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The Ultimate Guide of Gre Inequalities-CrackVerbal

Take away must-know inequality concepts and strategies that will help you navigate tricky questions on the GRE with limited information-The basic and advanced properties of Inequalities-<br>How to solve inequality-based questions on the GRE,<br>How to use the concepts of inequalities to solve Quantitative comparisons.If you are preparing for GRE test then do take up our free GRE online course at https://greonline.crackverbal.com/

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The Ultimate Guide of Gre Inequalities-CrackVerbal

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  1. THE MUST-HAVE GRE INEQUALITIES HANDBOOK

  2. O UR G RE INSTRUC TO R S Saikiran Dudyala Verbal Instructor Saikiran has six years of experience in the teaching industry and has trained hundreds of students for a plethora of courses ranging from spoken English and communication skills to GMAT, GRE, SAT, PSAT, ACT, IELTS, and TOEFL. An alumnus of the prestigious Guindy Engineering college in Chennai, and an engineer by qualification, he chose to heed to the call of passion and switched tracks to the education industry. Shirupa Gupta Verbal Instructor Shirupa is an alumna of the IIMB MPWE ( Management Program for Women Enterpreneurs). From teaching art to professors at ISB and IIM-B, to training students for standardized tests such as the GRE and GMAT, Shirupa truly is a multi-faceted instructor with a zest for teaching. She has the distinction of having represented India as a Youth Fellow at the United Nations ITU Telecom Asia 2002, and makes and sells paintings in her free time. Follow Shirupa on Mohammed Junaid Quant Instructor Junaid is the quintessential quants guy; a person who solves GMAT and GRE quant questions as part of his day-job, and solves puzzles and problems in his free time for fun. He has an M.Sc. in Mathematics, and 5 years of experience in training students across Test Prep courses for the GMAT, GRE, SAT, etc. When he? s not living and breathing Math, Junaid enjoys playing volleyball and cricket. Aditya Kumar Quant Instructor Aditya is an engineer by qualification, and an alumnus of the famous M.S Ramaih Institute of Technology. His foray into the education industry was a serendipitous accident; but teaching soon turned into a calling as he discovered how good he was at it. He has 3 years of experience training students for test-prep courses for the GMAT, GRE, and SAT. His super-power is his ability to uncomplicate a tangled mess of information into simple, byte-sized facts.

  3. INTRO DUC TIO N Hello there! You might be wondering, why did we write an entire E-book on Inequalities? Well, compared to other question types on the GRE, inequality questions are an especially slippery slope that send many test-takers tumbling down on the path to not-so-great Quant scores! This is because, while most equations indicate a clear relationship between two statements, inequality problems ask test takers to answer questions based on limited information. So, without further ado, let us examine some must-know inequality concepts and strategies that will help us navigate these tricky questions with limited information .

  4. We'll first start with the fundamental concept of inequalities, followed by basic properties and then move on to explore the complexities involved with some additional properties. Finally we will apply the key takeaways in quantitative comparison questions 1. Inequalities 2. Basic Properties 3. Advanced Properties 4. Inequalities in Quantitative Comparisons

  5. 1 . WHAT ARE INEQ U ALITIES? Equations and inequalities are both mathematical sentences formed by relating two expressions to each other. In an equation, the two expressions are deemed equal which is shown by the symbol =. Where as in an inequality, the two expressions are not necessarily equal - this is indicated by the symbols: >, <, ? or ?. x > y ----> x is greater than y x ? y ----> x is greater than or equal to y x < y ----> x is less than y x ? y ----> x is less than or equal to y

  6. INEQ U ALITIES O N A NUMBER LINE Number lines, such as those shown below, are an excellent way to visualize exactly what a given inequality means. A closed (shaded) circle at the endpoint of the shaded portion of the number line indicates that the graph is inclusive of that endpoint, as in the case of ? or ?. An open (unshaded) circle at the endpoint of the shaded portion of the number line indicates that the graph is not inclusive of that endpoint, as in the case of < or >

