# GRE Verbal reasoning format

The GRE Verbal Reasoning measure contains three types of questions:

• Text Completion
• Sentence Equivalence

Description: These are the traditional multiple-choice questions with five answer choices of which you must select one.

Strategy Tip:

Read all the answer choices before making your selection, even if you think you know what the answer is in advance. Don’ t be misled by answer choices that are only partially true or only partially answer the question. The correct answer is the one that most accurately and most completely answers the question posed.

Pay attention to context. When the question asks about the meaning of a word in the passage, be sure that the answer choice you select correctly represents the way the word is being used in the passage. Many words have quite different meanings in different contexts.

### Select One or More Answer Choices

Description: For the question type “Select One or More Answer Choices,” you are given three statements about a passage and asked to “indicate all that apply”. Either one, two, or all three can be correct (there is no “none of above” option). There is no partial credit; you must indicate all of the correct choices and none of the incorrect choices.

Strategy Tip: On “Select One or More Answer Choices,” don’ t let your brain be tricked into telling you, “Well, if two of them have been right so far, the other one must be wrong,” or any other arbitrary idea about how many of the choices should be correct. Make sure to consider each choice independently! You cannot use “process of elimination” in the same way as you do on normal multiple-choice questions.

### Select-in-Passage

Description: For the question type “Select-in-passage,” you are given an assignment such as “Select the sentence in the passage that explains why the experiment’ s results were discovered to be invalid”. Clicking anywhere on the sentence in the passage will highlight it. (As with any GRE question, you will have to click “Confirm” to submit your answer, so don’ t worry about accidentally selecting the wrong sentence due to a slip of the mouse.)

Strategy Tip: On “Select-in-Passage,” if the passage is short, consider numbering each sentence (i.e., writing 1 2 3 4 on your paper) and crossing off each choice as you determine that it is n’ t the answer. If the passage is long, you might write a number for each paragraph (1,2,3), and tick off each number as you determine that the correct sentence is not located in that paragraph.

### Questions 1 to 3 is based on the following reading passage.

Reviving the practice of using elements of popular music in classical composition an approach that had been in hibernation in the United States during the 1960s, composer Philip Glass (born 1937) embraced the ethos of popular music in his compositions. Glass based two symphonies on music by rock musicians David Bowie and Brian Eno, but the symphonies’ sound is distinctively his. Popular elements do not appear out of place in Glass’ s classical musical, which from its early days has shared certain harmonies and rhythms with rock music. Yet this use of popular elements has not made Glass a composer of popular music. His music is not a version of popular music packaged to attract classical listeners; it is not a version of popular music packaged to attract classical listeners; it is high art for listeners steeped in rock rather than the classics.

1. The passage addresses which of the following issues related to Glass’ s use of popular elements in his classical compositions?
2. How it is regarded by listeners who prefer rock to the classics.
3. How it has affected the commercial success of Glass’ s music.
4. Whether it has contributed to a revival of interest among other

Consider each of the three choices separately and select all that apply

1. The passage suggests that Glass’ s work displays which of the following qualities?
2. A return to the use of popular music in classical compositions.
3. An attempt to elevate rock music to an artistic status more closely approximating that of classical music
4. A long-standing tendency to incorporate elements from two apparently disparate musical styles.
5. Select the sentence that distinguishes two ways of integrating rock and classical music.

Explanation

The passage describes in general terms how Philip Glass uses popular music in his classical compositions and explores how Glass can do this without being imitative. Note that there are no opposing views discussed; the author is simply presenting his or her views.

Answer1: One of the important points that the passage makes is that when Glass uses popular elements in his music, the result is very much his own creation (it is “distinctively his”). In other words, the music is far from being derivative. Thus one issue that the passage addresses is the one referred to in answer Choice E—it answers it in the negative. The passage does not discuss the impact of Glass’ s use of popular elements on listeners, on the commercial success of his music, on other composers, nor on Glass’ s reputation, so none of Choices A through D is correct.

The correct answer is Choice E.

Answer2: To answer this question, it is important to assess each answer choice independently. Since the passage says that Glass revived the use of popular music in classical compositions, answer Choice A is clearly correct. On the other hand, the passage also denies that Glass composes popular music or packages it in a way to elevate its status, so answer Choice B is incorrect. Finally, since Glass’ s style has always mixed elements of rock with classical elements, Choice C is correct.

Thus, the correct answer is Choice A and Choice C.

Answer 3: Almost every sentence in the passage refers to incorporating rock music in classical compositions, but only the last sentence distinguishes two ways of doing so. It distinguishes between writing rock music in a way that will make it attractive to classical listeners and writing classical music that will be attractive to listeners familiar with rock.

Thus the correct answer is the last sentence of the passage.

## Verbal: Text completion questions

Text completions can consist of 1-5 sentences with 1-3 blanks. When Text completions have two or three blanks, you will select words or short phrases for those blanks independently. There is no partial credit; you must make every selection correctly.

Strategy Tip:

• Read the passage to get an overall sense of it.
• Identify words or phrases that seem particularly significant, either because they emphasize the structure of the passage (words like although or moreover) or because they are central to understanding what the passage is about.
• Think up your own words for the blanks. Try to fill in the blanks with words or phrases that seem to you to fit and then see if similar words are offered among the answer choices.
• Double-check your answers. When you have made your selection for each blank. Check to make sure that the passage is logically, grammatically, and stylistically coherent.

Sample Questions

Q 1) In parts of the Arctic, the land grades into the landfast ice so ______ that you can walk off the coast and not know you are over the hidden sea.

1. Permanently
2. Imperceptibly
3. Irregularly
4. Precariously
5. Relentlessly

Ans) “Relentlessly” means less severely or less harshly. “Precariously” means not safely or dangerously. “Imperceptibly” means impossible to see or notice.

The word that fills the blank has to characterize how the land grades into the ice in a way that explains how you can walk off the coast and over the sea without knowing it. The word that does that is “imperceptibly”; if the land grades imperceptibly into the ice, you might well not know that you had left the land.

Q 2) Leaders are not always expected to (i)_________ the same rules as are those they lead; leaders are often looked up to for a surety and presumption that would be viewed as (ii)__________ in most others.

a)Decree                    (d) Hubris

b) Proscribe               (e) Avarice

c) Conform to             (f) Anachronism

Ans) “Decree” means an official order. “Proscribe” means to make something illegal. “Hubris” means exaggerated pride or self-confidence. “Avarice” means a strong desire to have or get money. “Anachronism” means something (such as a word or an event) that is mistakenly placed with respect to time.

In the first blank, you need a word similar to “follow.” In the second blank, you need a word similar to “arrogance.” The correct answers are conform to and hubris.

Q 3) It is refreshing to read a book about our planet by an author who does not allow facts to be (i)______ by politics; well aware of the political disputes about the effects of human activities on climate and biodiversity, this author does not permit them to (ii)_________ his comprehensive description of what we know about our biosphere. He emphasizes the enormous gaps in our knowledge, the sparseness of our observations, and the (iii)_______, calling attention to the many aspects of planetary evolution that must be better understood before we can accurately diagnose the condition of our planet.

a) Overshadowed d) Enhance            g) Plausibility of our hypotheses

b) Invalidated        e) Obscure             h) Certainty of our entitlement

c) Illuminated         f) Underscore        i) Superficiality of our theories

Ans) The overall tone of the passage is clearly complimentary. To understand what the author of the book is being complimented on, it is useful to focus on the second blank. Here, we must determine what word would indicate something that the author is praised for not permitting. The only answer choice that fits the case is “obscure,” since enhancing and underscoring are generally good things to do, not things one should refrain from doing. Choosing “obscure” clarifies the choice for the first blank; the only choice that fits well with “obscure” is “overshadowed.” Notice that trying to fill blank (i) without filling blank (ii) first is very hard-each choice has at least some initial plausibility. Since the third blank requires a phrase that matches “enormous gaps” and “sparseness of our observations,” the best choice is “superficiality of our theories.”

Thus the correct answer is overshadowed (Choice A), Obscure (Choice E), and superficiality of our theories (Choice I).

## Verbal: Sentence equivalence questions

For this question type, you are given one sentence with a single blank. There are six answer choices, and you are asked to pick two choices that fit the blank and are similar in meaning.

No partial credit is given on Sentence Equivalence; both correct answers must be selected and no incorrect answers may be selected.

Here is a sample set of answer choices:

1. Tractable
2. Taciturn
3. Arbitrary
4. Tantamount
5. Reticent
6. Amenable

Tractable and amenable (tractable, amenable people will do whatever you want them to do) are similar in meaning. Taciturn and Reticent (both mean “not talkative”) also form a pair. Arbitrary (based on one’ s own will) and tantamount (equivalent) are not similar in meaning and therefore cannot be a pair. Therefore, the only possible correct answer pairs are (A) and (F), and (B) and (E). Now, you have to choose one-word group out of two.

The question is deliberately omitted here in order to illustrate how much you can do with the choices alone, if you have studied vocabulary sufficiently. “Learn words in groups” is a tip given by all GRE trainers especially to attempt sentence equivalence questions. Considering this, on our blog we have written forty-five chapters on GRE vocabulary using word groups to enhance vocabulary of students so that they can answer sentence equivalence questions comfortably.

Strategy tip: Analyze answer choices first and form possible word groups. If there is the only one-word group you can easily answer the question. If there are two or three-word groups, analyze the sentence- read for a textual clue that tells you what type of word must go in the blank. Then choose a suitable word group.

Sample Questions

Q 1) While athletes usually expect to achieve their greatest feats in their teens or twenties, opera singers don’ t reach the ________ of their vocal powers until middle age.

1. Harmony
2. Zenith
3. Acme
4. Terminus
6. Cessation

Ans) “Acme” means the highest point of something. “Zenith” means successful period or highest point. “Nadir” means lowest point of something. “Terminus” means end of something. “Cessation” means stopping of some action. “Harmony” means pleasing combination of different musical notes. “Zenith” and “Acme” form a pair. “Terminus” and “Cessation” also form a pair.

Now, analyze the sentence. We need a word similar in meaning to “greatest feats”. Therefore, correct answer choices are (B) and (C).

Most Popular Resources

GRE Argument Structure Passages

GRE Sentence Equivalence Tips

GRE Text Completion Tips

GRE Sentence Equivalence Test

# What is GRE?

The GRE (Graduate Record Examination) is an exam written and administered by the Educational Testing Service (ETS). The standardized test GRE is an admission requirement for masters and doctoral degree program from top universities of U.S.A.

The test which was completely revised in August 2011, is primarily delivered as a computer-based test and it is composed of Analytical Writing, Quantitative reasoning, and verbal reasoning.

Eligibility: Bachelor’ s degree holders and undergraduate students who are about to graduate are eligible for taking GRE. Fluency in English is assumed from students taking the GRE.

