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

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

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Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink] New post 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?
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Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink] New post 08 Jul 2018, 10:17
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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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Last edited by Carcass on 14 Jul 2018, 00:56, edited 1 time in total.
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Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink] New post 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?
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Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink] New post 14 Jul 2018, 00:56
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Sorry I made a mistake. \(xz > 0\) as the stems says.

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Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink] New post 14 Sep 2018, 10:19
What is the significance of this - |x| < |y + 2| < |z| -in the question ..?
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Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink] New post 15 Sep 2018, 06:39
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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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