  7. 2. BASIC PRO PER TIES There are 2 basic properties of inequalities which we can quickly prove using the example below. If we consider the true inequality 4 < 8 Adding 2 to both sides 6 < 10 (the inequality sign holds true) Subtracting 2 from both sides 2 < 6 (the inequality sign holds true) Multiplying both sides by +2 8 < 16 (the inequality sign holds true) Dividing both sides by +2 2 < 4 (the inequality sign holds true) Adding or subtracting the same expression to both sides of an inequality does not change the inequality . Multiplying or dividing the same positive number to both sides of an inequality does not change the inequality.

  8. Again considering the true inequality 4 < 8 Multiplying both sides by -2 -8 > -16 (the inequality sign reverses) Dividing both sides by -2 -2 > -4 (the inequality sign reverses) Multiplying or dividing the same negative number to both sides of an inequality reverses the inequality - this is also called the flip rule of inequalities.

  9. A LITTLE Q &A ANYO NE? Now that we are done with the basic properties of inequalities, here are a couple of questions to make you think. Question: Can we add or subtract a variable on both sides of an inequality? Answer: Yes, because adding or subtracting a variable is the same as adding or subtracting a number. Question: Can we multiply or divide both sides of an inequality by a variable? Answer: No, we cannot, if we do not know the sign of the number that the variable stands for. The reason is that you would not know whether to flip the inequality sign. Let us illustrate this with an example - If x/y > 1, most test-takers make the mistake of deducing that x>y, by multiplying both sides by y. But we haven? t been given any information about the sign of the number that the variable y stands for. If x = 3 and y = 2 then the above relation x/y > 1 will hold true and x will be greater than y. However if x = -3 and y= -2 then the above relation x/y > 1 will again hold true but x will not be greater than y. If x/y > 1, the only fact that can definitely be deduced is that both x and y are of the same sign .

  10. EXAMPLES Question: If a, b, c are non zero integers and a > bc, then which of the following must be true : I.a/b > c II.a/c > b III.a/bc > 1 A.I only B.II only C.III only D.I, II and III E.None of these Solution: Now the trap answer here will be D (I, II and III). The general tendency will be to multiply both sides of the first inequality a/b > c by b to get a > bc, both sides of the second inequality by c to get a > bc and both sides of the third inequality by bc to get a > bc. Remember we can never multiply or divide both sides of an inequality by a variable if the sign of the variable is not known and in the above problem the signs of b and c are not known. The above statements I, II and III can be true, if b and c are both positive but they will not be true if b and c are negative and since the question is of a "must be true" type, the answer here must be E.

  11. Solving an inequality means finding all of its solutions. A ? solution? of an inequality is a number which when substituted for the variable satisfies the inequality. The steps to solve a linear inequation are as follows: - - - Isolate the variable and always keep the variable positive Solve using the properties of inequalities Represent the inequality on a number line Solve: -6x + 4 ? -2 Isolating the variable by subtracting 4 from both sides we get -6x ? -6 Dividing both sides by -6 and flipping the inequality sign we get x ? 1

  12. 3. AD VANCED CO NCEPTS Well, so far, we saw how the basic operations are applied to inequalities. It is now time to delve into more complex properties of inequalities, dealing with : 1. Inequalities in fractions 2. Squaring Inequalities 3. Reciprocal of Inequalities 4. Like Inequalities 5. Max Min Concept of Inequalities 6. Quadratic Inequalities Let's get started!