Fee: Fee to take the exam is the US $205. Restriction on attempts: GRE exam can be attempted only after 21 days from the day of the previous exam and it cannot be attempted more than five times in a year. Exam Structure: The revised test has six sections. You will get a 10-minute break between the third and fourth sections and a 1-minute break between the others. The first section is analytical writing section. The other five sections can be seen in any order and will include: • Two verbal reasoning sections (20 questions each in 30 minutes per section) • Two Quantitative Reasoning sections (20 questions each in 35 minutes per section) • Either an unscored section or a research section. An unscored section will look just like a third Verbal or Quantitative Reasoning section, and you will not be told which of them doesn’t’ count. You have to attempt all these five sections seriously. If you get a research section, it will be identified as such, and will be the last section you get.  Section Section Type Questions Time Scored? 1 Analytical Writing 2 essays 30 minutes each Yes 2 Verbal #1 Quantitative #1 (order can vary) Approx. 20 30 minutes Yes 3 Approx. 20 35 minutes Yes 10-Minute Break 4 Verbal #2 Quantitative #2 Unscored Section (Verbal or Quant) (Order can vary) Approx. 20 30 minutes Yes 5 Approx. 20 35 minutes Yes ? Approx. 20 30 or 35 minutes No Last Research Section Varies Varies No Therefore, the GRE is very much a test of endurance, pacing and mental stamina play a big role in success on the exam. Multi-Stage Test: How the GRE Adapts The GRE is a multi-stage test (MST), meaning that it adapts at the section level. Within each section, you can freely move back and forth among the questions, so you can change answers, edit responses and flag questions to answer later. The adapting occurs from section to section rather than from question to question (e.g., if you do well on the first Verbal section, you will get a harder second verbal section and if you do well on the first quant section, you will get a harder second quant section). If you do poorly on the first section, the second section will be easier, but the range of scores for which you are eligible will be limited to the low end of the scoring range. If you perform well on the first section, the second question will contain more difficult questions, but you will be eligible for a higher score. Scoring: When you take the GRE, you will receive separate scores for the Analytical Writing, Quantitative Reasoning, and Verbal Reasoning sections. The scores for the revised GRE Quantitative Reasoning and Verbal Reasoning are reported on a 130-170 scale in 1-point increments. You will receive one 130-170 score for Verbal and a separate 130-170 score for Quant. Your performance in the first section is based solely on the number of questions you answer correctly. You only score points for questions that are answered correctly – there is no penalty for answering a question incorrectly, so it’ s to your benefit to attempt an answer to every question on the test. The Analytical Writing section is scored on a scale from 0 to 6, in 0.5-point increments. GRE Subject Test Whatever we have discussed so far is about GRE general test. There are also GRE subject tests. GRE subject tests are mandatory for students who are applying for masters or doctoral degree program in following disciplines: • Physics • Chemistry • Biology • Mathematics • English Literature • Psychology If a student is applying for masters or doctoral degree program in any other discipline then there is no need to have a GRE subject test score. How much emphasis do admission officers give to GRE score? The GRE is a key factor in the graduate school admission process. But, apart from GRE, following other credentials are required to get admitted to a top school for MS: • Academic CGPA/Percentage/Rank in your batch • Undergraduate University • SOP • LOR • Research papers • Work experience In 2010, Kaplan Test Prep surveyed admission officers from the top 50 graduate programs for education, psychology, and public administration (according to US News & World Report rankings) and found that these programs’ admissions officers cited the GRE as the most important factor in graduate school admissions more often than any other factor. Therefore, GRE score is an important factor but emphasis on GRE score vary not only from university to university, also from department to department and also from program to program. For example, most of the universities require high GRE score for business and economics programs and allows slightly low score for engineering based programs. In some universities, science and engineering programs may consider only GRE quant score while programs in liberal arts topics may consider only GRE verbal score of their interest. Some universities use GRE score for admission process but not for offering the scholarship, while some universities use GRE score for offering the scholarship but not for admissions. To find out how a particular university evaluates GRE score is to contact the person in charge of admissions for that specific program. Most Popular Resources GRE Verbal Reasoning Format GRE Quantitative Reasoning Format GRE Analytical Writing ## GRE: Statistics Measures of central tendency:- The main measure of central tendency are the arithmetic mean, mode, median. The average (arithmetic mean) of a group of number is defined as the sum of the values divided by the number of values. AVERAGE VALUE = Sum of values ÷ Number of values MEDIAN: When there are odd numbers of terms in a set, the median of a set is the middle of terms in a set, median of a set is the middle term.When all the terms in a set are listed in sequential order. When there are even number of terms in a set, the median of of a set is the average of two middle terms. MODE:- Mode is the number that appears most frequently in a list of numbers. MEASURES OF DISPERSION:- The main measures of dispersion are the range, interquartile range and standard deviation. First Quartile, Q is the median of all the numbers below the median. The second quartile, Q2 is the median of the entire data set. The third Quartile, Q3 is the median of all the numbers above the median. The interquartile range is the difference between Q3 and Q1. The range and interquartile range are often displayed on a Box AND Whisker plot also called a Box Plot. Problem 1) For the following Set : 2,2,3,4,5,5,6,21 Calculate Q1, Q2, Q3, interquartile range. Plot box and whisker plot for the following data? Sol. Median = Mean of 4th and 5th term= (4+s / 2) = 4.5 = Q Q1 is the median of 5,5,6, and 21 , Which is 5.5. Interquartile range :-Q3 – Q1 = 5.5 -2.5 = 3 RANGE = 21-2 = 19 Problem 2) Which of the following set of data applies to this box and whisker plot? a)-4, -4, -2, 0, 0, 5 b)-4, 1, 1, 3, 4, 4 c)-4, -4, -3, 1, 5 d)-5, 3, 4, 5 e)-4, -4, -2, -2, 0, 0, 0, 5 Sol. From this box and Whisker Plot, We can easily reckon that Q1 = – 3, Q2 =-1 and Q3 = 0 Median of ” -4,-4,-2,0,0,5” is -1 , Median of “ -4,-4,-2,-2,0,0,0,5” is -1 Q1 for Set “-4,-4,-2,0,0,5”= Median of “-4,-4,-2”= 4 Q1 for set “-4,-4,-2,-2,0,0,0,5” = Median of “-4,-4,-2,-2” = (-4-2 / 2) = ( – 3 ) “-4,-4,-2,-2,0,0,0,5” is the only answer choice left. We can choose it without checking if you are confident in your previous work. Standard Deviation:- Standard deviation is a measure of how spread out a set of numbers is. For a set (n1 , n2 , n3 , _ _ _ _ n ), in Which x is the mean. Standard deviation = $$\sqrt { \frac { { ({ n }_{ 1 }-x) }^{ 2 }+{ ({ n }_{ 2 }-x) }^{ 2 }———{ ({ n }_{ n }-x) }^{ 2 } }{ n } }$$ Problem 3) Calculate the following standard deviation for the data set 0,7,8,10 and 10? Ans) Mean = (0+7+8+10+10 / 5 ) = 7 Standard deviation = $$\sqrt { \frac { { (7-0) }^{ 2 }+{ (7-7) }^{ 2 }+{ (7-8) }^{ 2 }+{ (7-10) }^{ 2 }+{ (7-10) }^{ 2 } }{ 5 } } \\ =\sqrt { \frac { 49+1+9+9 }{ 5 } } =\sqrt { \frac { 68 }{ 5 } } =3.7$$ Normal Distribution:- In the normal distributions, data points tend to cluster around the mean properties of normal distribution curve:- 1)The mean is at the center and the highest point on the curve and because the distribution is symmetrical about the mean, the mean is also the median and mode. 2)The standard deviation since it is a measure of dispersion determines the width of normal distributors curve. The greater the standard deviation, the wider and flatter the normal distribution curve is. 3)The probability of a data point falling within one standard deviation above the mean or below the mean is 34% 4)The probability of a data point falling between (Mean-d ) and (Mean-2d) is 13.5%. 5)The probability of a data point falling between (Mean+d) and (Mean+2d) is 13.5% NOTE: The values shown in the diagram above for various key areas of the curve are approximate. You are likely to see slightly different value in GRE Questions. Problem 4) A Food manufacturer produces energy bars that have a mean weight of 50 grams. If given days production is 10,000 energy bars, how many of those bars would be expected to weight between 49.0 and 49.5 grams? (Assume that the weight are normally distributed) Sol. MEAN = 50 Gram 68% of the bar weight between 49.5 gm and 50.5 gm. Since 68% Corresponds to the area represented between (mean+d) and (mean-d). Therefore standard deviation= 0.5 gm 13.5% of these bars weigh between 49 and 49.5%. No. of bars weighing between 49 and 49.5 = $$=(\frac { 13.5 }{ 1000 } \times 10,000)={ 1,350 }^{ Ans }$$ Problem 5) How values among the 8000 homeowners of town X are normally distributed, with a standard deviation of$11,000 and a mean of $90,000. Quantity A The number of homeowners in Town X , Whose home value is greater than$ 112,000.

Quantity B     300

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given

Sol.

