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Suppose that |x| < |y + 2| < |z|. Suppose further that y >
[#permalink]
07 Aug 2017, 12:16

4

Expert Reply

1

Bookmarks

Question Stats:

Suppose that \(|x| < |y + 2| < |z|\). Suppose further that \(y > 0\) and that \(xz > 0\). Which of the following could be true ?

Indicate all statements that apply

❑ \(0 < y < x < z\)

❑ \(0 < x < y < z\)

❑ \(x < z < 0 < y\)

❑ \(0 < y + 1.5 < x < z\)

❑ \(z < x < 0 < y\)

_________________

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Questions' Banks and Collection:

ETS: ETS Free PowerPrep 1 & 2 All 320 Questions Explanation. | ETS All Official Guides

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Scores: The GRE average score at Top 25 Business Schools 2020 Ed. | How to study for GRE retake and score HIGHER - (2020)

How is the GRE Score Calculated -The Definitive Guide (2021)

Tests: GRE Prep Club Tests | FREE GRE Practice Tests [Collection] - New Edition (2021)

Vocab: GRE Prep Club Official Vocabulary Lists for the GRE (2021)

Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y >
[#permalink]
08 Jul 2018, 08:02

Carcass wrote:

Suppose that |x| < |y + 2| < |z|. Suppose further that y > 0 and that xz > 0. Which of the following could be true ?

Indicate all statements that apply

❑ 0 < y < x < z

❑ 0 < x < y < z

❑ x < z < 0 < y

❑ 0 < y + 1.5 < x < z

❑ z < x < 0 < y

Show: :: OA

A,B,D,E

Hello Carcass, i am a little not sure about the solutions, could you please just show us your strategy?

Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y >
[#permalink]
Updated on: 14 Jul 2018, 00:56

1

Expert Reply

Questions involving pure number properties and absolute value are really tough.

Start always from what you do know and the stem says to you.

Now, \(y > 0\) and \(xz > 0\), which means that y is positive and x and y can be both positive or negative.

At the same time\(y > x\) or\(y < x\). At this point, pick numbers that satisfy all the conditions you consider.

A) \(x=1,y=5,z=9\)

B)\(y=1,x=2,z=5\)

testing them in all answer choices and you can quickly see the only C is always false.

Hope this helps

_________________

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Questions' Banks and Collection:

ETS: ETS Free PowerPrep 1 & 2 All 320 Questions Explanation. | ETS All Official Guides

3rd Party Resource's: All Quant Questions Collection | All Verbal Questions Collection

Books: All GRE Best Books

Scores: The GRE average score at Top 25 Business Schools 2020 Ed. | How to study for GRE retake and score HIGHER - (2020)

How is the GRE Score Calculated -The Definitive Guide (2021)

Tests: GRE Prep Club Tests | FREE GRE Practice Tests [Collection] - New Edition (2021)

Vocab: GRE Prep Club Official Vocabulary Lists for the GRE (2021)

Start always from what you do know and the stem says to you.

Now, \(y > 0\) and \(xz > 0\), which means that y is positive and x and y can be both positive or negative.

At the same time\(y > x\) or\(y < x\). At this point, pick numbers that satisfy all the conditions you consider.

A) \(x=1,y=5,z=9\)

B)\(y=1,x=2,z=5\)

testing them in all answer choices and you can quickly see the only C is always false.

Hope this helps

_________________

New to the GRE, and GRE CLUB Forum?

GRE: All you do need to know about the GRE Test | GRE Prep Club for the GRE Exam - The Complete FAQ

Search GRE Specific Questions | Download Vault

Posting Rules: QUANTITATIVE | VERBAL

FREE Resources: GRE Prep Club Official LinkTree Page | Free GRE Materials - Where to get it!! (2020)

Free GRE Prep Club Tests: Got 20 Kudos? You can get Free GRE Prep Club Tests

GRE Prep Club on : Facebook | Instagram

Questions' Banks and Collection:

ETS: ETS Free PowerPrep 1 & 2 All 320 Questions Explanation. | ETS All Official Guides

3rd Party Resource's: All Quant Questions Collection | All Verbal Questions Collection

Books: All GRE Best Books

Scores: The GRE average score at Top 25 Business Schools 2020 Ed. | How to study for GRE retake and score HIGHER - (2020)

How is the GRE Score Calculated -The Definitive Guide (2021)

Tests: GRE Prep Club Tests | FREE GRE Practice Tests [Collection] - New Edition (2021)

Vocab: GRE Prep Club Official Vocabulary Lists for the GRE (2021)

Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y >
[#permalink]
14 Jul 2018, 00:11

Carcass wrote:

Questions involving pure number properties and absolute value are really tough.

Start always from what you do know and the stem says to you.

Now, \(y > 0\) and \(xy > 0\), which means that y is positive and x and y can be both positive or negative.

At the same time\(y > x\) or\(y < x\). At this point, pick numbers that satisfy all the conditions you consider.

A) \(x=1,y=5,z=9\)

B)\(y=1,x=2,z=5\)

testing them in all answer choices and you can quickly see the only C is always false.

Hope this helps

Start always from what you do know and the stem says to you.

