If –1

Question

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?

Bowtye · Accepted Answer

Answer is D. Let's look at the statements one by one Stmt I. x^3 < x if 0x 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(1-x)>x^2(x-1). Because lets concentrate on (1-x) and (x-1). We are given that -10 and (x-1)<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

Bunuel · Answer

Similar question to practice: if-0-x-1-which-of-the-following-inequalities-must-be-71301.html

viksingh15 · Answer

Are you sure answer is D, for what value a is invalid ?

EgmatQuantExpert · Answer

The algebraic approach to this question will be as follows:

I. x^3 < x

We 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)(x-1) < 0

Let's draw the Wavy Line for this inequality.

The expression x(x+1)(x-1) 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)(x-1) < 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 < |x|

Case 1: x is positive.

This means, |x| = x

So, Expression 2 becomes: x^2 < x
That is, x^2 - x < 0
Or x(x-1) < 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^2

We'll first find the range of values for which this expression holds true.

x^4(1-x) > x^2(x-1)

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(1-x) > x-1

Or, x^2(1-x) - (x-1) > 0
That is, x^2(1-x) + (1-x) > 0

(1-x)(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

Best Regards Japinder

sinhap07 · Answer

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

EMPOWERgmatRichC · Answer

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:

Show Spoiler

D

GMAT assassins aren't born, they're made,
Rich

sinhap07 · Answer

Hi Rich

Can you do this algebraically on the number line?

EgmatQuantExpert · Answer

Dear sinhap07 PFB my point-wise 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 sub-700 level questions before doing questions like this one. Step-by-step progression in difficulty level and complexity of questions is the way to feeling confident in a topic.

Best Regards Japinder

EMPOWERgmatRichC · Answer

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 long-winded 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

sinhap07 · Answer

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 sinhap07 PFB my point-wise 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 sub-700 level questions before doing questions like this one. Step-by-step progression in difficulty level and complexity of questions is the way to feeling confident in a topic.

Best Regards Japinder 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-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?

DavidSt · Answer

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!

EMPOWERgmatRichC · Answer

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 real-world 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

sananoor · Answer

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 (1-x) > x^2 (x- 1)
which is true as value of x is smaller than 1, so subtracting (x-1) will give -ve value

so D is the answer

Nanobotstv · Answer

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 < 1-x x²(1-x) < 1-x Since x is a negative or positive fraction, we can divide by 1-x, which also must be a positive value: x²(1-x)/(1-x) < (1-x)/(1-x) 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.

KSBGC · Answer

x can't be 0. x is either positive fraction ( 1/2) or negative fraction ( - 1/2) . If we square a fraction , it will be even smaller. I. x^3 < x . what if x is a negative fraction. put -1/2. II. x^2 < |x| . both sides will yield a positive result thus it will always be true. III. x^4 – x^5 > x^3 – x^2.........modify it. x^4 + x^2 > x^3 + x^5. .....one side is positive , another side is negative if we plug any negative fraction. Thus , it is true for all time. D is our choice.....

SolankiDas · Answer

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

By solving Statement III i am getting ((x-1)^2)(x^2)(x+1)<0 which implies x<-1 and doesnot statisfy the inequality. 
Can any explain

EMPOWERgmatRichC · Answer

Hi SolankiDas,

To start, we're told that -1 < X < 1 and that X ≠ 0. This means that X can be either a 'positive fraction' or a 'negative fraction.' This is an interesting piece of information - and it's worth noting that nothing about a GMAT question is ever 'random'; whoever wrote this prompt built the prompt around that concept (so we have to be careful to account for both of those options as we do our work). Based on how you simplified that inequality, I do not think that you were considering that X could be positive OR negative (and how that might impact how you manipulate the inequality.

We're asked which of the following MUST be true. While there's no harm in simplifying the inequality in the third Roman Numeral, that work is ultimately unnecessary. The prompt did NOT ask us to 'solve' that inequality; it asked if, given the restrictions about X, the inequality was TRUE? That's an important distinction. We're not asked to find every possible outcome here; we're asked whether positive fractions and negative fractions will create a 'true' statement based on that inequality.

III. X^4 - X^5 > X^3 - X^2

This is complex-"looking", but if you pay attention to the exponents it's not actually that tough.

IF...X = positive fraction....
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 fraction...
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

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: [email protected]

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