Number of homeowners in town X whose home value is greater than $112,000 =2.5% of 8000 = (2.5 / 100) × 8000 = 200 Therefore , Quantity B is Greater. Problem 6) On a particular test whose score is distributed normally, the 2nd percentile is 1720, while the 84th percentile if 1,990. What score round to the nearest 10, most closely corresponds to the 16th percentile? Sol. Mean -2x (deviation) = 1720 Mean + x(deviation) = 1990 3x(deviation) = 270 Deviation=90 Mean = 1720+2 (90)= 1900 Mean – deviation = 1900-90 = $${ 1810 }^{ Ans }$$ Problem 7) If a set of data consist of only the first ten positive multiples of 5, What is the interquartile range of the set? Sol. The first ten positive multiples of 5 are : 5,10,15,20,25,30,35,40,45,50. Q1 (Median of [5,10,15,20,25] ) = 15 Q3 (Median of [30,35,40,45,50] ) = 40 INTERQUARTILE Range = $${ Q }_{ 3 }-{ Q }_{ 1 }=40-15={ (25) }^{ Ans }$$ Problem 8) Exam grades among the students in Ms. Hashman’s class are normally distributed, & the 50th percentile is equal to a score of 77. Quantity A: The number of students who scored less than 80 on the exam. Quantity B: The number of students who scored greater than 74 on the exam. a) Quantity A is greater b) Quantity B is greater c) The two quantities are equal d) The relationship cannot be determined from the information given. Sol. The normal distribution is symmetrical around the mean. For any symmetrical distribution, the mean equals the median (also known as the 50th percentile). Thus the number of students who scored less than 3 points above the mean (77+3=80) must be the same as the number of students who scored greater than 3 points below the mean (77-3=74). Problem 9) Jane scored in the 68th percentile on a test & John scored in the 32nd percentile. Quantity A: The proportion of the class that received a score less than John’s score. Quantity B: The proportion of a class that scored equal to or greater than Jane’s score. a) Quantity A is greater b) Quantity B is greater c) The two quantities are equal d) The relationship cannot be determined from the information given. Sol. Percentiles define the proportion of a group that scores below a particular benchmark. Since John scored in the 32nd percentile, by definition, 32% of the class scored worse than John. Quantity A is equal to 32%. Jane scored in the 68th percentile, So 68% of class scored worse than She did. Since 100-68=32, 32% of the class scored equal to or greater than Jane. Quantity B is also equal to 32%. Problem 10) The lengths of a certain population of earthworms are normally distributed with a mean length of 30 cm & a standard deviation of 3 cm. One of the worm is picked at random. Quantity A: The probability that the worm is between 24 & 30 centimeters, inclusive. Quantity B: The probability that the worm is between 27 & 33 centimeters, inclusive. a) Quantity A is greater b) Quantity B is greater c) The two quantities are equal d) The relationship cannot be determined from the information given. Sol. Quantity A equals the area between (mean & Mean-2 (standard deviation)) of the distribution. Therefore, the probability that the worm is between 24 & 30 centimeters is about 48%. Quantity B equals the area between ((mean+standard deviation) & (mean-standard deviation)) of the distribution. Therefore, the probability that the worm is between 27 & 33 centimeters is 68%. Therefore, Quantity B is greater. Problem 11) A species of insect has an average means of 5.2 grams & a standard deviation of 0.6 grams. The mass of the insects follows a normal distribution. Quantity A: The percent of the insects that have a mass between 5.2 & 5.8 grams. Quantity B: The percent of the insects that have a mass between 4.9 & 5.5 grams. a) Quantity A is greater b) Quantity B is greater c) The two quantities are equal d) The relationship cannot be determined from the information given. Sol. Mean=5.2 grams, Standard Deviation=0.6 grams The percent of the insects that have a mass between 5.2 & 5.8 gm= 34% (approx.) However, Quantity B will require some estimating. Now that 4.9 is halfway between 4.6 & 5.2, while 5.5 is halfway between 5.2 & 5.8. The area between 4.9 & 5.5 is under the bigger part of the bell curve in the centre. Since the area under the centre is greater than the area between (mean & (mean-standard deviation)). Therefore, Quantity B is greater. Problem: Among the set {1, 2, 3, 4, 7, 7, 10, 10, 11, 14, 19, 19, 23, 24, 25, 26}, what is the ratio of the largest item in the 2nd Quartile to the average value in the 4th Quartile? Sol. Median = Mean of 18th & 19th term = $$=(\frac { 10+11 }{ 2 } )=10.5$$ First Quartile $${ Q }_{ 1 }$$ = Median of all numbers below the median = Median of {1 2 3 4 7 7 10 10} = Mean of {4 7} = $$=(\frac { 4+7 }{ 2 } )=5.5$$ Third Quartile $${ Q }_{ 3 }$$ = Median of all numbers above the median = = Median of {11 14 19 19 23 24 25 26} = Median of {19 23} $$=(\frac { 19+23 }{ 2 } )=21$$ Now, we have calculated $${ Q }_{ 1 },{ Q }_{ 2 }\quad and\quad { Q }_{ 3 }$$. Elements of first Quartile are {1 2 3 4} Elements of Second Quartile are {7 7 10 10} Elements of Third Quartile are {11 14 19 19} Elements of Fourth Quartile are {23 24 25 26} Ratio of largest item in the 2nd Quartile to the average value in the 4th Quartile = $$=\frac { 10 }{ 24.5 } =\frac { 100 }{ 245 }$$ DEVELOPED ## GRE Geometry: Circles A circle is a labeled by its center point: O means the circle with center point O Diameter A line segment, generally denoted by the variable, that connects two points on the circles and passes through the center of the circle Radius:- A line segment, generally denoted by the variable r from the center of the circle to any point on the circle. Chord: A line Segment joining two points on the segment of a circle is a chord. Tangent:- A line that touches only one point on the circumference of the circle is called Tangent. A line drawn a tangent to a circle is perpendicular to the radius at the point of contact. Circumference:- The distance around a circle is called the circumference. An Arc is a portion of the circumference of a circle. The shorter distance between A an B along the circle is called the minor arc. The longer distance between A and B is called the major arc. Arc length:- For an arc with a central angle measuring θ Degree: Length = ( θ / 360 )×(Circumference) = (θ / 360 )×( π d ) Problem 1) What is the length of arc AC of the circle with center O as shown in the figure? Sol. Arc length= (θ / 360 ) ( π d )= (60 / 360 ) (π (2)) 12 π / 6 = 2 π units. Area of circles: The area of circles is given by the formula A= π r2 A sector is a portion of the circle that is bounded by two radii and an arc. In a sector Whose central angle measures by Q degrees. Area of sector= (Q / 360) x (Area of circle) Problem 2) What is the area of sector AOC in the circle with the center O shown? Sol. Area of sector AOC= ( 60 / 360 ) (36π ) = ( 6 π ) units. PROPERTIES: 1. The perpendicular from the center of a circle to a chord bisects the chord and vice versa. If ∠OCB=90 °, THEN AC=BC If AC=BC , then ∠OCB= 90° 2) Equal chords of a circle are equidistant from the center. Conversely, chords equidistant from the center are always equal. 3) Any two angles in the same segment are equal. Thus ∠ACB = ∠ADB 4)The angle subtended by an arc at the center of the circle is twice the angle subtended by the same arc at any other point on the circle. ∠AOB= 2∠ACB 5)The angle subtended by a semicircle is a right angle. Conversely, the arc of a circle subtending a right angle at any point on the circle is a semi-circle. If AB is a diameter, then ∠ACB = 90° IF ∠ACB = 90° , then AB is a diameter. 6)Tangent drawn from common external point to a circle are equal. Problem 3) A 5 by 12 rectangle is inscribed in a circle. What is the circumference of the circle? Sol. ∠BCD= 90° BD is a diameter (Because The arc of a circle subtending a right angle at any point on the circle is a semi-circle) In BCD , BC2 + DC2 = BD2 (by Pythagoras theorem) BD = 13 Circumference of the circle = 2π (Radius) = 13 π units Problem 4) In the figure shown below, if the radius of a circle with centre P is three times the radius of circle with centre A, ∠BAC=∠QPR, and the shaded area of the circle with centre A is 3 π SQUARE units, then what is the area of the shaded part of circle with centre P? Sol. Let ∠BAC =∠QPR = Q° The shaded area of circle with centre at A = 3 π sq. units = ( Q / 360° ) π (AC)2 PQ = 3AC The shaded area of circle with centre at P = ( Q / 360° ) π (PQ)2 = ( Q / 360° ) π (3AC)2 = (3π x 9) = 27π sq. units Problem 5) Two congruent, adjacent circles are cut out of a 16 by 8 rectangle. The circles have the maximum diameter possible. What is the area of the paper remaining after the circles have been cut out? Sol. For the circles, the diameter of the circle is the same as the width of the rectangle. Remaining area = Area of rectangle – 2 x area of circle Area of rectangle = Length x Breadth = 8 x 16 = 128 Sq. units radius of circle = 4 units Area of circle = π (4)2 = 16 π Sq. units Remaining area = (128- 2x(16π) ) = ( 128 – 32 π) sq. units Problem 6) The Figure shows an equilateral triangle, where each vertex is the center of a circle. Each circle has a radius of 20. What is the area of the shaded region? Sol. Side of equilateral triangle =40 Area of equilateral triangle= √3/4 (40)2 =√3/4 (1600) = (400 √3) Area of three sectors subtending an angle of 60° = 3 x (60/360) (π) (r2) =(1/2) π (400) = 200π Area of shaded Region= Area of equilateral triangle- area of three sectors subtending an angle of 60° = (400√3 – 200 π ) sq. units Problem 7) If the diameter of the circle is 36, what is the length of arc ABC? Sol. Diameter =36, radius =18 Note that a minor arc is the “short way around” the circle from one point to another, & a major arc is the “long way around”. Arc is thus the same as major arc AC ∠AOC = 2∠ABC = 2(40°) = (80°) The angle subtended by an arc at the centre of the circle is twice the angle subtended by the same arc at any other point on the circle. Length of minor arc AC = $$=\quad (\frac { 80 }{ 360 } )\times (2\pi r)\\ =\quad (\frac { 4 }{ 18 } )(2\pi )(18)\\ =\quad (8\pi )$$ Circumference of circle = $$=2\pi (18)=36\pi$$ Length of major Arc AC = Circumference – Length of minor Arc AC = $$=36\pi -8\pi ={ (28\pi ) }^{ Ans }$$ Problem 8) AB is not a diameter of the circle. Quantity A: The area of the circle Quantity B: 9 a) Quantity A is greater b) Quantity B is greater c) The two quantities are equal d) The relationship cannot be determined from the information given. Sol. Since, a diameter is the longest straight line, you can draw from one point on a circle to another (that is, a diameter is the longest chord in a circle), the actual diameter must be greater than 6. If the diameter were exactly 6, the radius would be 3, & the area would be: Area $$={ (3) }^{ 2 }\pi =9\pi$$ However, since the diameter must actually be greater than 6, the area must be greater than $$9\pi$$. Therefore, Quantity A is greater. Problem 9) In the following figure, O is the centre of the circle & $$\angle ABO={ 30 }^{ 0 }$$. Find $$\angle ACB$$. Sol. In $$\triangle \quad ABO,\quad AO=OB$$ (radii of circle) $$\angle OBA=\angle BAO={ 30 }^{ 0 }$$ (Because Angles opposite to equal sides are equal) $$In\quad \triangle ABO,\quad \angle ABO+\angle BAO+\angle AOB={ 180 }^{ 0 }\\ \angle AOB={ 120 }^{ 0 }\\ \angle AOB=2\angle ADB$$ (Because the angle subtended by an arc at the canter of the circle is twice the angle subtended by the same arc at any other point on the circle) $${ 120 }^{ 0 }=2\angle ADB,\quad \angle ADB={ 60 }^{ 0 }$$ $$\angle ADB+\angle ACB={ 180 }^{ 0 }\\$$ (Opposite angles of cyclic quadrilateral are supplementary) $$\angle ACB={ 120 }^{ 0 }\\$$ Directions for questions 10 to 12: In the figure below, X & Y are circles with centres O & O’ respectively. MAB is a common target. The radii of X & Y are in the ratio 4:3 & OM=28cm. Problem: 10) What is that ratio of the length of OO’ to that of O’M? a) 1:4 b) 1:3 c) 3:8 d) 3:4 Sol. $$\triangle MAO’\sim \triangle MBO\\ (\angle OBM=\angle O’AM={ 90 }^{ 0 }(Radius\quad is\quad always\quad perpendicular\quad to\quad the\quad tangent\quad at\quad the\quad point\quad of\quad contact)\\ \angle BMO=\angle AMO'(Common))$$ The radii of X and Y are in the ratio 4:3. $$\frac { OB }{ O’A } =\frac { 4 }{ 3 } \\ \frac { OB }{ O’A } =\frac { OM }{ O’M } =\frac { BM }{ AM } \\ \frac { 4 }{ 3 } =\frac { OM }{ O’M } (Subtract\quad 1\quad from\quad both\quad sides)\\ \frac { 4 }{ 3 } -1=\frac { OM }{ O’M } -1=\frac { OM-O’M }{ O’M } =\frac { OO’ }{ O’M } \\ { (\frac { 1 }{ 3 } =\frac { OO’ }{ O’M } ) }^{ Ans }$$ d) is correct. Problem 11) What is the radius of circle with centre O? a) 2cm b) 3cm c) 4cm d) 5cm Sol. $$\frac { OB }{ O’A } =\frac { OM }{ O’M } =\frac { BA }{ AM } \\ \frac { 4 }{ 3 } =\frac { OM }{ O’M } =\frac { 28 }{ O’M } ,\quad O’M=21\\ OO’=OM-O’M=28-21=7cm\\$$ 7cm = (radius of circle with centre O) + (radius of circle with centre O’) Let radius of circle with centre O be $$4x$$ and radius of circle with centre O’ be $$3x$$. $$7cm=4x+3x\\ x=1$$ Therefore, radius of circle with centre O = 4cm c) is correct. Problem 12) The length of AM is a) $$8\sqrt { 3 }$$ b) 10cm c) 12cm d) 14cm Sol. Radius of circle X=OB=4cm OM=28cm $$In\quad \triangle OBM,\quad \angle OBM={ 90 }^{ 0 }\\ { OB }^{ 2 }+{ BM }^{ 2 }={ OM }^{ 2 }\\ OB=4cm,\quad OM=28cm\\ { 4 }^{ 2 }+{ BM }^{ 2 }={ 28 }^{ 2 },\quad { BM }^{ 2 }=768\\ BM=16\sqrt { 3 } \\ \frac { 4 }{ 3 } =\frac { BM }{ AM } =\frac { 16\sqrt { 3 } }{ AM } \\ AM=12\sqrt { 3 }$$ c) is correct. Problem 13) O is the centre of a circle of radius 5 units. the chord AB subtends an angle of $${ 60 }^{ 0 }$$ at the centre. Find the area of the shaded portion (approximate value). a) 50 Sq. units b) 75 Sq. units c) 88 Sq. units d) 67 Sq. units Sol. Area of circle = $$\pi { (5) }^{ 2 }=25\pi \quad Sq.units$$ In\quad \triangle AOB,\quad \angle AOB={ 60 }^{ 0 }\\ \angle OAB=\angle OBA (Angles opposite to equal sides are equal) $$\angle AOB+\angle OBA+\angle BAO={ 180 }^{ 0 }\\ \angle OAB=\angle OBA={ 60 }^{ 0 }$$ $$\triangle AOB$$ is an equilateral triangle with side 5 units. Area of equilateral $$\triangle AOB$$ = $$\frac { \sqrt { 3 } }{ 4 } ({ side }^{ 2 })=\frac { \sqrt { 3 } }{ 4 } { (5) }^{ 2 }=\frac { 25\sqrt { 3 } }{ 4 }$$ Area of shaded portion = $$25\pi -\frac { 25\sqrt { 3 } }{ 4 } =25(\pi -\frac { \sqrt { 3 } }{ 4 } )=67.67\quad Sq.units$$ ## GRE: Three Dimensional Figures (Uniform Solids) ## Volume The volume of a solid is the amount of space enclosed by the solid. ## Surface area In general, the surface area of a solid is equal to the sum of the areas of the solid’s faces. ## Rectangular Solid A solid with six rectangular faces ( all edges meet at right angles) Volume= Area of Base x Height = l x w x h Surface Area = Sum of areas of faces =2(lw + wh + hl ) ## Cube A special rectangular solid with all edges equal (l=w=h) is called a cube. For Example:- Die Volume = L3 Surface area = Sum of areas of faces = 6 x L2 ## Cylinder A uniform solid whose base is a circle is called a cylinder. Lateral Surface area (or area of rest of shell) = 2 π r h Volume = Area of base X Height = π r2h Total Surface area = (2 X Area of base) + Lateral Surface Area = 2 π r+ 2 π r h = 2 π r ( r + h) ## Sphere A sphere is made up of all the points in space at a certain distance from a center point, it is also called a three-dimensional circle. The difference from the center to a point on the sphere is the radius of the sphere. Volume of sphere = 4/3 π r3 Total Surface area = 4 π r2 Problem 1) The solid shown is half a rectangular solid. What is the volume of the solid shown? Sol. In Triangle ABC, AB2 + BC2 = AC2 AB+ 9= 25 AB=4 Area of base = (1/2) x 4 x 3 = 6 Sq. units Volume = 6 x 4 = 24 Cubic units Problem 2) The height of a cylinder is twice its radius. If the volume of the cylinder is 128 π. What is the radius? Sol. Let radius be r Height of cylinder (h) = 2 r Volume = π r2 h 128 π = π r2 ( 2 r ) 64= r3 r = 4 units. Problem 3) A cube of ice has edges of length 10. What is the volume of the largest cylinder that can be carved from the cube? Sol. The largest cylinder would have diameter 10 and height 10 ( each equal to the edge of the cube) Volume = π r2 h = π (100/4) (10) =1000 π / 4 = 250 π Cubic units Problem 4) A solid metal cylinder with a radius of 6 and a height of 3 is melted down and all of the metal is used to recast a new solid cylinder with a radius of 3. What is the height of the new cylinder? Sol. Volume of solid metal cylinder =Volume of new solid cylinder Let radius and height of solid metal cylinder be and new solid cylinder be π R12H1 = π R22H2 62. (3) = 32. h H = (36 x 3) / (3 x 3) = 12 units ## GRE: Data Interpretation Bar Graph:- These can be used to display the information that would otherwise appear in a table. On a bar Graph, the height of each column shows its value. Line Graph:- Line graph follow the same principle as bar graphs, except the values are presented as points. Pie Charts:- A Pie Charts shows how things are distributed; the fraction of a circle occupied by each piece of the “pie” indicates what fraction of the whole it represents. Pie chart showing John’s Expenditure, 2004 100% =$20,000

To Find the amt. John Paid in federal taxes in 2004, We find the slice labeled “Federal Taxes” and We see that federal taxes represented 20% of this expenditures. Since the whole is $20,000, We find that John’s Federal taxes for 2004 were 20% of$20,000 or 1/5 × $20,000 =$ 4,000

Problem 1): Number of hours worked per week per employee at Marshville Toy Company

Chart for problem 1 & problem 2

Number of hours worked per week per employee at Marshville toy company

No. of EmployeesHours worked per weak
415
925
1535
2740
550

What is the positive difference between the mode and the range of the numbers of hours worked per week per employee at Marshal Toy Company?

