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# Is x < y < z ? (1) |x - 1| < y < z + 1 (2) |x + 1| < y < z - 1

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Math Expert
Joined: 02 Sep 2009
Posts: 62554
Is x < y < z ? (1) |x - 1| < y < z + 1 (2) |x + 1| < y < z - 1  [#permalink]

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25 Feb 2020, 00:54
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Difficulty:

55% (hard)

Question Stats:

60% (02:19) correct 40% (01:59) wrong based on 65 sessions

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Is $$x<y<z$$ ?

(1) $$|x-1|<y<z+1$$

(2) $$|x+1|<y<z-1$$

Project DS Butler Data Sufficiency (DS3)

Are You Up For the Challenge: 700 Level Questions

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Intern
Joined: 22 Feb 2020
Posts: 5
Re: Is x < y < z ? (1) |x - 1| < y < z + 1 (2) |x + 1| < y < z - 1  [#permalink]

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25 Feb 2020, 06:24
1
1. do not need to worry about the negative case of x because the absolute value of the x case ensures that y will be greater than or equal to zero.
If x is 9, the y could be 8.5 and satisfy the equation. Additionally x could be 9, y could be 10 and z could be 11. This is insufficient

2. because we have demonstrated that the absolute value sign in this question leads us to be able to ignore the negative case, statement two is sufficient. y must be one greater than x and more than 1 less than z.

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Joined: 13 May 2018
Posts: 41
Re: Is x < y < z ? (1) |x - 1| < y < z + 1 (2) |x + 1| < y < z - 1  [#permalink]

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25 Feb 2020, 14:13
1
Bunuel wrote:
Is $$x<y<z$$ ?

(1) $$|x-1|<y<z+1$$

(2) $$|x+1|<y<z-1$$

Project DS Butler Data Sufficiency (DS3)

Are You Up For the Challenge: 700 Level Questions

1) Statement 1 implies two different things:

-y<x-1<y

which means

that y is greater than x-1 but we dont know whether y is greater than x.

hence insufficient, remove A.

2) Staement 2 implies that

-y+1<x<y-1

which means x+1<y which automatically means x<y.

Further the statement also mentions y<z-1

which means y<z.

Hence proved that x<y<z

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SACHIN SHARMA

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Joined: 03 Jun 2019
Posts: 2501
Location: India
GMAT 1: 690 Q50 V34
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Re: Is x < y < z ? (1) |x - 1| < y < z + 1 (2) |x + 1| < y < z - 1  [#permalink]

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28 Mar 2020, 03:45
1
Bunuel wrote:
Is $$x<y<z$$ ?

(1) $$|x-1|<y<z+1$$

(2) $$|x+1|<y<z-1$$

Project DS Butler Data Sufficiency (DS3)

Are You Up For the Challenge: 700 Level Questions

Asked: Is $$x<y<z$$ ?

(1) $$|x-1|<y<z+1$$
|x-1| may or may not be >x
y may or may not be < z
NOT SUFFICIENT

(2) $$|x+1|<y<z-1$$
|x+1|>x; z>z-1
x<|x+1|<y<z-1<z
x<y<z
SUFFICIENT

IMO B
Re: Is x < y < z ? (1) |x - 1| < y < z + 1 (2) |x + 1| < y < z - 1   [#permalink] 28 Mar 2020, 03:45
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