  13. # 1 : INEQ U ALITIES IN FRAC TIO NS - All proper fractions on the number line can be represented using the range -1 < x < 1 where x represents the proper fraction. - All positive proper fractions can be represented using the range 0 < x < 1 where x represents the positive proper fraction. For all proper fractions (0 < x < 1), ?x > x > x2 - If x = ¼ then ?x = ½ and x^2 = 1/16 Clearly here, ½ > ¼ > 1/16

  14. If x = 0.888, y = ?0.888 and z = (0.888)^2 which of the following is true A.x < y < z B.x < z < y C.y < x < z D.z < y < x E.z < x < y Solution: Since0.888 is a fraction, ?0.888 0.888 > (0.888)^2 y > x > z Reversing the inequality we get z < x < y Answer : E

  15. # 2 : SQ U ARING INEQ U ALITIES WE cannot square both sides of an inequality unless we know the signs of both sides of the inequality. If both sides are known to be negative then flip the inequality sign when you square. For instance, if a < -4, then the left hand side must be negative. Since both sides are negative, you can square both sides and reverse the inequality sign : a^2 > 16. However, if a > -4, then you cannot square both sides, because it is unclear whether the left side is positive or negative. If a is negative then a^2 < 16, but if x is positive then x^2 could be either greater than 9 or less than 9. If both sides are known to be positive, do not flip the inequality sign when you square. For instance, if a > 4, then the left side must be positive; since both sides are positive you can square both sides to yield a^2 > 16. However if a < 4 then you cannot square both sides, because it is unclear whether the left side is positive or negative.

  16. If one side is positive and one side is negative then you cannot be sure. For instance, if you know that a < b, a is negative, and b is positive, you cannot make any determination about x^2 vs. y^2. If for example, x = -2 and y = 2, then x^2 = y^2. If x = -2 and y = 3, then x^2 < y^2. If x = -2 and y = 1, then x^2 > y^2. It should be noted that if one side of the inequality is negative and the other side is positive, then squaring is probably not warranted . If signs are unclear, then you should not square. Put simply, we would not know whether to flip the sign of the inequality once you have squared it .

  17. # 3: RECIPRO C AL INEQ U ALITIES Taking the reciprocal of both a and b can change the direction of the inequality. The general rule is that when a < b then: - For (1/a ) > (1/b) when a and b are positive , flip the inequality. If 2 < 3, then ½ > 1/3 - For (1/a) > (1/b) when a and b are negative , flip the inequality. If -3 < -2, then 1/-3 > 1/-2 - For (1/a) < (1/b) when a is negative and b is positive , do not flip the inequality. If -3 < 2, then 1/-3 < 1/2 - If you do not know the sign of a or byou cannot take reciprocals. In summary, if you know the signs of the variables, you should flip the inequality unless a and b have different signs.

  18. If 3 ? 6/(x+1) ? 6, find the range of x Taking the reciprocal of the above range and flipping the inequality sign since the entire inequality is positive 1/3 ? (x + 1)/6 ? 1/6 Multiplying throughout by 6 2 ? (x + 1) ? 1 Subtracting 1 from all sides 1 ? x ? 0 --> 0 ? x ? 1

  19. # 4: LIKE INEQ U ALITIES The only mathematical operation you can perform between two sets of inequalities, provided the inequality sign is the same, is addition. If the signs are not the same then use the properties to flip the inequality sign and then add the two sets of inequalities . Like Inequalities can be added Adding the two inequalities 5 < 7 (TRUE) -8 < -5 (TRUE) -8 < 5 (TRUE) Subtracting the inequalities 1 < 0 (FALSE) 0 < -1 (FALSE) 0 < 1 (TRUE) Multiplying the inequalities 6 < 16 (TRUE) 16 < 6 (FALSE) 16 < 6 (FALSE) - The only mathematical operation you can safely perform between two sets of inequalities, provided the inequality sign is the same is addition. If the signs are not the same then use the properties to flip the inequality sign and then add the two sets of inequalities . -