Sol. The chart is a short way of showing a list of values that would begin: 15,15,15,15,25,25,25,25,25,25,25,25,25…..,.,…. (then the number 35 fifteen times, then the number 40 twenty- seven times, then the number 50 five times)

The mode is the number that appears in the list with the greatest frequency. Since 27 people worked 40 hours a weak. Therefore mode=40.

The range is the highest value in the list minus the lowest value in the list: 50-15=35. The positive difference between 40 and 35 = 5 Ans

Problem 2): Approximately What percent of the days with maximum temperature of  90 °F or more in St. Louis Occurred in July?

Ans:- From the grey bars on the “hot” day chart. St. Louis has a total of 1+8+15+12+4+4=44 days, when the temperature reached at least 90°F and 15 of these were in July.

These July days account for ( 15/44×100 )% = 34% Of all the hot days in St. Louis (approximately).

Use the following graph to answer problem 3-4

Problem 3)  If $720 million was spent on remodeling bedrooms in 2004, how many was spent remodeling kitchens? Sol. Let total amount spent on remodeling be x. $$\frac { 15 }{ 100 } \times x=720\\ x=\frac { 720\times 100 }{ 15 } =4800$$ Amount spent on remodeling kitchen = $$=\frac { 35 }{ 100 } \times 4800\\ =1680$$ Problem 4) On average, a homeowner will see a 75% return on investment for kitchen remodels. One homeowner increased the value of her home by$6000 by remodeling her kitchen. Assuming that gets the same percent return for bathroom remodels & that her spending follows 2004 averages, how much more would her home increase in value (to the nearest dollar) if she also remodeled the bathrooms?

Sol.Let amount spent on remodeling of kitchen be y

$$(y\times \frac { 75 }{ 100 } )=6000\\ y=(\frac { 6000\times 100 }{ 75 } )=8000$$

Amount spent on remodelling of kitchen is 35% of total amount.

Let total amount spent on remodeling be x.

$$(x\times \frac { 35 }{ 100 } )=8000\\ x=(\frac { 8000\times 100 }{ 35 } )=(\frac { 800000 }{ 35 } )$$

Amount spent on remodeling bathroom =

$$=(\frac { 25 }{ 100 } \times \frac { 800,000 }{ 35 } )=(\frac { 25\times 8000 }{ 35 } )$$

Increase in home’s value if he/she gets 25% return on remodelling of bathroom

$$=(\frac { 75 }{ 100 } \times \frac { 25\times 8000 }{ 35 } )=(\frac { 75\times 25\times 80 }{ 35 } )\\ =4285.7$$

Use the following graph to solve problem 5-6

Problem 5) In 2009, a homeowner put his house on the market. If the average drop from listing price to sales price is 6%. What was the minimum price (in whole dollars) that the house could have been listed at so that it sold at the price greater them or equal to the medium price?

Sol. Medium price in 2009 = $240,000 Average drop from listing price to sales price = 6% Let minimum price be x $$x-(\frac { x\times 6 }{ 100 } )=240,000\\ x=255,320$$ Problem 6) A real estate agent says the percent change between 2007 & 2008 is the same as between 2008 & 2009. Is this true or false? Sol. Percentage change = $$(\frac { Amount\quad of\quad change }{ Original\quad Amount } )\times 100$$ Percentage change from 2007 to 2008 = $$=(\frac { 230,000-240,000 }{ 240,000 } )\times 100\\ =-4.17%$$ Percentage change from 2008 to 2009 = $$=(\frac { 240,000-230,000 }{ 240,000 } )\times 100\\ =-4.35%$$ Because 4.35% is not equal to 4.17%, the statement is false. Number of hours worked per week per employee at Marshville toy company No. of EmployeesHours worked per weak 415 925 1535 2740 550 Problem 7) What is the average (arithmetic mean) number of hours worked per week per employee at Marshville Toy Company? a) 32 b) 33 c) 35 d) $$35\frac { 2 }{ 3 }$$ e) $$36\frac { 1 }{ 3 }$$ Sol. Total number of employees = 60 4 employees worked for 15 hours per week. 9 employees worked for 25 hours per week. 15 employees worked for 35 hours per week. 27 employees worked for 40 hours per week. 5 employees worked for 50 hours per week. This chart is a short way of showing a list of values that would begin: 15, 15, 15, 15, 25, 25, 25, 25, 25, 25, 25, 25, 25……. (then the number 35 fifteen times, then the number 40 twenty-seven times, then the number 50 five times). Arithmetic mean of hours worked per week per employees $$=(\frac { Total\quad no.\quad of\quad hours\quad worked }{ Total\quad no.\quad of\quad employees } )\\ =(\frac { 15\times 4+25\times 9+35\times 15+40\times 27+50\times 5 }{ 60 } )\\ =\frac { 2140 }{ 60 } =\frac { 214 }{ 6 } =\frac { 107 }{ 3 } \\ ={ (35\frac { 2 }{ 3 } ) }^{ Ans }$$ Use the following table to answer exercises 8-9  Number of pets 1 2 3 4 5 Number of household with this number of pets 70 246 183 49 9 Maximum monthly spending on pet supplies$57.32 $75.98$143.57 $140.48$170.23 Minimum monthly spending on pet supplies $6.34$35.47 $45.84$87.46 $111.20 Average monthly spending on pet supplies$31.25 $56.42$83.11 $127.74$147.38

Problem 8) The household group with which number of pets had the greatest average (arithmetic mean) monthly spending per pet?

Sol. This table tallies the number of households, according to number of pets in the household & each column captures information about these households. For example, the left most columns with numbers indicates that there are 70 households that have one pet & these households spend an average of $31.25 per month on pet supplies. In that group, maximum spending on pet supply was$57.32 and minimum spending on pet supply was $6.34. We have to find a group whose average monthly spending per pet was maximum. First group spent an average of$31.25 per pet.

Second group spent an average of $$(\frac { 56.42 }{ 2 } )=28.21$$ per pet.

Third group spent an average of $$(\frac { 83.11 }{ 3 } )=27.7$$ per pet.

Fourth group spent an average of $$(\frac { 56.42 }{ 2 } )=28.21$$ per pet.

Fifth group spent an average of $$(\frac { 56.42 }{ 2 } )=28.21$$ per pet.

Household group with 4 numbers of pets had the greatest average spending per pet.

Problem 9) Approximately what percent of the surveyed households have more than three pets?

a) 10%

b) 20%

c) 30%

d) 40%

e) 50%

Sol. There are 557 households of which 49 have four pets & 9 have five pets.

Percent of the surveyed households having more than three pets =

$$=(\frac { 58 }{ 557 } \times 100)\\ =10.41%\\ \sim 10%$$

Questions 10-12 refer to the following charts.

Problem 10) Imagine that at the beginning of 2014, Boston & Los Angeles implemented a public health program that reduced deaths of infants less than 1 year old by 20 percent, while the cities of Honolulu & Indianapolis terminated an identical program. What would have been the approximate total impact of these program changes on the number of infant deaths in these cities?

a) These would have been 600 more deaths.

b) These would have been 70 more deaths.

c) These would have been 30 fewer deaths.

d) These would have been 65 fewer deaths.

e) These would have been 150 fewer deaths.

Sol.

This question asks about deaths of infants under age 1 year in certain cities. That information is found in the bar graph, specifically in the left bar of each cluster. Note that this graph has two y-axes, one on the left and on the right. Deaths that occur before age 1 year are plotted against the y-axes on the left; be sure to read the correct bars against the correct axis.

Boston is represented by the bars on the far left and Los Angeles by the bars on the far right. Honolulu and Indianapolis are represented by the second and fourth pairs of bars, respectively.

Before doing any calculations, you might note that the numbers of infant deaths in both Boston and Los Angeles are significantly higher than in either Honolulu or Indianapolis. Therefore, the effect of introducing public health programs in Boston And Los Angeles will be greater than the effect of terminating programs in the other two cities. In other words, more deaths will be prevented than not prevented, and you can eliminate choices (A) and (B). On Test Day, even if you had no more time to invest in this problem, by using your critical thinking skills to eliminate answer choices, you would improve your odds of guessing the correct answer.

In 2014, Boston had just under 200 infant deaths, and Los Angeles had just over 200. That’ s about 400 infant deaths total for these cities. The question wants you to imagine that a public health program was introduced in these cities, dropping infant deaths by 20%. In that case, there would be (0.2)(400) = 80 fewer infant deaths in those two cities, or 320 infant deaths instead of the actual 400.

Honolulu had about 60 infant deaths, and Indianapolis had not quite 150-call it 140. That’ s 60+140=200 deaths, and you’ re told to imagine what would have happened had a program reducing infant deaths by 20% been terminated. Without such a program acting to reduce infant deaths , the number would go up, but by how much? Call the number of infant deaths in Honolulu and Indianapolis d. The current 200 infant deaths represent 80% of what the number would be without the program, so you can write the following equation and solve: 0.8d = 200; d = 250, There would have been 50 more deaths.

Therefore, had the stated program changes occurred, there would have been about 320+250=570 deaths. There were actually about 400+200 = 600 deaths. That means about 30 fewer deaths would have occurred. Alternatively, you could add the differences in the two cities to get the net difference: -80 + 50 = 30. Either way, the correct answer is (C).

Problem 11) If in 2014, the population of Los Angeles was 75 percent greater than the population of Houston, What is the the ratio of the incidence of pneumonia & influenza related deaths (expressed as a percent of the city’s population) in Houston to the incidence in Los Angeles in that year?

a) $$\frac { 1 }{ 2 }$$

b) $$\frac { 7 }{ 8 }$$

c) $$\frac { 15 }{ 16 }$$

d) $$\frac { 8 }{ 7 }$$

e) $$\frac { 4 }{ 3 }$$

Sol.

This question concerns deaths from pneumonia/influenza, which you find in the pie chart.

The pie chart represents the 3,532 pneumonia/influenza deaths that occurred in five cities. Of these, 19% occurred in Houston, and 38% occurred in Los Angeles. That’ s a ratio of 19:38 or 1:2 . However, the question is asking for the ratio of the rate of incidence of these deaths.

If Los Angeles’ s population were twice the size of Houston’ s, then the rate of incidence in the two cities would be the same, giving a ratio of 1:1. Because Los Angeles’ s population is less than twice Houston’ s, you know that the rate of incidence is higher in Los Angeles than in Houston. The ratio of Houston to Los Angeles must be less than 1. On the basis of this critical thinking, you can eliminate (D) and (E).

The question does not give you the actual population of the cities, but it does tell you their relative populations. If you call the population of Houston p, then the population of Los Angeles is 1.75p or (7/4)p. As shown in the pie chart, for every 1 death from pneumonia/influenza in Houston, there were 2 in Los Angeles. In Houston, therefore, the rate of incidence of deaths can be represented by the ratio of 1 to p, or (1/p). In Los Angeles, the rate of deaths can be represented as 2 to (7/4)p, or

$$\frac { 2 }{ \frac { 7p }{ 4 } } =2\times \frac { 4 }{ 7p } =\frac { 8 }{ 7p }$$

Now find the ratio of the rates of incidence:

$$\frac { \frac { 1 }{ p } }{ \frac { 8 }{ 7p } } =\frac { 1 }{ p } \times \frac { 7p }{ 8 } =\frac { 7 }{ 8 }$$

The correct answer is (B), $$\frac { 7 }{ 8 }$$.

Problem 12) Assuming that 80% of all deaths due to pneumonia or influenza occurs among the elderly, defined as those ages 65 & over, in which city was the least proportion of all deaths among the elderly attributed to pneumonia or influenza?

a) Houston

b) Boston

c) Indianapolis

d) Honolulu

e) Los Angeles

Sol.

This question asks about deaths due to pneumonia/influenza, so look at the pie chart for those data. It also concerns people who died at the age 65 or over, and that information is in the bar graph, specifically in the right-hand bar of each pair of bars. Remember that the deaths occurring at age 65 and over are plotted against the y-axis on the right side of the bar graph.

Only 80% of the pneumonia/influenza deaths occurred among the elderly. However, because this 80% applies equally to each city, you do not have to take into account as you compare the percentages of deaths in each city attributed to pneumonia/influenza. If a city has the most such deaths when 100% of pneumonia/influenza deaths are counted, then it will also have the most such deaths when the number of pneumonia/influenza deaths is reduced by 20% for all cities.

If you compare cities using estimation, you thinking might go like this: Boston and Houston have about the same number of P/I deaths, making them easy to compare. Houston has far more deaths from all causes, so Houston’ s proportion of P/I deaths is lower than Boston’ s. Houston has about 50% more P/I deaths than Honolulu or Indianapolis, and it has more than four times (Honolulu) and three times (Indianapolis) the deaths from all causes. Therefore, Houston’ s proportion of P/I deaths is lower than Honolulu’ s or Indianapolis’ s. Finally, compare Houston to Los Angeles, Houston has more total deaths and far fewer P/I deaths than Los Angeles has, so its proportion of P/I deaths is definitely lower than Los Angeles’ s. Houston is the winner; choice (A) is correct.

