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Suppose that x < y + 2 < z. Suppose further that y > [#permalink]
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 ❑ \(0 < y < x < z\) ❑ \(0 < x < y < z\) ❑ \(x < z < 0 < y\) ❑ \(0 < y + 1.5 < x < z\) ❑ \(z < x < 0 < y\)
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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 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]
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.
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?



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Re: Suppose that x < y + 2 < z. Suppose further that y > [#permalink]
14 Jul 2018, 00:56
Sorry I made a mistake. \(xz > 0\) as the stems says. Regards
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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 ..?



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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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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 ❑ \(0 < y < x < z\) ❑ \(0 < x < y < z\) ❑ \(x < z < 0 < y\) ❑ \(0 < y + 1.5 < x < z\) ❑ \(z < x < 0 < y\) 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 >
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