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If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t

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If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t [#permalink]

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New post Updated on: 28 May 2015, 07:11
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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?

Originally posted by sinhap07 on 28 May 2015, 07:06.
Last edited by Bunuel on 28 May 2015, 07:11, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t [#permalink]

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New post 28 May 2015, 07:11
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?


Similar question to practice: if-0-x-1-which-of-the-following-inequalities-must-be-71301.html
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Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t [#permalink]

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New post 28 May 2015, 11:05
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 ?
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Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t [#permalink]

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New post 28 May 2015, 11:28
1
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(1-x)>x^2(x-1).
Because lets concentrate on (1-x) and (x-1). We are given that -1<x<1 which means that (1-x)>0 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
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Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t [#permalink]

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New post 28 May 2015, 12:43
2
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 < 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.

Image

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.

Image

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

Image

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
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Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t [#permalink]

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New post 30 May 2015, 23:32
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 < 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.

Image

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.

Image

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

Image

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


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
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Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t [#permalink]

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New post 01 Jun 2015, 22:14
2
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:

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Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t [#permalink]

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New post 01 Jun 2015, 22:19
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?
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If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t [#permalink]

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New post 01 Jun 2015, 22:43
sinhap07 wrote:
EgmatQuantExpert wrote:

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.

Image

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.

Image

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

Image

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


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
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Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t [#permalink]

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New post 02 Jun 2015, 21:21
1
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,
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Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t [#permalink]

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New post 03 Jun 2015, 00:16
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 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[/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?
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Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t [#permalink]

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New post 03 Jun 2015, 01:42
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!
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Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t [#permalink]

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New post 03 Jun 2015, 10:23
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 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.

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Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t [#permalink]

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New post 08 Feb 2016, 23:23
1
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
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Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t [#permalink]

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New post 11 Oct 2016, 01:40
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 < 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.
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Re: If –1 < x < 1 and x ≠ 0, which of the following inequalities must be t [#permalink]

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