You can also solve using calculation. The pie chart represents a total of 3,532 deaths, and it gives the percentage distribution among the five cities. Determine the number of pneumonia/influenza deaths for each city:

Deaths from P/I

Boston     $$3,532\times 0.18=636$$

Honolulu     $$3,532\times 0.13=459$$

Houston      $$3,532\times 0.19=671$$

Indianapolis    $$3,532\times 0.12=424$$

Los Angeles    $$3,532\times 0.38=1,342$$

Now find the number of deaths among the elderly from the bar graph. Boston is between 4,000 and 5,000 or say 4,500. Honolulu is at about 3,000, Houston is the tallest bar at about 13,000, Indianapolis is at 4,000 and the bar for Los Angeles is at about 9,000.

## GRE: Number System

Natural Numbers:- These are the numbers (1,2,3 etc) that are used for counting. In other words, all positive integers are natural numbers.

Prime Numbers:- A natural number larger than unity is a prime number if it does not have other divisors except for itself & unity.

Whole Numbers:- The set of numbers that includes all natural numbers and the number zero are called whole numbers.

Composite Numbers:- It is a natural number that has at least one divisor different from unity and itself.

Real Numbers:- All numbers between negative infinity & positive infinity that can be represented on a number line are called real numbers.

Rational Numbers:- A rational numbers is defined as a number of the form (a/b) where a and b are integers and b ≠ 0.

Rational numbers that are not integers have decimal values, can be of two types:

1.Terminating Decimal Fraction

2.Non- Terminating Decimal Fractions

Terminating decimal fractions :- For example  17/4 = 4.24  ,  21/5 = 4.2

Non-Terminating decimal fractions: Non-terminating decimal fractions are of two types:

Non-Terminating Repeating Fractions: These are a non terminating repeating fraction. For Example 5.333…..,   6.161616…..

Neither terminating nor repeating fractions: For example 5.273612…..

## Concept of GCD (Greatest Common Divisor or Height Common Factor)

Consider two natural numbers n1 and n2.

If the numbers n1 and n2 are exactly divisible by same number x, then x is a common divisor of n1 and n2.

The highest of all the common divisors of n1 and n2 is called as the GCD or the H.C.F:-

### Steps for calculating HCF

Suppose we have to find H.C.F of 150,210 and 375.

Step 1:- Writing down the standard form of numbers

150=5×5×3×2

210=5×2×7×3

375=5×5×5×3

Step 2:- Writing prime factors common to all the three numbers are taking their product is 51  ×  31

Step 3:- H.C.F = 5×3= 15

## Concept of LCM (Least common multiple)

Let n1 and n2 be two natural numbers distinct from each other. The smallest natural number n that is exactly divisible by n1 and n2 is called the least common multiple (LCM)

### Steps for calculation of LCM:-

Suppose we have to find LCM of 150,210,375.

Step 1:- Write down the standard form of numbers

150=5×5×3×2

210=5×2×7×3

375=5×5×5×3

Step 2:-  Write down all the prime factors that appear at least once in any of the numbers: 5,3,2,7.

Step3: -Raise each of the prime factor to their highest available power

The LCM= 2 × 3 × 53 × 7 = 5270

HCF OF TWO OR MORE FRACTION IS GIVEN BY :

=HCF of Numerators Divided By LCM Of Denomination

LCM OF TWO OR MORE FRACTIONS IS GIVEN BY:

=LCM of Numerators  Divided By HCF of Denominators

Problem 1) Three traffic signals change after 36,42 and 72 seconds respectively.If the lights are first switched on at 9:00 sharp. At What time will they change simultaneously?

Sol. In order to distribute $4000 and 180 pencils evenly, the numbers of employees must be a factor of each of these two numbers. We have to find the greatest number of employees possible therefore, we have to find HCF of$4000 and 180.

4000=2×2×2×5×2×5×2×5 =25  ×  53

180= 2×9×2×5 = 22  ×  32 × 5

HCF = 22 × 5  = 20

Problem 2) The number of students who attended a school could be divided among 10,12 or 16 buses, such that each bus transports an equal number of students that could attend the school?

Sol.  No, of students must be divisible by 10,12 and 16. We have to find a minimum number of students that could attend the school or we have to find a minimum number that is divisible by 10,12 and 16.

LCM of 10,12,16                            10= 2×5

=24 × 3×5                                       12=2×2×3

=16×15                                          16=2×2×2×2

ANS:-  (240)

Problem 3) There are 576 boys and 448 girls in a school that are to be divided into sections containing equal no. of students of either boys or girls alone. Find the minimum total number of sections thus formed

a) 24

b) 32

c) 16

d) 20

Sol.  We have to find HCF of 576 and 448

576= 32  × 26

448= 26  × 7

HCF= 26 = 64

Number of sections = 576/64 + 448/64  = (9+7) = $${ 16 }^{ Ans }$$

Divisibility:- Any integer I (Commonly called dividend)

When divided by a natural number N (Called Divisor), there exist a unique pair of number Q and R ,Which are called the quotient and remainder respectively.

Dividend=Divisor × Quotient + Remainder

Suppose We divide 42 by 5. The result has a quotient of 8 and remainder of 2.

$$Let\quad N={ p }_{ 1 }^{ { \alpha }_{ 1 } }.{ p }_{ 2 }^{ { \alpha }_{ 2 } }.{ p }_{ 3 }^{ { \alpha }_{ 3 } }—–{ p }_{ k }^{ { \alpha }_{ k } }\quad Where\quad { p }_{ 1 },{ p }_{ 2 },{ p }_{ 3 }—–{ p }_{ k }\\ are\quad different\quad primes\quad and\quad { \alpha }_{ 1 },{ \alpha }_{ 2 },{ \alpha }_{ 3 }——{ \alpha }_{ k }\quad are\quad natural\quad numbers.\\ Total\quad number\quad of\quad divisors\quad of\quad N\quad =\\ =\quad (1+{ \alpha }_{ 1 })(1+{ \alpha }_{ 2 })(1+{ \alpha }_{ 3 })——-(1+{ \alpha }_{ k })\\ Sum\quad of\quad these\quad divisors\quad =\\ =\quad ({ p }_{ 1 }^{ 0 }+{ p }_{ 1 }^{ 1 }+{ p }_{ 1 }^{ 2 }—–{ p }_{ 1 }^{ { \alpha }_{ 1 } })({ p }_{ 2 }^{ 0 }+{ p }_{ 2 }^{ 1 }+{ p }_{ 2 }^{ 2 }——{ p }_{ 2 }^{ { \alpha }_{ 2 } })——-({ p }_{ k }^{ 0 }+{ p }_{ k }^{ 1 }—-{ p }_{ k }^{ { \alpha }_{ k } })\\$$

Problem 4) Find the sum of factors and the number of factor of 240?

Sol.  240 = 24 × 41 × 51

Total Number of divisor = (1+4) (1+1) (1+1)

=5×2×2= 20

Sum of these divisors

= (20 + 21 + 22 + 24 ) (30 + 31) (S0 + S1)

= (1+2+4+8+16) (4) (6)

=24 (31) = (744)

Problem 5)

Quantity A  The number of positive factor of 20.

Quantity B  The number of positive factor of 32.

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

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

Sol.  32 = 25                                        ,                               20= 22  51

Number of distinct positive factors of 32 = (6)

Number of distinct positive factors of 20 = 3×2 = 6

The two quantities are equal.

The highest power of a prime number p, which divides n! exactly is given by

= [n/p] + [n/p2] +[n/p3]……..

Where [x] denotes the greatest integer less than or equal to x

Remainder Theorem:-

The remainder of the expression

[A1 × B1 × C1 ………………+ A2 × B2 × C2] / m

will be the same as the remainder of the expression

[A1R × B1R × C1R ………………+ A2R × B2R × C2R] / m

Where A1R  is the remainder when A1 is divided by M. B1R is the remainder when B1 is divided by M and so on.

Problem 6) Find the remainder when 17 ×23×126×38 divided by 8

Sol.     ( 17 ×23×126×38 / 8 )

= (1×7×6×6 / 8 ) = ( 42×6 / 8) = (2*6 / 8 )     =    ( 12 / 8 )

Remainder=   ( 4 )

Problem 7) If a=16 b and b is a prime number greater than 2, How many positive distinct factors does a have?

Sol.   a =16b = 24. b1

No. of positive distinct factors = ( 4 + 1 ) ( 1 + 1 )    = (10)

Problem 8) If x is a positive integer, which one of the following could be the remainder. When 73x  is divisible by

a) 0

b) 1

c) 2

d) 3

e) 4

f) 5

g) 6

h) 7

i) 8

j) 9

Indicate all such remainders

Sol.  We have to find the unit digit of 73x

As with multiplication, when an integer is raised to a power, the unit digit is determined solely by the product of units digits.

Those products will form a repeating pattern.

31= 3. 32 =9. 33 =27. 34 = 81 and 35 =243

Here  the pattern returns to its original value of 3 and we have got a repeating pattern

3,9,7,1,3,7,1,3 —–

Thus the unit digit of  73x  must be 1,3,7,9

### Divisibility Rules:

Divisibility by 2-A number is divisible by 2 if the last digit is divisible by 2.

Divisibility by 5 – A number is divisible by 5 if the last digit is 5 or 0

Divisibility by 4 – A number is divisible by 4 if the last  two digit  are divisible by 4

Divisibility by 8 – A number is divisible by 8 if the last  3 digit  are divisible by 8

Divisibility by 11 – A number is divisible by 11 if the difference of the sum of the digits in the odd places and the sum of the digits in the even places is zero or divisible by 11.

Problem 9) w, x, y and z are consecutive odd integers such that $$w<x<y<z.$$ which of the following statements must be true?

Indicate all such statements.

a) wxyz is odd.

b) w+x+y+z is odd

c) w+z=x+y

Sol.  Odd×odd = odd

wx is odd. yz is odd and wxyz is also odd.

Odd+odd = even

Even+even=even

w+x=even, y+z=even

Therefore,w+x+y+z is also even.

w,x,y and z are consecutive odd integers, all can be defined in terms of w.

w=w, y=w+4

x=w+z, z=w+6

w+z=2w+6, x+y=2w+6

Therefore, w+z=x+y

Problem 10)

Quantity A Remainder when $${ (67) }^{ 99 }$$ is divided by 7.

Quantity B Remainder when $${ (75) }^{ 80 }$$ is divided by 7.

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

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

Sol.  Using remainder theorem

$$(\frac { { 67 }^{ 99 } }{ 7 } )\rightarrow (\frac { { 4 }^{ 99 } }{ 7 } )=\frac { { { (4 }^{ 3 }) }^{ 33 } }{ 7 } =\frac { { 64 }^{ 33 } }{ 7 } \\ Remainder\quad =\quad 1\\ \frac { { 75 }^{ 80 } }{ 7 } \rightarrow \frac { { 5 }^{ 80 } }{ 7 } \rightarrow \frac { { ({ 5 }^{ 6 }) }^{ 13 }\times { 5 }^{ 2 } }{ 7 } \rightarrow \frac { { 1 }^{ 13 }\times 25 }{ 7 } \\ Remainder\quad =\quad 4$$

Quantity B is greater.

Problem 11) If set S consists of all positive integers that are multiples of both 2 and 7, how many numbers in set S are between 140 and 240, inclusive?

A positive integer that is a multiple of both 2 and 7 is a multiple of 14. Since 140 is a multiple of 14, start listing there and count the terms in the range:

140, 154, 168, 182, 196, 210, 224, 238

Problem 12) If then which of the following must be true?

a) $${ a }^{ 2 }>a>b>{ b }^{ 2 }$$

b) $$b>a>{ a }^{ 2 }>{ b }^{ 2 }$$

c) $${ b }^{ 2 }>a>{ a }^{ 2 }>b$$

d) $${ b }^{ 2 }>{ a }^{ 2 }>b>a$$

e) $${ b }^{ 2 }>b>a>{ a }^{ 2 }$$

Sol.  The correct answer is $$({ b }^{ 2 }>b>a>{ a }^{ 2 })$$

The goal in this question is to order $$a,{ a }^{ 2 },b\quad and\quad { b }^{ 2 }$$ by magnitude. Based on the original inequality

$$0<a<\frac { 1 }{ b } <1$$, several things are true.

$$\frac { 1 }{ b } <1\quad \& \quad b>1$$

Therefore, $${ b }^{ 2 }>b$$

Second, note that $$a<1$$ in the given inequality.

a is a positive number less than 1,

$${ a }^{ 2 }<a<1\\ { b }^{ 2 }>b>1\\ { ({ b }^{ 2 }>b>a>{ a }^{ 2 }) }^{ Ans }$$

Problem 13) On a number line, the distance from A to B is 4 and the distance from B to C is 5.

Quantity A  The distance from A to C

Quantity B   9

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

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

Sol.  If the points A, B and C are in alphabetical order from left to right, then the distance from A to C will be 9. However, alphabetical order is not required. If the points are in the order C, A and B from left to right,then the distance from A to C is 5-4=1. Therefore, the relationship cannot be determined.

Problem 14) The integer a is even and the integer b is odd. Which of the following statements must be true?

