima, nós já falamos: a probabilidade disto acontecer é de 1/4.
P(ganhar em uma jogada) = 1/4
Ganhar em duas jogadas significa: minha primeira jogada era uma carta “não copas”, P = 3/4, E a segunda jogada é de copas, P = 1/4. Porque nós repomos e embaralhamos, as jogadas são independentes, então o E significa multiplicação.
P(ganhar em duas jogadas) =(3/4)*(1/4) = 3/16
P(ganhar em duas ou menos jogadas) =P(ganhar em uma jogada) + P(ganhar em duas jogadas)
= 1/4 + 3/16 = 7/16
P(pelo menos três jogadas para ganhar) = 1 – P(ganhar em duas ou menos jogadas)
= 1 – 7/16 = 9/16
Resposta = B
Esta postagem apareceu originalmente em inglês no Magoosh blog e foi traduzida por Jonas Lomonaco.
The post Matemática GMAT: a Questão de “Pelo Menos” de Probabilidade appeared first on Magoosh GMAT Blog.
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