  20. If 4a + 2b < n and 4b + 2a > m, then b ? a must be A.< (m ? n)/2 B.? (m ? n)/2 C.> (m ? n)/2 D.? (m ? n)/2 E.? (m + n)/2 Given 4a + 2b < n and 4b + 2a > m. Multiplying the second inequality 4b + 2a > m by -1 we get -4b ? 2a < -m. We can always add like inequalities. Now adding the two inequalities 4a + 2b < n and -4b ? 2a < -m 4a + 2b < n -4b - 2a < -m ________________ 2a ? 2b < n ? m Dividing both sides by 2 a ? b < (n ? m)/2 Multiplying both sides by -1 b ? a > (m ? n )/2 Answer : C

  21. # 5: MIN AND MAX INEQ U ALITIES Problems involving optimization: specifically, minimization or maximization problems are a common occurrence on the GRE . In these problems, you need to focus on the largest and smallest possible values for each of the variables, as some combination of them will usually lead to the largest or smallest possible result. Read on to learn from an example!

  22. If -7 ? x ? 6 and -7 ? y ? 8, what is the maximum possible value for xy? To find the maximum and minimum possible values for xy, place the inequalities one below the other and make sure the inequality signs are the same. You need to test the extreme values for x and for y to determine which combinations of extreme values will maximize ab. -7 ? x ? 6 -7 ? y ? 8 The four extreme values of xy are 49, 48, -56 and -42. Out of these the maximum possible value of xy is 49 and the minimum possible value is -56. Whenever two ranges of inequalities are given in x and y and you need to evaluate the value of x + y , x * y, and x ? y then use the max-min concept 1. Place the two inequality ranges one below the other 2. Make sure the inequality signs are the same in both cases 3. If the signs are not the same use the properties we have discussed before to make them the same 4. Now add/multiply/subtract both in a straight line and diagonally to get 4 values 5. The greatest value will be max and the lowest value will be min

  23. 1/2 < x < 2/3 , and y^2 < 100 QUANTITY A QUANTITY B xy 6 Since y^2 < 100 ---> -10 < y < 10 Now placing the two ranges one below the other and finding out the extreme values of xy 1/2 < x < 2/3 -10 < y < 10 The four extreme values of xy here are -5, -20/3 , 5, 20/3. Out of these the maximum value of xy is 20/3 and the minimum value of xy is -20/3. Now since Quantity A can take values from -20/3 to 20/3 a definite relationship cannot be determined with Quantity B. Answer : D

  24. # 6: Q U ADRATIC INEQ U ALITIES 3x^2 ? 7x + 4 ? 0 Factorizing the above quadratic inequation 3x^2 ? 7x + 4 ? 0 ---> 3x^2 ? 3x ? 4x + 4 ? 0 ---> 3x(x - 1) - 4(x - 1) ? 0 ---> (3x - 4)(x - 1) ? 0 we get 1 and 4/3 as critical points. We place them on number line. Since the number line is divided into three regions, now we can get 3 ranges of x i) x < 1 (all values of x when substituted in (3x ? 4)(x ? 1) makes the product positive) ii) 1 ? x ? 4/3 (all values of x when substituted in (3x ? 4)(x ? 1) makes the product negative iii) x > 4/3 (all values of x when substituted in (3x ? 4)(x ? 1) makes the product positive)

  25. At this point we should understand that for the inequality (3x-4)(x-1) ? 0 to hold true, exactly one of (3x-4) and (x-1) should be negative and other one be positive. Let? s examine 3 possible ranges one by one. i) If x > 4/3, obviously both the factors i.e. (3x-4) and (x-1) will be positive and in that case inequality would not hold true. So this cannot be the range of x. ii) If x is between 1 and 4/3 both inclusive, (3x-4) will be negative or equal to zero and (x-1) will be positive or equal to zero. Hence with this range inequality holds true. Correct. iii) If x < 1, both (3x-4) and (x-1) will be negative hence inequality will not hold true. So the range of x that satisfies the inequality 3x^2 ? 7x + 4 ? 0 is 1 ? x ? 4/3 The steps to solve a quadratic inequation are as follows: 1.Isolate the variable and always keep the variable positive. 2.Maintain the Inequation in the form ax^2 + bx + c > 0 or < 0. 3.Obtain the factors of Inequation. 4.Place them on number line. The number line will get divided into the three regions. 5.Mark the rightmost region with + sign, the next region with a ? sign and the third region with a + sign (alternating + and ? starting from the rightmost region). 6.If the Inequation is of the form ax^2 + bx + c < 0, the region having the ? sign will be the solution of the given quadratic inequality. 7. If the Inequation is of the form ax^2 + bx + c > 0, the region having the + sign will be the solutions of the given quadratic inequality.