Indicate all such statements

a) a+2b is even

b) 2a+b is even

c) ab is even

d) $${ (a) }^{ b }$$ is even

e) $${ (a+b) }^{ 2 }$$ is even

f) $${ a }^{ 2 }-{ b }^{ 2 }$$ is odd

Sol.  If b is odd, then 2b must be even.

even+even = even

(a+2b) must be even

even+odd = odd

2a+b must be odd.

even×odd = even

ab must be even.

$${ (even) }^{ odd }$$ must be even.

$${ (a) }^{ b }$$ must be even

$${ (odd) }^{ even }$$ must be odd

$${ (odd) }^{ odd }$$ must be odd

(a+b) is odd

Therefore, $${ (a+b) }^{ 2 }$$ must be odd.

$${ (a) }^{ 2 }$$ must be even and $${ (b) }^{ 2 }$$  must be odd.

$$({ a }^{ 2 }-{ b }^{ 2 })$$ must be odd.

Problem 15) Find the number of zeros in 137!?

Sol.  The highest power of a prime number p, which divides n! exactly is given by

$$\left[ \frac { n }{ p } \right] +\left[ \frac { n }{ { p }^{ 2 } } \right] +\left[ \frac { n }{ { p }^{ 3 } } \right] ——-$$

Where $$\left[ x \right]$$ denotes the greatest integer less than or equal to x.

$$\left[ \frac { 137 }{ 5 } \right] +\left[ \frac { 137 }{ 25 } \right] +\left[ \frac { 137 }{ 125 } \right] \\ =1+5+27=\quad 33\quad zeros$$

Since the restriction on the number of zeroes is due to the number of fives.

Therefore, 137! Has 33 zeros.

Problem 16) Find the number of numbers from 1 to 100 which are not divisible by any one of 2 & 3.

a) 16

b) 17

c) 18

d) 33

Sol.  Number of numbers not divisible by 2 & 3 = Total no. of numbers – number of numbers divisible by either 2 or 3.

Number of numbers divisible by 2 = 50

Numbers divisible by 3 (but not by 2, as it has already been counted) are 3,9, 15, 21, 27, 33—–93, 99.

Number of numbers divisible by 3 (but not by 2)

$$=\frac { (99-3) }{ 6 } +1=17$$

Hence, number of numbers divisible by either 2 or 3

= 50+17 = 67

Number of numbers not divisible by any of 2 & 3 =

Total number of numbers – no. of numbers divisible by either 2 or 3

= 100 – 67

= $${ 33 }^{ Ans }$$

## Conditions of similarity:-

• AAA Similarity:- If in two triangles, corresponding angles are equal, then triangles are similar.

If ∠A = ∠P , ∠B = ∠Q , ∠C = ∠R

Then △ABC ~ △ PQR

• SSS Similarity:- If the corresponding sides of two triangles are proportional then they are similar.

If AB/PQ = BC/QR = AC/PR

Then   △ABC ~ △ PQR

• SAS Similarity: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.

If AB/PQ = BC/QR  &  ∠B = ∠Q

Then △ABC ~ △ PQR

The similarity of Triangles:- Similarity of triangles is a special case where if either of the conditions of similarity holds, The other will hold automatically.

Congruency of Triangles:-If two triangles are congruent, then corresponding angles are equal.

## Conditions of Congruency:-

1.SAS Congruency:- If two sides and an included angle of one triangle are equal to two sides and an included angle of another, the two triangles are congruent.

In  △ABC and △ PQR

IF AB=PQ,

BC=QR,

∠ABC = ∠PQR

Then △ABC~△ PQR

2.ASA Congruency:- If two angles and the included side of one triangle is equal to two angles and the included side of another, The two triangles are congruent.

In  △ABC and △ DEF

If  ∠A =∠D

AB = DE

∠B = ∠E, Then △ABC ≅ △ PQR

3. AAS Congruency:- If two angles and the side opposite to one of the angles is equal to the corresponding angles and the side of another triangle, the triangles are congruent.

If  In △ABC and △ DEF

∠A =∠D

∠B=∠E

AC=DF

Then △ABC≅△ DEF

4.SSS Congruency:- If three sides if one triangle is equal to three sides of another triangle, the two triangles are congruent.

In △ABC and △ DEF

If AB=DE, BC=EF and AC=DF

THEN  △ABC ≅ △ DEF

5.SSA Congruency:- If two sides and the angle opposite to any of the sides are equal to corresponding sides and the angle opposite to any of the sides, Then the triangles are congruent.

In △ABC and △ DEF

AB=DF

BC= EF

∠BAC = ∠EDF

Then △ABC ≅ △ DEF

## Right Angled Triangle:

Pythagoras theorem: In the case of a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In right △ABC, AB2 + BC2 = AC2

## Special figure:

Note: A lot of questions are based on the figure

In this figure, △ABC ~△ ADB ~ △BDC

• △ABC ~△ BDC

AB/BD = BC/DC = AC/BC

## Area Theorem:

According to area theorem:The ratio of areas of two similar triangles is the square of the ratio of corresponding sides.

If △ABC ~ △ DEF

Ar(ABC): Ar(DEF) = (AB/DE)2 = (BC/EF)2 = (AC/DF)2

Problem 1) For similar triangles, the ratio of their corresponding sides is 2:3. What is the ratio of their areas?

Sol. According to area theorem :The ratio of areas of two similar triangles is the square of the ratio of corresponding sides.

Ratio of areas= (2/3)2  =  4/9

Problem 2) What is the area of the triangle shown?

Sol.

∠B = ∠C = 60

∠A= 60

(Sum of angles of a triangle is equal to 180°)

∠A = ∠B = ∠C = 60°

(Angles of an equilateral triangle are equal to 60° & triangle having each angle equal to 60° is an equilateral triangle )

:. △ABC is an equilateral △

In △ABD

BD2 +AD2 = AB2 (Pythagoras theorem)

Area of Triangle = ½ × Base × Height

=1/2 × 10 × 5√3

= 25√3  sq. unit

Problem 3) What is the area of the triangle Shown?

Sol.

x2 +( x + 1)2 =25 (by Pythagoras theorem)

2x2 +2x =24

2x2 + 2x -24 =0

x2 + x – 12 =0

x = 3 or -4

x= 3 (Because length cannot be negative)

Area = ½ (base) x (height) = ½(7)(8)   = 28 sq. units

Problem 4) What is the value of x in the triangle shown?

Sol.      In △ABC

4 + 3 = AC  (BY Pythagoras theorem)

AC = 5

△ABC~ △BDC

AB/BD =BC/DC = AC/BC

4/x  =  5/3

12 = 5x          ,      x =(12/5)  = Ans

## Polygons

A polygon is a closed figure whose sides are straight line segments.

The perimeter of the polygon is the sum of the lengths of the sides. A vertex of a polygon is the point where two adjacent sides meet.

A diagonal of a polygon is a line segment connecting two nonadjacent vertices.

A regular polygon has sides of equal length and interior angles of equal measure.

Sum of all angles of a polygon with n sides

= (n-2) π radians = (n-2) (180°)

Area of Regular polygon = (ns2/4) x Cot (180/n)

Where     s = length of side

n = No. of sides

## Parallelogram (IIgm)

A quadrilateral with two pairs of parallel sides.

Properties of parallelogram:

Area of Parallelogram = Base x Height

Diagonals of Parallelogram bisect each other.

The opposite angles in a Parallelogram are equal.

## Rectangle

A Parallelogram with four equal angles, each at right angle is a rectangle.

∠ABC = ∠BCD = ∠CDA = ∠BAD

Properties of a Rectangle:

Diagonals of a rectangle are equal and bisect each other.

## Rhombus

A Parallelogram having all the sides equal is a rhombus.

Area of Rhombus = ½ x product of diagonals

Properties:-

Diagonals of a rhombus bisect each other at right angle.

## Square

A Square is a rectangle with adjacent sides equal or rhombus with angles equal to 90°.

Properties:

Diagonals of square are equal and bisect each other.

Area = ½ x (diagonal) 2

Diameter of circle circumscribing a square also acts as a diagonal of square.

BD=Diameter of Circle

= Diagonal of Square = a √2

## Trapezium

A Trapezium is a quadrilateral with only two sides parallel to each other.

Area of Trapezium = ½ (Sum of Parallel sides)x height

## Rectangular Hexagon

A Rectangular Hexagon is a actually a combination of 6 equilateral triangles all of side “a”

Area of hexagon = 6(3/4 a2) = 3√3 / 2 a2

Problem 5) Right △ABC and rectangle EFGH have the same perimeter. What is the value of x?

Sol. In △ABC

AB2 + BC2 = AC2 (BY Pythagoras theorem)

AC=5

Perimeter of △ABC = 3+4+5 = 12

Perimeter of rectangle EFGH =2(2+x)

12 = 2 (2+x)

2 + x = 6

x = 4

Problem 6) What is the area of the square in the figure above?

Sol. (Side2 + Side2 = Diagonal2) (by Pythagoras theorem)

2 Side2  = 10= 100

Side = 5 √2

Area = Side2 = 50 sq units

Problem 7) The area of a rectangle can be represented by the expression 2x2+9x+10. The length is 2x+5 and the width is 6. What is the value of the area of the rectangle?

Sol. Area = 2x2 +9x 10 = Length x Breadth

2x2 +9x 10 = (2x +5)6

2x2  – 3x – 20 = 0

x = 4 or -2.5

If x = -2.5, then length = 2(-2.5) + 5 = 0 (Length can’ t be zero,∴ Negative value of x is discarded)

x = 4

Area = 2 (16) +9 (4) +10

= 32 +36+ 10   = 78 sq Units

Problem 8) In the figure shown, ABCD is a rectangle. AB=8 and BC=6. R,S,T, AND Q are midpoints of the sides of rectangle ABCD. What is the perimeter of RSTQ?

Sol. AB = CD = 8, BC = AD = 6

AR = RB = DT = CT

AQ = QC = BS = SD = 3

AR2 + AQ2 = RQ2  (By Pythagoras theorem)

32 + 42 = RQ2

RQ = 5

RQ=RS=ST=QT=5

Perimeter of RSTQ = 20

Problem 9) In the figure shown ABCD is a square with a side length of 16. R,S,T AND Q are midpoints of the sides of ABCD .

What is the area of RSTQ?

Sol. AB=16

AR=RB=BS=SC=TC=DT=QD=AQ=8

AR2 + AQ2 = RQ2  (By Pythagoras Theorem)

RQ=8 √2

Similarly RQ=RS=QT=ST= 8 √2

In △ARQ, ∠QAR = 90°, AR = AQ

∠AQR = ∠DQT (Angles opposite to equal sides are equal)

∠AQR = ∠DQT = 45°

∠RQT=90(Linear pair)

Therefore, RSTQ is square

Area (8 √2)2= 128 sq units

Problem 10) In the given figure, AD||BC. Find the value of x.

a) x=8, $$\frac { 15 }{ 3 }$$

b) x=8, $$\frac { 11 }{ 3 }$$

c) x=8, $$\frac { 13 }{ 3 }$$

d) x=7, $$\frac { 13 }{ 3 }$$

Sol.

$$In\quad \triangle ODA\quad and\quad \triangle OBC,\quad AD\parallel BC\\ \angle ODA=\angle OBC$$ (Alternate interior Angles are equal)

$$\angle OAD=\angle OCB$$ (Alternate interior angles are equal)

$$\triangle ODA\sim \triangle OBC\\ \frac { OD }{ OB } =\frac { DA }{ BC } =\frac { OA }{ OC } \\ \frac { OD }{ OB } =\frac { AO }{ OC } \\ \frac { x-5 }{ x-3 } =\frac { 3 }{ 3x-19 } \\ (x-5)(3x-19)=3x-9\\ 3{ x }^{ 2 }-19x-15x+95-3x+9=0\\ 3{ x }^{ 2 }-37x+104=0\\$$

$$x=\frac { 37\pm \sqrt { { 37 }^{ 2 }-4(3)(104) } }{ 6 } =\frac { 37\pm 11 }{ 6 } \\ =8,\frac { 11 }{ 3 }$$

c) is correct.

Problem 11) In the right angle $$\triangle PQR$$, find RS?

Sol.

$$In\quad \triangle QRP\quad and\quad \triangle RSP\\ \angle QRP=\angle RSP\quad (both\quad are\quad right\quad angles)\\ \angle QPR=\angle SPR\quad (common)\\ Therefore,\quad \triangle QRP\sim \triangle RSP\\ \frac { QR }{ RS } =\frac { RP }{ SP } =\frac { QP }{ RP } \\ \frac { 5 }{ RS } =\frac { QP }{ 12 } \\ In\quad \triangle QRP\\ { QR }^{ 2 }+{ RP }^{ 2 }={ QP }^{ 2 }(By\quad pythagoras\quad theorem)\\ { 12 }^{ 2 }+{ 5 }^{ 2 }={ QP }^{ 2 }\\ QP=13\\ \frac { 5 }{ RS } =\frac { 13 }{ 12 } \\ RS={ (\frac { 60 }{ 13 } ) }^{ Ans }$$

Problem 12)

The area of semicircle O is $$16\pi$$.

$$\angle CDE=\angle ABE={ 30 }^{ 0 }\\ AC=\sqrt { 2 }$$

Quantity A:  DE

Quantity B:  $$3\sqrt { 5 }$$

Sol.

$$In\quad \triangle EDC\quad and\quad \triangle EBA\\ \angle EDC=\angle EBA\\ \angle CED=\angle AEB={ 90 }^{ 0 }\\ \triangle EDC\sim EBA\\ \frac { EC }{ AE } =\frac { DC }{ BA } \\ Let\quad EC\quad be\quad x\\ In\quad \triangle EDC\\ \angle CED={ 90 }^{ 0 },\quad \angle CDE={ 30 }^{ 0 },\quad \angle DCE={ 60 }^{ 0 }$$

Recall that the sides of a 30-60-90 triangle are in the proportion

$$x:x\sqrt { 3 } :2x$$.

