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

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Bunuel
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Is x < y < z ? (1) |x - 1| < y < z + 1 (2) |x + 1| < y < z - 1 [#permalink] New post   25 Feb 2020, 01:54
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60% (02:05) correct 40% (01:56) wrong based on 168 sessions
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Is \(x<y<z\) ?


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

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


Are You Up For the Challenge: 700 Level Questions
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B
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B
Kody
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Re: Is x < y < z ? (1) |x - 1| < y < z + 1 (2) |x + 1| < y < z - 1 [#permalink] New post   25 Feb 2020, 07:24
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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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sachinsavailable
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Is x < y < z ? (1) |x - 1| < y < z + 1 (2) |x + 1| < y < z - 1 [#permalink] New post   25 Feb 2020, 15:13
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Bunuel wrote:
Is \(x<y<z\) ?


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

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


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


hence the answer IS B.­
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Kinshook
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Is x < y < z ? (1) |x - 1| < y < z + 1 (2) |x + 1| < y < z - 1 [#permalink] New post   28 Mar 2020, 04:45
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Bunuel wrote:
Is \(x<y<z\) ?


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

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


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
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bumpbot
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Re: Is x < y < z ? (1) |x - 1| < y < z + 1 (2) |x + 1| < y < z - 1 [#permalink] New post   16 Nov 2022, 11:01
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Is x < y < z ? (1) |x - 1| < y < z + 1 (2) |x + 1| < y < z - 1 [#permalink]
16 Nov 2022, 11:01
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