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

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

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New post 25 Feb 2020, 00:54
11
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

59% (02:13) correct 41% (01:54) wrong based on 82 sessions

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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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New post 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.

Answer is B.
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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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New post 25 Feb 2020, 14:13
1
Bunuel wrote:
Is \(x<y<z\) ?


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

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


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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


hence the answer IS B.
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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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New post 28 Mar 2020, 03:45
2
Bunuel wrote:
Is \(x<y<z\) ?


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

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


Project DS Butler Data Sufficiency (DS3)


For DS butler Questions Click Here


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

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

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