.Or if you are familiar with trigonometry, just recall that

$$\sin { { 30 }^{ 0 } } =\frac { 1 }{ 2 } =\frac { perpendicular }{ hypotenuse } \\ Therefore,\quad \frac { CE }{ CD } =\frac { 1 }{ 2 } =\frac { x }{ CD } ,\quad CD=2x\\ \frac { EC }{ AE } =\frac { DC }{ BA } \\ \frac { x }{ x+\sqrt { 2 } } =\frac { 2x }{ BA }$$

Area of semicircle = $$\frac { \pi { r }^{ 2 } }{ 2 } =16\pi$$ $${ r }^{ 2 }=36,\quad r=4\sqrt { 2 } =OB\\ \frac { x }{ x+\sqrt { 2 } } =\frac { 2x }{ 8\sqrt { 2 } } \\ x+\sqrt { 2 } =4\sqrt { 2 } \\ x=3\sqrt { 2 } \\ Therefore,\quad CE=3\sqrt { 2 } ,\quad CD=6\sqrt { 2 }$$

Side of $$\triangle CDE$$ are in ratio $$x:x\sqrt { 3 } :2x$$ $$CE=3\sqrt { 2 } ,\quad CD=6\sqrt { 2 } \\ ED=3\sqrt { 6 } \\ Therefore,\quad DE=3\sqrt { 6 }$$

(Quantity A is greater than Quantity B).

Problem 13) What is the area of a triangle that has two sides that each have a length of 10, & whose perimeter is equal to that of a square whose area is 81?

a) 30

b) 48

c) 36

d) 60

e) 42

Sol.

Area of square = 81

Side of square = $$\sqrt { 81 } =9$$

Perimeter of square = 4×9 = 36

Perimeter of triangle = 36

Two sides of triangle have a length of 10.

Third side of triangle = 36-10-10 = 16

Any triangle that has equal sides is an isosceles triangle. Drop a perpendicular to divide it into two individual right triangles.

$$In\quad \triangle ABC,\quad AD\bot BC\\ In\quad \triangle ABD,\quad { AB }^{ 2 }={ BD }^{ 2 }+{ AD }^{ 2 }\quad (by\quad pythagoras\quad theorem)\\ { 10 }^{ 2 }={ 8 }^{ 2 }+{ AD }^{ 2 }\\ AD=6\\ Area\quad of\quad \triangle ABC=\frac { 1 }{ 2 } \times Base\times Height=\frac { 1 }{ 2 } \times 16\times 16=48\quad Sq.units$$

Problem 14) Michael wants to split his rhombus-shaped garden into two triangular plots, one for planting strawberries & one for planting vegetables, by erecting a fence from one corner of the garden to the opposite corner. If one corner of the garden measures $${ 60 }^{ 0 }$$ & one side of the garden measures $$x$$ meters, which of the following could be the area of the vegetable plot?

Indicate all such value

a) $$\frac { { x }^{ 2 }\sqrt { 3 } }{ 2 }$$

b) $$\frac { { x }^{ 2 }\sqrt { 3 } }{ 4 }$$

c) $$\frac { { x }^{ 2 } }{ 2 }$$

d) $$\frac { { x }^{ 2 }\sqrt { 3 } }{ 4 }$$

e) $$\frac { { x }^{ 2 } }{ 4 }$$

Sol.  Whenever a geometry problem comes without a figure, start by drawing one yourself:

Michael wants to split this garden into two triangles.

Try it this way first:

$$\triangle ACD$$ is an equilateral. (Because all angles are equal to $${ 60 }^{ 0 }$$)

Area of triangle ACD = Area of vegetable plot =

$$=\frac { \sqrt { 3 } { x }^{ 2 } }{ 4 }$$

This is one of the choices, but the garden might be split the other way:

$$Draw\quad AE\bot BD,\quad In\triangle AEB\\ \angle AEB={ 90 }^{ 0 },\quad \angle ABE={ 30 }^{ 0 },\quad \angle EAB={ 60 }^{ 0 }$$

Recall that the sides of a 30-60-90 triangle are in the proportion $$x:\sqrt { 3 } x:2x$$

Or if you are familiar with basic trigonometry

$$In\quad \triangle AEB,\quad \sin { { 30 }^{ 0 } } =\frac { 1 }{ 2 } =\frac { AE }{ AB } =\frac { AE }{ x } \\ AE=\frac { x }{ 2 } ,\quad EB=\frac { x\sqrt { 3 } }{ 2 }$$

Therefore, we can easily derive that

$$AE=\frac { x }{ 2 } ,\quad EB=\frac { x\sqrt { 3 } }{ 2 }$$ (by using properties of triangle or by using basic trigonometry)

Similarly, we can find that $$ED=\frac { \sqrt { 3 } x }{ 2 } ,\quad BD=\sqrt { 3 } x$$ $$Ar(\triangle ABD)=\frac { 1 }{ 2 } \times base\times height\\ =\frac { 1 }{ 2 } \times \sqrt { 3 } x\times \frac { x }{ 2 } \\ =\frac { \sqrt { 3 } { x }^{ 2 } }{ 4 }$$

This is same as the previous area, so only one answer choice is B.

If you set the polynomial $$a{ x }^{ 2 }+bx+c$$ equal to zero, where a, b and c are constants and

a ≠ 0 , there is a special name for it. It is called a quadratic, You can find the values for x that make the equation true.

Example:     $${ x }^{ 2 }-3x+2=0$$

To find the solutions, also called roots of an equation, start by factoring whenever possible. You can factor $${ x }^{ 2 }-3x+2$$ into (x-2)  (x-1) , making the quadratic equation:

(x-2) (x-1) = 0

To find the roots, set each binomial equal to 0. That gives you:-

(x-2) = 0      or     (x-1) =0

Solving for x, you get x=2 or x=1

The solutions to a quadratic in the form $$a{ x }^{ 2 }+bx+c=0$$ can also be found using the quadratic formula provided a, b and c are real numbers and a ≠ 0 then:

$$x\quad =\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$$

EXAMPLE:- Find the Solution of 2x2 + 9x +9 = 0

Solution :- a=2 , b=9 , c=9

$$x\quad =\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$$ $$x\quad =\quad \frac { -9\pm \sqrt { 81-4(2)(9) } }{ 2(2) }$$

=  (-9±3) / 4

ANS= -3     or -3/2

Sum of the roots of a quadratic equation =  -b/a

Product of the roots of a quadratic equation = c/a

• $$D\quad =\quad { b }^{ 2 }-4ac$$ is the discriminant of a quadratic equation.

If D < 0 (i.e. the discriminant is negative) then the equation has no real roots.

If D= 0 then the quadratic equation has two equal roots.

If D> 0 (i.e. the discriminant is positive) then the quadratic equation has two distinct roots.

Let f(x)= ax2 + bx +c , Where a, b, c are real and a ≠ 0, Then y=f(x) represents a parabola, Whose axis is parallel to y- axis. This gives the following cases:

1) a> 0 and D = (b2 -4ac) < 0 (Roots are imaginary).

$$f(x)>0\quad \forall \quad x\in R$$

2) a > 0 and D = 0 (The roots are real and equal)

f(x) will be positive for all values of x except at the vertex where f(x) = 0

3) When a> 0 and D > 0 (The roots are real and distinct)

F(x) will be equal to zero when x is equal to either of $$\alpha \quad or\quad \beta$$.

$$f(x)>0\quad \forall x\in (-\infty ,\alpha )\bigcup (\beta ,\infty )\\ and\quad f(x)<0\quad \forall x\in (\alpha ,\beta )$$

4) When a < 0 and D=0 (Roots are imaginary)

$$f(x)<0\quad \forall x\in R$$

5) When a<0 and D=0 (Roots are real and equal)

F(x) is negative for all values of x except at the vertex Where f(x)= 0

6) When a<0 and D>0 (Roots are real and distinct)

$$f(x)<0\quad \forall x\in (-\infty ,\quad \alpha )\bigcup (\beta ,\quad \infty )\\ f(x)>0\quad \forall x\in (\alpha ,\quad \beta )$$

## Sample Problems

Problem 1)  If -4 is a solution for x in the equation $${ x }^{ 2 }+kx+8=0$$, what is k?

Sol.  If -4 is a solution for x in the equation.

Then,$${ (-4) }^{ 2 }+k(-4)+8=0\\ 16-4k+8=0\\ -4k=-24\\ k={ (6) }^{ Ans }$$

Problem 2) Given that $${ x }^{ 2 }-13x=30$$, what is x?

Sol.

$${ x }^{ 2 }-13x=30\\ { x }^{ 2 }-13x-30=0\\ x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ =(\frac { 13\pm \sqrt { 169-4(-30) } }{ 2 } )\\ =\frac { 13\pm \sqrt { 289 } }{ 2 } =\frac { 13\pm 17 }{ 2 } \\ =\frac { -4 }{ 2 } \quad or\quad \frac { 30 }{ 2 } \\ =-2\quad or\quad 15$$

Problem 3) If the area of a certain square (expressed in square meters) is added to its perimeter (expressed in meters), the sum is 77. What is the length of a side of the square?

Sol.  Let side of square be x.

Area if square $$={ x }^{ 2 }$$

Perimeter of square = 4x

$${ x }^{ 2 }+4x=77\\ { x }^{ 2 }+4x-77=0\\ \alpha ,\beta =\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2 } \\ \alpha =\frac { -4+\sqrt { 16+4(77) } }{ 2 } ,\quad \beta =\frac { -4-\sqrt { 16+4(77) } }{ 2 } \\ \alpha =\frac { -4+18 }{ 2 } ,\quad \beta =\frac { -4-18 }{ 2 } \\ \alpha =7,\quad \beta =-11$$

Since the ride of a square must be positive, negative value (-11) has been discorded.

Side of the square = 7

Problem 4)  If $${ x }^{ 2 }-6x-27=0$$ and $${ y }^{ 2 }-6y-4=0$$, what is the maximum value of x+y?

Sol.

$${ x }^{ 2 }-6x-27=0\\ \alpha =\frac { -b+\sqrt { { b }^{ 2 }-4ac } }{ 2a } ,\quad \beta =\frac { -b-\sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \alpha =\frac { 6+\sqrt { 36-4(-27) } }{ 2 } ,\quad \beta =\frac { 6-\sqrt { 36-4(-27) } }{ 2 } \\ \alpha =\frac { 6+\sqrt { 144 } }{ 2 } ,\quad \beta =\frac { 6-\sqrt { 144 } }{ 2 } \\ \alpha =\frac { 6+12 }{ 2 } ,\quad \beta =\frac { 6-12 }{ 2 } \\ \alpha =9,\quad \beta =-3$$

x may be 9 or -3.

$${ y }^{ 2 }-6y-40=0\\ \alpha =\frac { -b+\sqrt { { b }^{ 2 }-4ac } }{ 2a } ,\quad \beta =\frac { -b-\sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \alpha =\frac { 6+\sqrt { 36-4(-40) } }{ 2 } ,\quad \beta =\frac { 6-\sqrt { 36-4(-40) } }{ 2 } \\ \alpha =\frac { 6+\sqrt { 196 } }{ 2 } ,\quad \beta =\frac { 6-\sqrt { 196 } }{ 2 } \\ \alpha =\frac { 6+14 }{ 2 } ,\quad \beta =\frac { 6-14 }{ 2 } \\ \alpha =10,\quad \beta =-4$$

y may be 10 or -4.

For maximum value of (x+y),

x should be maximum & y should be maximum.

Maximum value of (x+y) = 9+10 = 19

Problem 5)  xy>0

Quantity A: $${ (x+y) }^{ 2 }$$

Quantity B: $${ (x-y) }^{ 2 }$$

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

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

Sol.  Quantity A: $${ (x+y) }^{ 2 }$$

Quantity B: $${ (x-y) }^{ 2 }$$

Now subtract $${ x }^{ 2 }+{ y }^{ 2 }$$ from both columns.

Then, Quantity A:       2xy

Quantity B:         -2xy

xy>0

xy is positive, so the value in quantity A will be positive, regardless of the values of x and y. Similarly, the value in Quantity B will always be negative, regardless of the values of x & y.

Quantity A is larger.

