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Is x > y? (1) x - y - 1 > 0 (2) x - y + 1 > 0

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Joined: 08 Sep 2016
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Is x > y? (1) x - y - 1 > 0 (2) x - y + 1 > 0  [#permalink]

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Updated on: 11 Jul 2017, 07:10
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Difficulty:

25% (medium)

Question Stats:

76% (01:25) correct 24% (01:11) wrong based on 65 sessions

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Is x > y?

(1) x - y - 1 > 0

(2) x - y + 1 > 0

Originally posted by shinrai15 on 11 Jul 2017, 06:31.
Last edited by Bunuel on 11 Jul 2017, 07:10, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Is x > y? (1) x - y - 1 > 0 (2) x - y + 1 > 0  [#permalink]

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11 Jul 2017, 06:50
1
shinrai15 wrote:
Is x > y?

(1) x - y - 1 > 0

(2) x - y + 1 > 0

(1) \(x - y - 1 > 0 \implies x-y > 1 > 0 \implies x >y\). Sufficient.

(2) \(x - y + 1 > 0 \implies x-y > - 1\)

If \(x-y=0 \implies x=y\)
If \(x-y=1 \implies x>y\).

Hence insufficient.

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Re: Is x > y? (1) x - y - 1 > 0 (2) x - y + 1 > 0  [#permalink]

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12 Jul 2017, 17:14
shinrai15 wrote:
Is x > y?

(1) x - y - 1 > 0

(2) x - y + 1 > 0

Statement One Alone:

x - y - 1 > 0

We can manipulate the inequality to read:

x > y + 1

Since x is always greater than y + 1, x will always be greater than y itself. Statement one alone is sufficient to answer the question.

Statement Two Alone:

x - y + 1 > 0

We can manipulate the inequality to read:

x > y - 1

The information in statement two does not answer the question. For instance, if x = 10 and y = 8, then x is greater than y. However, if x = 10 and y = 10, then x is not greater than y, even though the inequality holds.

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Re: Is x > y? (1) x - y - 1 > 0 (2) x - y + 1 > 0   [#permalink] 12 Jul 2017, 17:14
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