  26. Question: Will the above procedure hold good even for a cubic or a fourth degree equation? Answer: YES. For a cubic inequality we get 3 critical points which when plotted on the number line divides the number line into 4 regions. Mark the rightmost region as +ve and alternate the sign as shown below Now based on whether the right hand side of the cubic inequality is < 0 or > 0 we get the solution to lie in 2 of the 4 regions.

  27. 4. Q U ANTIT ATIVE CO MP ARISO NS O N THE G RE Now that we are through with the properties of inequalities, lets see how we can make use of these properties in quantitative comparisons. A quantitative comparison question is a big inequality in itself since it asks you to compare and determine which of the two quantities is greater. So the rules of inequalities can be used here, provided the initial comparison is not tampered with. For e.g. If we consider a basic quantitative comparison question where quantity B is clearly greater than quantity A, Quantity A Quantity B 4 6 Adding 2 6 8 (Quantity B is still greater) Subtracting 2 2 4 (Quantity B is still greater) Multiplying by +2 8 Dividing by +2 12 (Quantity B is still greater) 3 (Quantity B is still greater) 2 Multiplying by -2 -8 -12 (Quantity A is greater) Dividing by -2 -2 -3 (Quantity A is greater) It is very evident that if we multiply or divide by a negative number the comparison will never be consistent with the initial comparison.

  28. Here are a few things you need to remember when you are using the properties of inequalities to simplify complex quantitative comparison questions: 1.Add or subtract any value to both quantities 2.Multiply or divide by a positive value 3.Square both sides only when the quantities are both positive 4.Never multiply or divide both quantities by a negative number 5.Never multiply or divide both quantities by a variable if the sign of the variable is unknown 6.If the sign of the variable is always positive then it is possible to multiply or divide both quantities by the positive variable (for e.g. x2 ,since x2 is always positive)

  29. Quantity A Quantity B b ? 3a2 ? 7 6 + b ? 3a2 Instead of plugging in values for a and b we can simplify the quantities by adding 3a2 to both quantities and then subtracting b from both quantities. Solution: Quantity A Quantity B b ? 3a^2 ? 7 6 + b ? 3a^2 Adding 3a^2 b ? 3a^2 ? 7 + 3a^2 6 + b ? 3a^2 + 3a^2 b ? 7 6 + b Subtracting b b ? 7 ? b 6 + b ? b ? 7 6 Answer: B (Quantity B is greater)

  30. Question: Quantity A Quantity B 1/x (x + 1)/x^2 Solution: Instead of plugging in values for x we can simplify the quantities by multiplying x^2 to both quantities since x^2 is always going to be positive Quantity A Quantity B 1/x (x + 1)/x^2 Multiplying x^2 x x + 1 Subtracting x 0 1 Answer: B (Quantity B is greater)

  31. CO NCLUSIO N After reading our simple guide, you should now know what strategies you must employ for inequality questions on the GRE! We hope this guide helps you along the way to a 170 on GRE Quant! CRACKVERBAL GRE COURSE OFFERINGS - - - - - GRE Classroom ( Bangalore and Chennai) GRE Online GRE Personal Tutoring GRE Flashcards GRE Wordtoonz App Would you like to know how we can help you achieve a 330+ on the GRE?

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