January 26, 2019 January 26, 2019 07:00 AM PST 09:00 AM PST Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions. January 27, 2019 January 27, 2019 07:00 AM PST 09:00 AM PST Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 27 Aug 2014
Posts: 74

If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t
[#permalink]
Show Tags
Updated on: 28 May 2015, 06:11
Question Stats:
55% (02:19) correct 45% (02:07) wrong based on 221 sessions
HideShow timer Statistics
If –1 < x < 1 and x ≠ 0, which of the following inequalities must be true? I. x^3 < x II. x^2 < x III. x^4 – x^5 > x^3 – x^2 (A) I only (B) II only (C) III only (D) II and III only (E) I, II, and III Can someone try doing this algebraically?
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by sinhap07 on 28 May 2015, 06:06.
Last edited by Bunuel on 28 May 2015, 06:11, edited 1 time in total.
Renamed the topic and edited the question.



Math Expert
Joined: 02 Sep 2009
Posts: 52451

Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t
[#permalink]
Show Tags
28 May 2015, 06:11



Manager
Joined: 17 Oct 2013
Posts: 54

Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t
[#permalink]
Show Tags
28 May 2015, 10:05
sinhap07 wrote: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be true?
I. x^3 < x II. x^2 < x III. x^4 – x^5 > x^3 – x^2
(A) I only (B) II only (C) III only (D) II and III only (E) I, II, and III
Can someone try doing this algebraically? Are you sure answer is D, for what value a is invalid ?



Current Student
Joined: 13 Nov 2014
Posts: 108

Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t
[#permalink]
Show Tags
28 May 2015, 10:28
viksingh15 wrote: sinhap07 wrote: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be true?
I. x^3 < x II. x^2 < x III. x^4 – x^5 > x^3 – x^2
(A) I only (B) II only (C) III only (D) II and III only (E) I, II, and III
Can someone try doing this algebraically? Are you sure answer is D, for what value a is invalid ? Answer is D. Let's look at the statements one by one Stmt I. x^3 < x if 0<x<1 then x^3< x but if 1<x<0 then x^3>x So this statement is not always true Stmt II. x^2 < x Because we know that x is a number less than one but not equal to zero then x^2 will always be less than x. Why? think of positive fractions (and you can think in terms of positive fractions because the inequality is in regards to x). Lets set x = 1/2, then x^2 = 1/4 and 1/4<1/2 So Stmt II is always true Stmt III. x^4 – x^5 > x^3 – x^2 This one may seem tricky but lets break it down. x^4 – x^5 > x^3 – x^2 = x^4(1x)>x^2(x1). Because lets concentrate on (1x) and (x1). We are given that 1<x<1 which means that (1x)>0 and (x1)<0. x^4 will always be positive and x^2 will always be positive so without doing any math we are looking at positive > negative... which is always true. So Stmt III is always true Answer is D
_________________
Gmat prep 1 600 Veritas 1 650 Veritas 2 680 Gmat prep 2 690 (48Q 37V) Gmat prep 5 730 (47Q 42V) Gmat prep 6 720 (48Q 41V)



