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# Suppose that |x| < |y + 2| < |z|. Suppose further that y >

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

36% (01:45) correct 63% (01:31) wrong based on 46 sessions

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

❑ $$0 < y < x < z$$

❑ $$0 < x < y < z$$

❑ $$x < z < 0 < y$$

❑ $$0 < y + 1.5 < x < z$$

❑ $$z < x < 0 < y$$
[Reveal] Spoiler: OA

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Intern
Joined: 14 Jun 2018
Posts: 36
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Kudos [?]: 7 [0], given: 100

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

[Reveal] Spoiler: OA
A,B,D,E

Hello Carcass, i am a little not sure about the solutions, could you please just show us your strategy?
Founder
Joined: 18 Apr 2015
Posts: 6901
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Kudos [?]: 1343 [1] , given: 6316

Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink]  08 Jul 2018, 10:17
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KUDOS
Expert's post
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
_________________

Last edited by Carcass on 14 Jul 2018, 00:56, edited 1 time in total.
typo
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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

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?
Founder
Joined: 18 Apr 2015
Posts: 6901
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Kudos [?]: 1343 [0], given: 6316

Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink]  14 Jul 2018, 00:56
Expert's post
Sorry I made a mistake. $$xz > 0$$ as the stems says.

Regards
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Intern
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Kudos [?]: 3 [0], given: 0

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 ..?
Director
Joined: 20 Apr 2016
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WE: Engineering (Energy and Utilities)
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Kudos [?]: 670 [3] , given: 148

Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink]  15 Sep 2018, 06:39
3
KUDOS
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

❑ $$0 < y < x < z$$

❑ $$0 < x < y < z$$

❑ $$x < z < 0 < y$$

❑ $$0 < y + 1.5 < x < z$$

❑ $$z < x < 0 < y$$

[Reveal] Spoiler: 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] 15 Sep 2018, 06:39
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