Problem 6)  Find the value of $$\sqrt { 6+\sqrt { 6+\sqrt { 6+——— } } }$$

a) 4

b) 3

c) 5

d) 5

Sol.  Let

$$y\quad =\quad \sqrt { 6+\sqrt { 6+\sqrt { 6+——— } } } \\ y\quad =\quad \sqrt { 6+y } \quad (Squaring\quad both\quad sides)\\ { y }^{ 2 }=6+y\\ { y }^{ 2 }-y-6=0\\ \alpha =\frac { 1+\sqrt { 1-4(-6) } }{ 2 } ,\quad \beta =\frac { 1-\sqrt { 1-4(-6) } }{ 2 } \\ \alpha =\frac { 1+5 }{ 2 } ,\quad \beta =\frac { 1-5 }{ 2 } \\ \alpha =3,\quad \beta =-2\\ Therefore,\quad y=3\quad (Since\quad \sqrt { 6+\sqrt { 6+\sqrt { 6—— } } } cannot\quad be\quad negative)$$

Problem 7)  If the roots of the equation , differ by 2, then which of the following is true?

a) $${ c }^{ 2 }=4(c+1)$$

b) $${ b }^{ 2 }=4(c+1)$$

c) $${ c }^{ 2 }=b+4$$

d) $${ b }^{ 2 }=4(c+2)$$

Sol.

$${ x }^{ 2 }-bx+c=0$$

Let $$\alpha \quad and\quad \beta$$ be the roots of equation.

$$\alpha +\beta =b\quad \quad \alpha \beta =c\\ \alpha -\beta =2\\ { (\alpha +\beta ) }^{ 2 }={ \alpha }^{ 2 }+{ \beta }^{ 2 }+2\alpha \beta \\ (Subtract\quad 4\alpha \beta \quad from\quad both\quad sides)\\ { (\alpha +\beta ) }^{ 2 }-4\alpha \beta ={ \alpha }^{ 2 }+{ \beta }^{ 2 }+2\alpha \beta -4\alpha \beta \\ { \alpha }^{ 2 }-2\alpha \beta +{ \beta }^{ 2 }\\ ={ (\alpha -\beta ) }^{ 2 }\\ \\ { b }^{ 2 }-4c={ 2 }^{ 2 }=4\\ { b }^{ 2 }=4c+4\\ Therefore,\quad { ({ b }^{ 2 }=4(c+1)) }^{ Ans }$$

## GRE: Solving Inequalities

Inequalities may be written with the symbols Shown here:-

 < Less than > Greater than ≤ Less than or equal to ≥ Greater than or equal to

A range of values is often expressed on a number line.

Two ranges are shown below:-

a) Represents the set of all numbers between -4 and 0 excluding the endpoints -4 and 0 or -4< x < 0

b) Represents the set of all numbers greater than -1, upto and including 3, or -1 < x ≤ 3

Rules for working with inequalities:-

• Treat inequalities like equations when adding or subtracting terms or when multiplying/Dividing by a positive number on both sides of the inequality.
• Flip the inequality sign if you multiply or divide both sides of an equality by a negative number.,

Problem 1) Solve for x and represent the solution set on a number line.

$$3-\frac { x }{ 4 } \ge 2$$

Sol.     Multiply both sides by 4                              12-x  ≥ 8

Subtract 12 from both sides                        -x ≥ -4

Multiply both sides by -1 and flip the inequality symbol      x ≤ 4

Problem 2) Solve the inequality

$$\frac { 3x }{ 2 } -\frac { 3x }{ 4 } <\frac { 7x }{ 4 } -1$$

Sol.

$$\frac { 3x }{ 2 } -\frac { 3x }{ 4 } -\frac { 7x }{ 4 } <-1\quad (Subtract\quad \frac { -7x }{ 4 } \quad from\quad both\quad sides)\\ \frac { 3x }{ 4 } -\frac { 7x }{ 4 } <-1\\ \frac { -4x }{ 4 } <-1\\ -x<-1\quad (Multiply\quad both\quad sides\quad by\quad -1)\\ x>1$$

Absolute Value: The absolute value of a number describes how far that number is away from 0 on a number line. The symbol of absolute value is |number|.

STEPS TO SOLVE ABSOLUTE VALUE EQUATION:-

• Take what’s inside the absolute value sign and set up two equations.First sets the positive value equal to the other side of the equation, and the second sites the negative value equal to the other side.
• Solve both equations:

Problem 3)        $$|2x+4|=30$$

Sol.

$$2x+4=30\quad or\quad -(2x+4)=30\\ x=13\quad \quad \quad \quad \quad \quad \quad +2x+4=-30\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x=-17\\ x=13\quad or\quad (-17)$$

Inequalities and absolute value:-

### Steps for solving questions involving both inequalities and absolute value

• Set up Two Equations:- The first inequality replaces the absolute value with the positive of what’s inside and the second replaces the absolute value with the negative of what’s inside.

Problem 4)  | x | ≥ 4

Sol.

+( x ) ≥ 4 or –x ≥ 4

x ≥ 4 or x ≤ -4

Problem 5) | x+3 | ≤ 5

Sol.

+( x+3) < 5 and        -( x+3 ) < 5

x < 2                        x > -8

The only numbers that make the original inequality true are those that are true for both inequalities. X should be greater than -8 and less than 2.

NOTE:-   In conclusion, for solving absolute value expressions that are greater than some quantity set up two equations & after sol; having two equations.We have to take real numbers that are satisfied by either equation. However, for solving absolute value expressions that are less than some quantity set up two equations and after solving two equations, We have to take real numbers that are satisfied by both the equations.

Problem 6) If |3x + 7|  ≥ 2x + 12 , then which of the following is TRUE?

a) $$x\le \frac { -19 }{ 5 }$$

b) $$x\ge \frac { -19 }{ 5 }$$

c) $$x\ge 5$$

d) $$x\le \frac { -19 }{ 5 } \quad or\quad x\ge 5$$

Ans:- ( 3x + 7 ) ≥ 2x +12                               -( 3x + 7 ) ≥ 2x + 12

x ≥ 5                                                    -3x ≥ 2x +19

-5x ≥ 19

x ≤  -19/5

$${ (x\le \frac { -19 }{ 5 } \quad or\quad x\ge 5) }^{ Ans }$$

a), c) and d) are correct answers.

Using extreme values: Extreme values are helpful where the questions involve the potential range of value for variables in the problem.

Problem 7) If  -7  ≤ a  ≤ 6   and  -7  ≤ b  ≤ 8, What is the maximum possible value for ab?

Sol. Let us consider the different extreme value scenarios for a, b and ab.

abab
Min -7Min -749
Min -7Max 8-56
Max 6Min -7-42
Max 6Max 848

We can easily reckon that ab is maximized when we take the negative extreme values for both a and b. Therefore maximum value of  ab = 49

Problem 8) If $$|y|\le -4x\quad and\quad |3x-4|=2x+6$$,what is the value of x?

Sol.      $$|y|\le -4x$$

Any absolute value cannot be negative. |y| must be positive & positive value cannot be lesser than any negative value. Therefore, -4x must be positive.

For -4x to be positive, x must be negative.

Now solve the absolute value equation by using the identity that |a|=a when a is positive or zero & |a|=-a when a is negative.

$$|3x-4|=2x+6\\ 3x-4=2x+6\quad \quad or\quad -(3x-4)=2x+6\\ 3x-4-2x-6=0\quad \quad \quad \quad -3x+4=2x+6\\ x=10\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -5x=2\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x=(\frac { -2 }{ 5 } )$$

x must be negative. Therefore, x can only be $$\frac { -2 }{ 5 } [latex]. Problem 9) If [latex]2{ (x-1) }^{ 3 }+3\le 19$$, which of the following must be true?

a) $$x\ge 3$$

b) $$x\le 3$$

c) $$x\ge -3$$

d) $$x\le -3$$

e) $$x<-3\quad or\quad x>3$$

Sol.

$$2{ (x-1) }^{ 3 }+3\le 19\\ 2{ (x-1) }^{ 3 }\le 16\\ { (x-1) }^{ 3 }\le 8$$

Taking the cube root of an inequality is permissible here, because cubing a number, while squaring it, does not change its sign.

$$x-1\le 2\\ { (x\le 3) }^{ Ans }$$

Problem 10) Which of the following could be the graph of all values of $$x$$ that satisfy the inequality

$$4-11x\ge (\frac { -2x+3 }{ 2 } )$$

Sol.First solve the inequality,

$$4-11x\ge \frac { -2x+3 }{ 2 } \\ 8-22x\ge 20x\\ \frac { 1 }{ 4 } \ge x\\ x\le \frac { 1 }{ 4 }$$

Thus, the correct choice should show the gray line beginning to the right of zero (in the positive zone), & continuing indefinitely into the negative zero.

Problem 11) $$p+|k|>|p|+k$$

Quantity A:         p

Quantity B:         k

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

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

Sol.  In general, there are four cases for the signs of p & k, some of which can be ruled out by the constraints of this equation:

 p k p+|k|>|p|+k + + In this case, both sides should be positive. Therefore, this case is not true. + - This case is true. (positive value)+absolute value>(positive value)+(negative value) - - It may between. Both sides are positive plus a negative. For this condition to the true (|k|>|p|) - + This case is not true. k+(negative value)

Additionally, check whether p or k could be zero.

If p=0, p+|k|>|p|+k is equivalent to |k|>k.

This is true when k is negative.

If k=0, p+|k|>|p|+k (when k=0)

p>|p|

This is not true for any p value. So, there are three possible cases for p & k.

Let’s interpret this more: –

Use identity that |a|=-a, when a is negative.

 p k Interpret + - p=positive >k=negative - - p+|k|>|p|+k p-k>-p+k 2p>2k Therefore, p>k 0 - p=0>k=negative Therefore, p>k

In all cases, p is greater than k.

Therefore, Quantity A is greater.

Problem 12) $$|x|y>x|y|$$

Quantity A: $${ (x+y) }^{ 2 }$$

Quantity B: $${ (x-y) }^{ 2 }$$

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

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

Sol.  In general, there are four cases for the signs of x & y, some of which can be ruled out by the constraint in the question using identity that |a|=-a when a is negative.

 x y |x|y>x|y| is equivalent to True or False? + + xy>xy False: xy=xy + - xy>x(-y) False: xy is negative & (-xy) is negative. - + (-x)y>(-xy) True: (-xy) is positive & xy is negative - - (-x)y>(-xy) False: -xy=-xy

Note that if either x or y equals 0, that case would also fail the constraint.

Quantity A: $${ (x+y) }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }+2xy$$

Quantity B: $${ (x-y) }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }-2xy$$

X is negative & y is positive.

Therefore, xy is negative.

$${ x }^{ 2 }+{ y }^{ 2 }-2xy>{ x }^{ 2 }+{ y }^{ 2 }+2xy\\ { (x-y) }^{ 2 }>{ (x+y) }^{ 2 }$$

Therefore, Quantity B is greater.

Problem 13) If y<0 & 4x>y, which of the following could be equal to

$$\frac { x }{ y }$$  ?

a) 0

b) $$\frac { 1 }{ 4 }$$

c) $$\frac { 1 }{ 2 }$$

d) 1

e) 4

Sol.  4x>y (divide both sides by y)

$$\frac { 4x }{ y } <1$$

Remember to switch the direction of the inequality sign when multiplying or dividing by a negative Quantity.

$$\frac { 4x }{ y } <1$$ (Now divide both sides by 4)

$$\frac { x }{ y } <\frac { 1 }{ 4 }$$

The only answer choice less than $$\frac { 1 }{ 4 }$$ is 0.

Problem 14) If |1-x|=6 & |2y-6|=10, which of the following could be the value of xy?

Indicate all such value?

a) -40

b) -10

c) -14

d) 56

Sol. Remember that to solve the absolute value equation,

First set the positive value equal to the other side of the equation & the second set the negative value equal to the other side of the equation.

$$|1-x|=6,\quad -(1-x)=6\\ -1+x=6\\ x=7\\ \\ |1-x|=6,\quad +(1-x)=6\\ 1-x=6\\ x=-5\\ Therefore,\quad x=-5\quad or\quad 7\\ \\ |2y-6|=10,\quad 2y-6=10\\ 2y=16,\quad y=8\\ \\ 2y-6=-10,\quad 2y=-10+6=-4\\ y=-2\\ Therefore,\quad y=8\quad or\quad y=-2$$

Now calculate all four possible combinations for xy:

(-5) (8) =-40

(-5) (-2) =10

7(8) =56

7(-2) =-14

The correct answer are -40, -14 & 56 only.

Problem 15) If & x<0, which of the following could be the value of x?

Indicate all such values:

a) -6

b) -14

c) -18

d) -22

e) 36

Sol.  Remember that for solving absolute value expressions that are greater than some quantity, set up two equations (the first inequality replaces the absolute value with the positive of what’s inside & the second replaces the absolute value with the negative of what’s inside) & we have to take real numbers that are satisfied by either equation.

$$\frac { |x+4| }{ 2 } >5,\quad |x+5|>10\\ x+4>10\quad \quad \quad -(x+4)>10\\ x>6\quad \quad \quad \quad \quad \quad -x>14\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x<-14$$

x>6 is not a valid solution range, as the other inequality indicates that x is negative.

x<-14, note that this fits the other inequality which states that x<0.

If x<-14, only -18 & -22 are correct answers.