eGMAT Representative
Joined: 04 Jan 2015
Posts: 2466

Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t
[#permalink]
Show Tags
28 May 2015, 11:43
Quote: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be true?
I. x^3 < x II. x^2 < x III. x^4 – x^5 > x^3 – x^2
(A) I only (B) II only (C) III only (D) II and III only (E) I, II, and III
Can someone try doing this algebraically? The algebraic approach to this question will be as follows: I. x^3 < xWe will first find the range of values of x for which this inequality holds true. So, let's solve this inequality: x^3  x < 0 => x(x^2  1) < 0 => x(x+1)(x1) < 0 Let's draw the Wavy Line for this inequality. The expression x(x+1)(x1) will be less than zero for those values of x where the Wavy Line goes below the number line. So, we can say that the inequality x(x+1)(x1) < 0 holds true for x < 1 or for 0 < x < 1 But, we know that x lies between 1 and 1. This means, that for some possible values of x (values that lie between 0 and 1), Expression 1 will hold true and for other possible values of x (values that lie between 1 and 0), Expression 1 will NOT hold true. So, it's not a MUST BE TRUE expression. II. x^2 < xCase 1: x is positive.This means, x = x So, Expression 2 becomes: x^2 < x That is, x^2  x < 0 Or x(x1) < 0 Again, by drawing the Wavy Line, we can see that the values of x that satisfy Case 1 are 0 < x < 1. Case 2: x is negative.This means, x = x So, Expression 2 becomes: x^2 < x That is, x^2 + x < 0 Or x(x+1) < 0 From the Wavy Line, it's clear that the values of x that satisfy Case 2 are 1 < x < 0 Now, we are given that the range of possible values of x are between 1 and 1, excluding 0. This means, that x will either lie between 1 and 0, exclusive, in which case, Case 2 above applies and Expression 2 holds true. OR x will lie between 0 and 1, exclusive, in which case, Case 1 above applies and Expression 2 holds true. This means, Expression 2 holds true for all possible values of x. So, it is a MUST BE TRUE expression. III. x^4 – x^5 > x^3 – x^2We'll first find the range of values for which this expression holds true. x^4(1x) > x^2(x1) Since x is not equal to zero, it's safe to divide both sides by x^2, which being a positive number, will not impact the sign of inequality. We get: x^2(1x) > x1 Or, x^2(1x)  (x1) > 0 That is, x^2(1x) + (1x) > 0 (1x)(x^2+1) > 0 Since x^2 + 1 will always be positive, the above inequality will hold true when 1  x > 0. That is, 1 > x Thus we see that Expression 3 will hold true for all values of x that are less than 1. Now, we are given that the only possible values of x are 1 < x < 1. Since all these possible values fall within the range in which Expression 3 holds true, we can conclude that Expression 3 will hold true for all possible values of x. So, it is a MUST BE TRUE expression. Hope this solution was useful! Best Regards Japinder
_________________
Register for free sessions Number Properties  Algebra Quant Workshop
Success Stories Guillermo's Success Story  Carrie's Success Story
Ace GMAT quant Articles and Question to reach Q51  Question of the week
Must Read Articles Number Properties – Even Odd  LCM GCD  Statistics1  Statistics2  Remainders1  Remainders2 Word Problems – Percentage 1  Percentage 2  Time and Work 1  Time and Work 2  Time, Speed and Distance 1  Time, Speed and Distance 2 Advanced Topics Permutation and Combination 1  Permutation and Combination 2  Permutation and Combination 3  Probability Geometry Triangles 1  Triangles 2  Triangles 3  Common Mistakes in Geometry Algebra Wavy line  Inequalities Practice Questions Number Properties 1  Number Properties 2  Algebra 1  Geometry  Prime Numbers  Absolute value equations  Sets
 '4 out of Top 5' Instructors on gmatclub  70 point improvement guarantee  www.egmat.com