Now, \(y > 0\) and \(xy > 0\), which means that y is positive and x and y can be both positive or negative.

At the same time\(y > x\) or\(y < x\). At this point, pick numbers that satisfy all the conditions you consider.

A) \(x=1,y=5,z=9\)

B)\(y=1,x=2,z=5\)

testing them in all answer choices and you can quickly see the only C is always false.

Hope this helps

How y > 0 and xy > 0, means that both could be positive or negative? For y>0, it means that x must be positive in order for xy to be greater that 0?

Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y >
[#permalink]
14 Jul 2018, 00:56

Expert Reply

Sorry I made a mistake. \(xz > 0\) as the stems says.

Regards

_________________

GRE Prep Club OFFICIAL Android App

New to the GRE, and GRE CLUB Forum?

GRE: All you do need to know about the GRE Test | GRE Prep Club for the GRE Exam - The Complete FAQ

Search GRE Specific Questions | Download Vault

Posting Rules: QUANTITATIVE | VERBAL

FREE Resources: GRE Prep Club Official LinkTree Page | Free GRE Materials - Where to get it!! (2020)

Free GRE Prep Club Tests: Got 20 Kudos? You can get Free GRE Prep Club Tests

GRE Prep Club on : Facebook | Instagram

Questions' Banks and Collection:

ETS: ETS Free PowerPrep 1 & 2 All 320 Questions Explanation. | ETS All Official Guides

3rd Party Resource's: All Quant Questions Collection | All Verbal Questions Collection

Books: All GRE Best Books

Scores: The GRE average score at Top 25 Business Schools 2020 Ed. | How to study for GRE retake and score HIGHER - (2020)

How is the GRE Score Calculated -The Definitive Guide (2021)

Tests: GRE Prep Club Tests | FREE GRE Practice Tests [Collection] - New Edition (2021)

Vocab: GRE Prep Club Official Vocabulary Lists for the GRE (2021)

Regards

_________________

New to the GRE, and GRE CLUB Forum?

GRE: All you do need to know about the GRE Test | GRE Prep Club for the GRE Exam - The Complete FAQ

Search GRE Specific Questions | Download Vault

Posting Rules: QUANTITATIVE | VERBAL

FREE Resources: GRE Prep Club Official LinkTree Page | Free GRE Materials - Where to get it!! (2020)

Free GRE Prep Club Tests: Got 20 Kudos? You can get Free GRE Prep Club Tests

GRE Prep Club on : Facebook | Instagram

Questions' Banks and Collection:

ETS: ETS Free PowerPrep 1 & 2 All 320 Questions Explanation. | ETS All Official Guides

3rd Party Resource's: All Quant Questions Collection | All Verbal Questions Collection

Books: All GRE Best Books

Scores: The GRE average score at Top 25 Business Schools 2020 Ed. | How to study for GRE retake and score HIGHER - (2020)

How is the GRE Score Calculated -The Definitive Guide (2021)

Tests: GRE Prep Club Tests | FREE GRE Practice Tests [Collection] - New Edition (2021)

Vocab: GRE Prep Club Official Vocabulary Lists for the GRE (2021)

Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y >
[#permalink]
14 Sep 2018, 10:19

What is the significance of this - |x| < |y + 2| < |z| -in the question ..?

Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y >
[#permalink]
15 Sep 2018, 06:39

4

Carcass wrote:

Suppose that \(|x| < |y + 2| < |z|\). Suppose further that \(y > 0\) and that \(xz > 0\). Which of the following could be true ?

Indicate all statements that apply

❑ \(0 < y < x < z\)

❑ \(0 < x < y < z\)

❑ \(x < z < 0 < y\)

❑ \(0 < y + 1.5 < x < z\)

❑ \(z < x < 0 < y\)

Show: :: OA

A,B,D,E

Given y > 0 and xz > 0)

Therefore : y must be positive,

Either x and z are both positive or they are both negative

If x and z are positive,

then y > x or y < x

Let us take two possibilities such as:

1. x = 1, y = 3, z = 6,

2. y = 2, x = 3, z = 6.

Then we can see that Option A and Option B are true

For option C,

It cannot be true because If x and z are both negative, then x must be greater than z, since |z| > |x|.

Option D

It can be true if we consider fraction let say,

y = 1, x = 2.5, and z = 4. (As nothing is mentioned for the quantities to be integer, it can be fraction as well)

Option E

Similarly if x and z are both negative,

x = -1, y = 2, and z = -4 such that |z| > |x|, y > 0

so z < x

_________________

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Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y >
[#permalink]
18 Jul 2020, 07:25

1

well need to increase speed on this!

if I take y =2 ,x = 1, z = 3 --> the condition is not even upheld

This means y has to be less than 1 but positive

so y =1, x = 2, z = 3 --> so D holds true and so does E. A and B are possible but not a 'must'. Hence a,b,d,e.

Kudos if you like this explanation

if I take y =2 ,x = 1, z = 3 --> the condition is not even upheld

This means y has to be less than 1 but positive

so y =1, x = 2, z = 3 --> so D holds true and so does E. A and B are possible but not a 'must'. Hence a,b,d,e.

Kudos if you like this explanation

gmatclubot

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