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Is x > y? (1) |x| > y. (2) x + y > 0.

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Is x > y? (1) |x| > y. (2) x + y > 0.  [#permalink]

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New post 24 Jul 2017, 23:27
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

55% (01:50) correct 45% (01:35) wrong based on 136 sessions

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

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New post 25 Jul 2017, 01:12
St 1. x >y or x < - y NS

St 2 x > - y NS

Ans E


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

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New post 25 Jul 2017, 01:47
1
Bunuel wrote:
Is x > y?

(1) |x| > y .
(2) x + y > 0.


(1) if x>0 then x>y
if x<0 then x>-y
Insufficient

(2) x>-y insufficient

on combining we get x>-y..insufficient

Hence E
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Is x > y? (1) |x| > y. (2) x + y > 0.  [#permalink]

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New post 25 Jul 2017, 09:02
Bunuel wrote:
Is x > y?

(1) |x| > y .
(2) x + y > 0.



1) x>y (when x>0) and x<-y when (x<0).....Insufficient
2) x>-y, Clearly insufficient

Combining we get x > y answer C
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Re: Is x > y? (1) |x| > y. (2) x + y > 0.  [#permalink]

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New post 25 Jul 2017, 14:16
mynamegoeson wrote:
Bunuel wrote:
Is x > y?

(1) |x| > y .
(2) x + y > 0.


(1) if x>0 then x>y
if x<0 then x>-y
Insufficient

(2) x>-y insufficient

on combining we get x>-y..insufficient

Hence E


Why do we get x>-y not x>y?
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Is x > y? (1) |x| > y. (2) x + y > 0.  [#permalink]

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New post 26 Jul 2017, 15:12
2
1
Is x > y?

(1) |x| > y

Let x = 10 & y = 1...........Answer is Yes

Let x = -10 & y =1...........Answer is No

Insufficient


(2) x + y > 0

Case 1: Let x = 10 & y = 1...........Answer is Yes

Case 2: Let x = 10 & y = -1...........Answer is Yes

Case 3: Let x = 1 & y = 10...........Answer is No

Case 4: Let x = -1 & y = 10...........Answer is No

Insufficient

Combining 1 & 2

|x| - y > 0

x + y > 0
----------------- Sum both inequalities as both are in same direction

x + |x| >0 ........The only way to prove this inequality is that x is POSITIVE. If x is negative then result is zero which is invalid.


x > 0 & |x| > y ...........then x > y

Answer: C

Another Approach:

Combine 1 & 2 together.

We need to match cases from Statement 2 that fit into Statement 1

Case 3 & 4 are NOT valid.

Case 1 & 2 are valid

based on above x > y

Answer: C
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Re: Is x > y? (1) |x| > y. (2) x + y > 0.  [#permalink]

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New post 26 Jul 2017, 15:50
Bunuel wrote:
Is x > y?

(1) |x| > y .
(2) x + y > 0.



St 1: Insuff
6,3 - Yes
-6,3 - No

St 2: Insuff
-2,4 - No
4,-2 - Yes

Combining Yes & Yes - Suff
Answer C
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Re: Is x > y? (1) |x| > y. (2) x + y > 0.  [#permalink]

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New post 28 Jul 2017, 10:59
For me its E
is x> y?

1) |x| >y
so two cases: a) x > y b) x< -y Not suff
2) x + y> 0
x> - y
Not suff

1) + 2): we have 3 equations --> x> y; x < -y; x> -y So both not suff. So E
OA and OE awaited!
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Re: Is x > y? (1) |x| >  [#permalink]

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New post 31 Jul 2017, 06:15
Madhavi1990 wrote:
For me its E
is x> y?

1) |x| >y
so two cases: a) x > y b) x< -y Not suff
2) x + y> 0
x> - y
Not suff

1) + 2): we have 3 equations --> x> y; x < -y; x> -y So both not suff. So E
OA and OE awaited!


x> y; x < -y; x> -y. x>y is as good as x>-y

So x>y, hence C
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Is x > y? (1) |x| > y. (2) x + y > 0.  [#permalink]

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New post 13 Dec 2017, 14:16
\(x > y\) ?

Statement 1: \(|x| > y\) ?
Insufficient
case 1: x = -16, y = 8
case 2: x = 16, y = 8

Statement 2: \(x + y > 0\)
InSufficient
case 1: x = 3, y = 2
case 2: x = 2, y = 3

(1) + (2)

adding two inequalities, \(x + |x| + y > y\) => \(x + |x| > 0\) => x is positive
since x is positive => statement 1 : \(|x| > y\) => \(x > y\) => answers our question => (C)
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Is x > y? (1) |x| > y. (2) x + y > 0.   [#permalink] 13 Dec 2017, 14:16
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