Manager
Joined: 27 Aug 2014
Posts: 74

Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t
[#permalink]
Show Tags
30 May 2015, 22:32
EgmatQuantExpert wrote: Quote: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be true?
I. x^3 < x II. x^2 < x III. x^4 – x^5 > x^3 – x^2
(A) I only (B) II only (C) III only (D) II and III only (E) I, II, and III
Can someone try doing this algebraically? The algebraic approach to this question will be as follows: I. x^3 < xWe will first find the range of values of x for which this inequality holds true. So, let's solve this inequality: x^3  x < 0 => x(x^2  1) < 0 => x(x+1)(x1) < 0 Let's draw the Wavy Line for this inequality. The expression x(x+1)(x1) will be less than zero for those values of x where the Wavy Line goes below the number line. So, we can say that the inequality x(x+1)(x1) < 0 holds true for x < 1 or for 0 < x < 1 But, we know that x lies between 1 and 1. This means, that for some possible values of x (values that lie between 0 and 1), Expression 1 will hold true and for other possible values of x (values that lie between 1 and 0), Expression 1 will NOT hold true. So, it's not a MUST BE TRUE expression. II. x^2 < xCase 1: x is positive.This means, x = x So, Expression 2 becomes: x^2 < x That is, x^2  x < 0 Or x(x1) < 0 Again, by drawing the Wavy Line, we can see that the values of x that satisfy Case 1 are 0 < x < 1. Case 2: x is negative.This means, x = x So, Expression 2 becomes: x^2 < x That is, x^2 + x < 0 Or x(x+1) < 0 From the Wavy Line, it's clear that the values of x that satisfy Case 2 are 1 < x < 0 Now, we are given that the range of possible values of x are between 1 and 1, excluding 0. This means, that x will either lie between 1 and 0, exclusive, in which case, Case 2 above applies and Expression 2 holds true. OR x will lie between 0 and 1, exclusive, in which case, Case 1 above applies and Expression 2 holds true. This means, Expression 2 holds true for all possible values of x. So, it is a MUST BE TRUE expression. III. x^4 – x^5 > x^3 – x^2We'll first find the range of values for which this expression holds true. x^4(1x) > x^2(x1) Since x is not equal to zero, it's safe to divide both sides by x^2, which being a positive number, will not impact the sign of inequality. We get: x^2(1x) > x1 Or, x^2(1x)  (x1) > 0 That is, x^2(1x) + (1x) > 0 (1x)(x^2+1) > 0 Since x^2 + 1 will always be positive, the above inequality will hold true when 1  x > 0. That is, 1 > x Thus we see that Expression 3 will hold true for all values of x that are less than 1. Now, we are given that the only possible values of x are 1 < x < 1. Since all these possible values fall within the range in which Expression 3 holds true, we can conclude that Expression 3 will hold true for all possible values of x. So, it is a MUST BE TRUE expression. Hope this solution was useful! Best Regards Japinder Japinder two concerns: 1. why have we not flipped the inequality sign for stmt 2 when changing the sign to negative? 2. for stmt 3, we have x values less than 1 till negative infinity. how can this be sufficient as the range we want it to be is from 1 to +1



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13378
Location: United States (CA)

Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t
[#permalink]
Show Tags
01 Jun 2015, 21:14
Hi All, Most Roman Numeral questions on the Official GMAT are based on Number Properties and this question is no exception. You might find that TESTing VALUES helps you to better understand the patterns involved here... We're told that 1 < X < 1 and that X ≠ 0. We're asked which of the following MUST be true. RN 1: X^3 < X IF...X = 1/2....1/8 < 1/2 IF...X = 1/2....1/8 is NOT < 1/2 Roman Numeral 1 is NOT always TRUE. Eliminate Answers A and E RN 2: X^2 < X Since X is either a positive fraction or a negative fraction, X^2 will be 'closer to 0' than X (so X^2 will be LESS than X. For example.... IF...X = 1/2....1/4 < 1/2 IF...X = 1/2....1/4 < 1/2 Roman Numeral 2 IS ALWAYS TRUE Eliminate Answer C RN 3: X^4  X^5 > X^3  X^2 This one is complex"looking", but if you pay attention to the exponents it's not actually that tough. IF...X = positive.... X^4 > X^5 and X^2 > X^3 So.... X^4  X^5 = positive X^3  X^2 = negative a positive > a negative, so this option is ALWAYS TRUE IF....X = negative X^4 and X^2 = positive X^5 and X^3 = negative X^4  X^5 = (+)  () = positive X^3  X^2 = ()  (+) = negative Again, a positive > a negative, so this option is ALWAYS TRUE Roman Numeral 3 IS ALWAYS TRUE Eliminate Answer B. Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****



