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Is x less than y? (1) x-y+1<0 (2) x-y-1<0

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Is x less than y? (1) x-y+1<0 (2) x-y-1<0 [#permalink]

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New post 26 Feb 2011, 16:40
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Is x less than y?

(1) x-y+1<0
(2) x-y-1<0
[Reveal] Spoiler: OA

Last edited by Bunuel on 07 Aug 2012, 00:30, edited 1 time in total.
Edited the question and added the OA.
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Re: Is x less than y? (1) x-y+1<0 (2) x-y-1<0 [#permalink]

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New post 26 Feb 2011, 16:47
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Re: Is x less than y? (1) x-y+1<0 (2) x-y-1<0 [#permalink]

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New post 28 Feb 2011, 18:32
yes A it is

1.) x-y<-1 This is satisfied when x<y

2.) x-y<+1 Not sufficient

example x=1.5 and y=1; 1.5-1 < 1
x = - 2 and y= -1; -2 - (-1) < 1

A it is.
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Re: Can someone help on a tricky DS question on Inequalities [#permalink]

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New post 06 Aug 2012, 19:08
gagan_g2k wrote:
Is x less than y?
(1) x-y+1<0
(2) x-y-1<0

The official answer is a but as per my solution it should be d.


That's a very easy trap to fall for, I to fell for it.
Added the two inequalities and got x-y <0 and condluded both are necessary
But if you look at x-y +1 < 0 this means that x+y < -1
which means that X-Y < 0 as -1 < 0 , so we can get the answer from one only.

Now looking at 2) X-Y -1< 0 this means that x-y < 1 which doesn,t clearly mean that x<y as something less than 1 can be positive and negative both.

So A is the answer.
Damn i need more practice !
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Re: Can someone help on a tricky DS question on Inequalities [#permalink]

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New post 06 Aug 2012, 22:26
gagan_g2k wrote:
Is x less than y?
(1) x-y+1<0
(2) x-y-1<0

The official answer is a but as per my solution it should be d.


(1) \(x-y+1<0\) implies \(x-y<-1<0\), which means \(x-y<0\) or \(x<y\).
Hence sufficient.

(2) Take \(x = y = 0\) for which obviously x is not less than y, but (2) holds. For x = 0 and y = 1, (2) still holds, but now x < y.
Not sufficient.

Answer A.
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Re: Is x less than y? (1) x-y+1<0 (2) x-y-1<0 [#permalink]

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Is x>y? [#permalink]

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New post 15 Jul 2016, 01:23
Is x>y?

Statement 1: x-y-1>0

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

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New post 15 Jul 2016, 01:51
Silviax wrote:
Is x>y?

Statement 1: x-y-1>0

Statement 2: x-y+1>0



Given x>y

Then we need to find x-y > 0 ?

Stat 1: x-y-1>0
=> x-y > 1...Sufficient.

Stat 2: x-y + 1 > 0.
=> x-y > -1... Insufficient ( Since we need to seek x-y > 0 )

Thus A is correct answer.
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Re: Is x>y? [#permalink]

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New post 15 Jul 2016, 02:01
The way you explained makes a lot more sense than the official explanation of Magoosh.

I can now see that statement 1 is sufficient by itself.
Thanks!
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Re: Is x>y? [#permalink]

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New post 15 Jul 2016, 02:10
Perfect solution given....

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

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New post 15 Jul 2016, 02:18
Re: Is x less than y? (1) x-y+1<0 (2) x-y-1<0   [#permalink] 15 Jul 2016, 02:18
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