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Is x + y > 0 ?

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Is x + y > 0 ? [#permalink]

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New post 15 Jul 2014, 13:58
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Is x + y > 0 ?

(1) x - y > 0
(2) x^2 - y^2 > 0
[Reveal] Spoiler: OA

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Last edited by Bunuel on 15 Jul 2014, 14:14, edited 2 times in total.
Edited the question.

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Is x + y > 0 ? [#permalink]

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New post 15 Jul 2014, 14:18
Is x + y > 0 ?

(1) x - y > 0 --> x > y. One number is greater than another. From this we cannot say whether their sum is positive. Not sufficient.

(2) x^2 - y^2 > 0 --> x^2 > y^2 --> |x| > |y|. One number is further from 0 than another. From this we cannot say whether their sum is positive. Not sufficient.

(1)+(1) From (2) (x - y)(x + y) > 0 (x + y and x - y have the same sign) and from (1) x - y > 0, thus x + y > 0. Sufficient.

Answer: C.
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Re: Is x + y > 0 ? [#permalink]

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achintsodhi wrote:
Is x + y > 0?

(1) x – y > 0

(2) \(x^2 – y^ 2 > 0\)


Statement 1: x – y > 0
This means x > y
Case 1: x = 2, y = 1
Here x + y > 0
Case 2: x = -10, y = -11
Here x + y < 0
Insufficient

Statement 2: \(x^2 – y^ 2 > 0\)
Or \(x^2 > y^ 2\)
From this too, we cannot say anything about x + y
Case 1: x = 4, y = 1
x + y > 0
Case 2: x = -4, y = 1
x + y < 0
Insufficient

Statement 1 and 2 Combined:
From statement 1, x - y > 0
From statement 2, \(x^2 – y^ 2 > 0\) or (x+y)(x-y) >0
Since we already know that (x-y) > 0 from statement 1,
Therefore x + y > 0
Sufficient

Correct Option: C
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Re: Is x + y > 0 ? [#permalink]

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New post 07 Nov 2016, 17:03
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felippemed wrote:
Is \(x + y >0\)?

(I) \(x - y > 0\)
(II) \(x^2 - y^2 > 0\)



Statement 1:

x>y

If both x and y are positive then YES

If both x and y are negative then NO

Insufficient

Statement 2:

(x+y)(x-y)>0

If x-y>0, then YES

If x-y<0, then NO

Insufficient

Statement 1&2:

x-y>0,

then x+y>0

Sufficient

C


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Re: Is x + y > 0 ? [#permalink]

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New post 07 Nov 2016, 18:49
Posted from my mobile device

Maybe a better approach is theoretically.

Posted from my mobile device

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Re: Is x + y > 0 ? [#permalink]

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New post 07 Nov 2016, 18:52
acegmat123 wrote:
felippemed wrote:
Is \(x + y >0\)?

(I) \(x - y > 0\)
(II) \(x^2 - y^2 > 0\)



Statement 1:

x>y

If both x and y are positive then YES

If both x and y are negative then NO

Insufficient

Statement 2:

(x+y)(x-y)>0

If x-y>0, then YES

If x-y<0, then NO

Insufficient

Statement 1&2:

x-y>0,

then x+y>0

Sufficient

C


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I am not sure your approach is correct.

You could have a gigantic positive x e a tiny negative y and still hold the conditions true.

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Re: Is x + y > 0 ? [#permalink]

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New post 07 Nov 2016, 19:08
felippemed wrote:
acegmat123 wrote:
felippemed wrote:
Is \(x + y >0\)?

(I) \(x - y > 0\)
(II) \(x^2 - y^2 > 0\)



Statement 1:

x>y

If both x and y are positive then YES

If both x and y are negative then NO

Insufficient

Statement 2:

(x+y)(x-y)>0

If x-y>0, then YES

If x-y<0, then NO

Insufficient

Statement 1&2:

x-y>0,

then x+y>0

Sufficient

C


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I am not sure your approach is correct.

You could have a gigantic positive x e a tiny negative y and still hold the conditions true.



Is it in Statement 1 or 2 or 1&2 taken together?

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Re: Is x + y > 0 ? [#permalink]

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New post 07 Nov 2016, 22:19
1)x-y>0 is insufficient as it will show x>y but if x and y are negative then x+y can't be greater than 0. (Insufficient)
2)(x+y)(x-y)>0 is insufficient as we can't say whether x+y>0 or not

Combining we can say x+y>0.. So Answer is C

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Re: Is x + y > 0 ? [#permalink]

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New post 08 Nov 2016, 04:45
Quote:
Is it in Statement 1 or 2 or 1&2 taken together?


The answer is C, but the approach is a bit more complex. Otherwise the right choice is by luck.

Here is a conceptual solution to the problem
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Is x + y > 0 ? [#permalink]

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New post 29 Jan 2017, 02:17
sunita123 wrote:
Is \(x + y > 0\)?

(1) \(x - y > 0\)
(2) \(x^{2} - y^{2} > 0\)


We can rephrase the question by subtracting y from both sides of the inequality: Is \(x > -y\) ?
 
(1) INSUFFICIENT: If we add y to both sides, we see that x is greater than y. We can use numbers here to show that this does not necessarily mean that \(x > -y\). If \(x = 4\) and \(y = 3\), then it is true that \(x\) is also greater than \(-y\). However if \(x = 4\) and \(y = -5\), \(x\) is greater than \(y\) but it is NOT greater than \(-y\). 
 
(2) INSUFFICIENT:  If we factor this inequality, we come up \((x + y)(x – y) > 0\). 
For the product of \((x + y)\) and \((x – y)\) to be greater than zero, the must have the same sign, i.e. both negative or both positive. 
This does not help settle the issue of the sign of \(x + y\). 

(1) AND (2) SUFFICIENT: From statement 2 we know that \((x + y)\) and \((x – y)\) must have the same sign, and from statement 1 we know that \((x – y)\) is positive, so it follows that \((x + y)\) must be positive as well.
 
The correct answer is C.
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Re: Is x + y > 0 ? [#permalink]

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New post 16 Oct 2017, 01:59
Bunuel wrote:
Is x + y > 0 ?

(1) x - y > 0 --> x > y. One number is greater than another. From this we cannot say whether their sum is positive. Not sufficient.

(2) x^2 - y^2 > 0 --> x^2 > y^2 --> |x| > |y|. One number is further from 0 than another. From this we cannot say whether their sum is positive. Not sufficient.

(1)+(1) From (2) (x - y)(x + y) > 0 (x + y and x - y have the same sign) and from (1) x - y > 0, thus x + y > 0. Sufficient.

Answer: C.


Hi Bunuel

A little help plz.
This is how i did statement 2.
x^2 - y^2 > 0 --> (x+y) (x-y) > 0 --> if we divide both sides by x-y we get x+y > 0
Hence sufficient.
Can you please help me fill gaps in my understanding here ?

Thanks in advance.
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Re: Is x + y > 0 ?   [#permalink] 16 Oct 2017, 01:59
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