Manager
Joined: 27 Aug 2014
Posts: 74

Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t
[#permalink]
Show Tags
01 Jun 2015, 21:19
EMPOWERgmatRichC wrote: Hi All, Most Roman Numeral questions on the Official GMAT are based on Number Properties and this question is no exception. You might find that TESTing VALUES helps you to better understand the patterns involved here... We're told that 1 < X < 1 and that X ≠ 0. We're asked which of the following MUST be true. RN 1: X^3 < X IF...X = 1/2....1/8 < 1/2 IF...X = 1/2....1/8 is NOT < 1/2 Roman Numeral 1 is NOT always TRUE. Eliminate Answers A and E RN 2: X^2 < X Since X is either a positive fraction or a negative fraction, X^2 will be 'closer to 0' than X (so X^2 will be LESS than X. For example.... IF...X = 1/2....1/4 < 1/2 IF...X = 1/2....1/4 < 1/2 Roman Numeral 2 IS ALWAYS TRUE Eliminate Answer C RN 3: X^4  X^5 > X^3  X^2 This one is complex"looking", but if you pay attention to the exponents it's not actually that tough. IF...X = positive.... X^4 > X^5 and X^2 > X^3 So.... X^4  X^5 = positive X^3  X^2 = negative a positive > a negative, so this option is ALWAYS TRUE IF....X = negative X^4 and X^2 = positive X^5 and X^3 = negative X^4  X^5 = (+)  () = positive X^3  X^2 = ()  (+) = negative Again, a positive > a negative, so this option is ALWAYS TRUE Roman Numeral 3 IS ALWAYS TRUE Eliminate Answer B. Final Answer: GMAT assassins aren't born, they're made, Rich Hi Rich Can you do this algebraically on the number line?



eGMAT Representative
Joined: 04 Jan 2015
Posts: 2466

If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t
[#permalink]
Show Tags
01 Jun 2015, 21:43
sinhap07 wrote: EgmatQuantExpert wrote: The algebraic approach to this question will be as follows: I. x^3 < xWe will first find the range of values of x for which this inequality holds true. So, let's solve this inequality: x^3  x < 0 => x(x^2  1) < 0 => x(x+1)(x1) < 0 Let's draw the Wavy Line for this inequality. The expression x(x+1)(x1) will be less than zero for those values of x where the Wavy Line goes below the number line. So, we can say that the inequality x(x+1)(x1) < 0 holds true for x < 1 or for 0 < x < 1 But, we know that x lies between 1 and 1. This means, that for some possible values of x (values that lie between 0 and 1), Expression 1 will hold true and for other possible values of x (values that lie between 1 and 0), Expression 1 will NOT hold true. So, it's not a MUST BE TRUE expression. II. x^2 < xCase 1: x is positive.This means, x = x So, Expression 2 becomes: x^2 < x That is, x^2  x < 0 Or x(x1) < 0 Again, by drawing the Wavy Line, we can see that the values of x that satisfy Case 1 are 0 < x < 1. Case 2: x is negative.This means, x = x
So, Expression 2 becomes: x^2 < xThat is, x^2 + x < 0 Or x(x+1) < 0 From the Wavy Line, it's clear that the values of x that satisfy Case 2 are 1 < x < 0 Now, we are given that the range of possible values of x are between 1 and 1, excluding 0. This means, that x will either lie between 1 and 0, exclusive, in which case, Case 2 above applies and Expression 2 holds true. OR x will lie between 0 and 1, exclusive, in which case, Case 1 above applies and Expression 2 holds true. This means, Expression 2 holds true for all possible values of x. So, it is a MUST BE TRUE expression. III. x^4 – x^5 > x^3 – x^2We'll first find the range of values for which this expression holds true. x^4(1x) > x^2(x1) Since x is not equal to zero, it's safe to divide both sides by x^2, which being a positive number, will not impact the sign of inequality. We get: x^2(1x) > x1 Or, x^2(1x)  (x1) > 0 That is, x^2(1x) + (1x) > 0 (1x)(x^2+1) > 0 Since x^2 + 1 will always be positive, the above inequality will hold true when 1  x > 0. That is, 1 > x Thus we see that Expression 3 will hold true for all values of x that are less than 1. Now, we are given that the only possible values of x are 1 < x < 1. Since all these possible values fall within the range in which Expression 3 holds true, we can conclude that Expression 3 will hold true for all possible values of x. So, it is a MUST BE TRUE expression. Hope this solution was useful! Best Regards Japinder Japinder two concerns: 1. why have we not flipped the inequality sign for stmt 2 when changing the sign to negative? 2. for stmt 3, we have x values less than 1 till negative infinity. how can this be sufficient as the range we want it to be is from 1 to +1 Dear sinhap07PFB my pointwise response to your queries: 1. Not sure what you mean by 'changing the sign to negative'. If you're referring to the highlighted part of my solution above, then please note that this is the expression for the inequality in Case 2. in Case 2 (when x is negative), x = x, and so, the expression for the inequality x^2 < x becomes x^2 < x. We flip the inequality sign only when we multiply both sides of an inequality with the same negative number. For example, if I were to multiply both sides of the above inequality with 1, THEN yes, I would flip the inequality sign. The inequality would then become: x^2 > x 2. Let me explain with an easy example. Suppose in a community college, all students whose family income is less than $30000 per annum get a scholarship. If Alex is a student of this college and his family's annual income is between $20,000 $25,000, can you say with confidence that Alex gets this scholarship? Sure you can Let's now apply this analogy to the question at hand. By solving St. III, we saw that all values of x that are less than 1 (till negative infinity) satisfy this statement. The question statement tells us that x can only lie between 1 and +1. So, for all these possible values of x, can we be sure that St. III will be satisfied? By the parallel with the Alex example, you know that the answer is 'Yes.' If you are still unsure, think this way: all the possible values of x (values between 1 and +1) are less than 1. Therefore, they WILL satisfy St. III. So, St. III will be satisfied by all possible values of x. So, it is a MUST BE TRUE statement. Hope this helped. I would suggest getting more practice with basic conceptual questions and then sub700 level questions before doing questions like this one. Stepbystep progression in difficulty level and complexity of questions is the way to feeling confident in a topic. Best Regards Japinder
_________________
Register for free sessions Number Properties  Algebra Quant Workshop
Success Stories Guillermo's Success Story  Carrie's Success Story
Ace GMAT quant Articles and Question to reach Q51  Question of the week
Must Read Articles Number Properties – Even Odd  LCM GCD  Statistics1  Statistics2  Remainders1  Remainders2 Word Problems – Percentage 1  Percentage 2  Time and Work 1  Time and Work 2  Time, Speed and Distance 1  Time, Speed and Distance 2 Advanced Topics Permutation and Combination 1  Permutation and Combination 2  Permutation and Combination 3  Probability Geometry Triangles 1  Triangles 2  Triangles 3  Common Mistakes in Geometry Algebra Wavy line  Inequalities Practice Questions Number Properties 1  Number Properties 2  Algebra 1  Geometry  Prime Numbers  Absolute value equations  Sets
 '4 out of Top 5' Instructors on gmatclub  70 point improvement guarantee  www.egmat.com



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13378
Location: United States (CA)

Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t
[#permalink]
Show Tags
02 Jun 2015, 20:21
Hi sinhap07, You only seem interested in learning an algebraic approach to this question, so I have to ask WHY? Assuming that your goal is to score at a high level on the GMAT, you have to be ready to take advantage of the fact that most GMAT questions can be solved in a variety of ways. Certain Quant questions on the GMAT are actually DESIGNED to reward Test Takers who don't try to take a longwinded math approach. So you're not only looking to answer questions correctly, but you should also be looking to learn other (possibly more efficient/faster) methods to do so besides just "doing math." I used an approach to answering this question that a smart child could use, which means that you can use it too; there's something to be said for getting away from complex math (unless the question gives you no choice). GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****



Manager
Joined: 27 Aug 2014
Posts: 74

Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t
[#permalink]
Show Tags
02 Jun 2015, 23:16
Japinder two concerns: 1. why have we not flipped the inequality sign for stmt 2 when changing the sign to negative? 2. for stmt 3, we have x values less than 1 till negative infinity. how can this be sufficient as the range we want it to be is from 1 to +1[/quote] Dear sinhap07PFB my pointwise response to your queries: 1. Not sure what you mean by 'changing the sign to negative'. If you're referring to the highlighted part of my solution above, then please note that this is the expression for the inequality in Case 2. in Case 2 (when x is negative), x = x, and so, the expression for the inequality x^2 < x becomes x^2 < x. We flip the inequality sign only when we multiply both sides of an inequality with the same negative number. For example, if I were to multiply both sides of the above inequality with 1, THEN yes, I would flip the inequality sign. The inequality would then become: x^2 > x 2. Let me explain with an easy example. Suppose in a community college, all students whose family income is less than $30000 per annum get a scholarship. If Alex is a student of this college and his family's annual income is between $20,000 $25,000, can you say with confidence that Alex gets this scholarship? Sure you can Let's now apply this analogy to the question at hand. By solving St. III, we saw that all values of x that are less than 1 (till negative infinity) satisfy this statement. The question statement tells us that x can only lie between 1 and +1. So, for all these possible values of x, can we be sure that St. III will be satisfied? By the parallel with the Alex example, you know that the answer is 'Yes.' If you are still unsure, think this way: all the possible values of x (values between 1 and +1) are less than 1. Therefore, they WILL satisfy St. III. So, St. III will be satisfied by all possible values of x. So, it is a MUST BE TRUE statement. Hope this helped. I would suggest getting more practice with basic conceptual questions and then sub700 level questions before doing questions like this one. Stepbystep progression in difficulty level and complexity of questions is the way to feeling confident in a topic. Best Regards Japinder[/quote] Thanks Japinder. Stmt 3 is clear so is stmt 2. Just to confirm on the inequality sign in stmt 2, my version would be that usually when dealing with modulus, we flip signs. Eg. x<mod5 so x<5 or x=>5. But in this case, because we are dealing with a variable x and not an integer 5, we dont know what value the variable can take and hence, the inequality sign doesnt flip. Is my line of thought correct?



Intern
Joined: 28 May 2015
Posts: 8
Concentration: Finance, Economics
GPA: 4

Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t
[#permalink]
Show Tags
03 Jun 2015, 00:42
EMPOWERgmatRichC wrote: Hi All, Most Roman Numeral questions on the Official GMAT are based on Number Properties and this question is no exception. You might find that TESTing VALUES helps you to better understand the patterns involved here... We're told that 1 < X < 1 and that X ≠ 0. We're asked which of the following MUST be true. RN 1: X^3 < X IF...X = 1/2....1/8 < 1/2 IF...X = 1/2....1/8 is NOT < 1/2 Roman Numeral 1 is NOT always TRUE. Eliminate Answers A and E RN 2: X^2 < X Since X is either a positive fraction or a negative fraction, X^2 will be 'closer to 0' than X (so X^2 will be LESS than X. For example.... IF...X = 1/2....1/4 < 1/2 IF...X = 1/2....1/4 < 1/2 Roman Numeral 2 IS ALWAYS TRUE Eliminate Answer C RN 3: X^4  X^5 > X^3  X^2 This one is complex"looking", but if you pay attention to the exponents it's not actually that tough. IF...X = positive.... X^4 > X^5 and X^2 > X^3 So.... X^4  X^5 = positive X^3  X^2 = negative a positive > a negative, so this option is ALWAYS TRUE IF....X = negative X^4 and X^2 = positive X^5 and X^3 = negative X^4  X^5 = (+)  () = positive X^3  X^2 = ()  (+) = negative Again, a positive > a negative, so this option is ALWAYS TRUE Roman Numeral 3 IS ALWAYS TRUE Eliminate Answer B. Final Answer: GMAT assassins aren't born, they're made, Rich Thanks for this Rich! Is testing values always the way you would approach these inequalities problems? I really liked this approach and I was just wondering if this is an effective way to solve these issues!



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13378
Location: United States (CA)

Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t
[#permalink]
Show Tags
03 Jun 2015, 09:23
Hi DavidSt, As a tactical approach, TESTing VALUES will work on many Quant questions on Test Day, so it's important to build up those skills now (during your training/practice) so you can more readily use this approach when the clock is ticking. Using realworld numbers instead of abstract variables often makes dealing with the concepts easier and helps to more readily spot patterns; I use this approach often. GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****



Senior Manager
Joined: 23 Jun 2012
Posts: 380
Location: Pakistan
Concentration: Strategy, International Business
GPA: 3.76

Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t
[#permalink]
Show Tags
08 Feb 2016, 22:23
its an easy one! as question says that x isnt equal to 0 so let suppose x = 1/4 0r 1/4 now one must Know that squaring (1/4) will make it more close to 0. Option 1) If the value of X is positive then squaring this value will make it more close to 0, thus x^2 < x Option 2) On the other hand if the value of x is ve then squaring this value will make it close to 0 which will be bigger than the value of x; thus, x^ 2 > x Now choice 1 says X^3 < x which negates Option 2...so choice 1 will be wrong Choice 2 says X^2 < x which means 1/4 ^2 < 1/4...no matter what the value of X is, (ve or +ve), X will be greater than its square value. Now choice 3 says: x^4 – x^5 > x^3 – x^2 x^4 (1x) > x^2 (x 1) which is true as value of x is smaller than 1, so subtracting (x1) will give ve value so D is the answer
_________________
Push yourself again and again. Don't give an inch until the final buzzer sounds. Larry Bird Success isn't something that just happens  success is learned, success is practiced and then it is shared. Sparky Anderson S



Manager
Joined: 13 Sep 2016
Posts: 120

Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t
[#permalink]
Show Tags
11 Oct 2016, 00:40
DavidSt wrote: EMPOWERgmatRichC wrote: Hi All, Most Roman Numeral questions on the Official GMAT are based on Number Properties and this question is no exception. You might find that TESTing VALUES helps you to better understand the patterns involved here... We're told that 1 < X < 1 and that X ≠ 0. We're asked which of the following MUST be true. RN 1: X^3 < X IF...X = 1/2....1/8 < 1/2 IF...X = 1/2....1/8 is NOT < 1/2 Roman Numeral 1 is NOT always TRUE. Eliminate Answers A and E RN 2: X^2 < X Since X is either a positive fraction or a negative fraction, X^2 will be 'closer to 0' than X (so X^2 will be LESS than X. For example.... IF...X = 1/2....1/4 < 1/2 IF...X = 1/2....1/4 < 1/2 Roman Numeral 2 IS ALWAYS TRUE Eliminate Answer C RN 3: X^4  X^5 > X^3  X^2 This one is complex"looking", but if you pay attention to the exponents it's not actually that tough. IF...X = positive.... X^4 > X^5 and X^2 > X^3 So.... X^4  X^5 = positive X^3  X^2 = negative a positive > a negative, so this option is ALWAYS TRUE IF....X = negative X^4 and X^2 = positive X^5 and X^3 = negative X^4  X^5 = (+)  () = positive X^3  X^2 = ()  (+) = negative Again, a positive > a negative, so this option is ALWAYS TRUE Roman Numeral 3 IS ALWAYS TRUE Eliminate Answer B. Final Answer: GMAT assassins aren't born, they're made, Rich Thanks for this Rich! Is testing values always the way you would approach these inequalities problems? I really liked this approach and I was just wondering if this is an effective way to solve these issues! I found this technique as well it might be helpful to you. Since 1 < x < 1 and x ≠ 0, x must a NEGATIVE OR POSITIVE FRACTION. Statement I: x³ < x If x = 1/2, then x³ = 1/8. In this case, x³ > x. Since it does not have to be true that x³ < x, eliminate A and E. Statement II: x² < x Since x is nonzero, x² > 0 and x > 0. Since both sides of the inequality are positive, we can square the inequality: (x²)² < (x)² x⁴ < x². Since x² > 0, we can divide both sides by x²: x⁴/x² < x²/x² x² < 1. Since the square of a negative or positive fraction must be less than 1, statement II must be true. Eliminate C. Statement III: x⁴  x⁵ < x²  x³ Since x is nonzero, we can divide by x², which must be a positive value: (x⁴  x⁵)/x² < (x²  x³)/x² x²  x³ < 1x x²(1x) < 1x Since x is a negative or positive fraction, we can divide by 1x, which also must be a positive value: x²(1x)/(1x) < (1x)/(1x) x² < 1. Since the square of a negative or positive fraction must be less than 1, statement III must be true. Eliminate B. The correct answer is D.



NonHuman User
Joined: 09 Sep 2013
Posts: 9464

Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t
[#permalink]
Show Tags
06 Apr 2018, 05:02
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t &nbs
[#permalink]
06 Apr 2018, 05:02






