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# Is x > y ? (1) x + y > 0 (2) x^2 – y^2 > 0

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Math Expert
Joined: 02 Sep 2009
Posts: 46207
Is x > y ? (1) x + y > 0 (2) x^2 – y^2 > 0 [#permalink]

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15 Jan 2018, 08:27
00:00

Difficulty:

35% (medium)

Question Stats:

67% (01:07) correct 33% (00:53) wrong based on 46 sessions

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Is x > y ?

(1) x + y > 0
(2) x^2 – y^2 > 0

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Math Expert
Joined: 02 Aug 2009
Posts: 5898
Re: Is x > y ? (1) x + y > 0 (2) x^2 – y^2 > 0 [#permalink]

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15 Jan 2018, 08:43
Bunuel wrote:
Is x > y ?

(1) x + y > 0
(2) x^2 – y^2 > 0

Is $$x>y$$? or is $$x-y>0$$

(1) $$x + y > 0$$
nothing much
insuff

(2) $$x^2 – y^2 > 0$$..
$$x^2 – y^2 > 0.........(x-y)(x+y)>0$$
so both x-y and x+y are of SAME sign
if x+y>0, x-y>0..
If x+y<0, x-y<0
insuff

combined..
x+y>0, so x-y>0
suff
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Is x > y ? (1) x + y > 0 (2) x^2 – y^2 > 0 [#permalink]

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15 Jan 2018, 23:39
Bunuel wrote:
Is x > y ?

(1) x + y > 0
(2) x^2 – y^2 > 0

Check Points : Positive , negative and neutral test cases.
Assumption : Lets assume condition 1 to be true . We have x+y > 0

Negative test case : x=-5, y=7. x+y>0 Yes
Is X > y . Answer No

Positive test case : x=7, y=5. x+y>0 Yes

We do not have a definite answer to this answer choice. Hence A is not sufficient.

Assumption : Lets assume condition 2 to be true . We have x^2 – y^2 > 0. Breaking it down leads to (x+y)(x-y) > 0
=>if ( x+y ) is positive, (x-y) has to be positive
=> if (x+y) is negative, (x-y) has to be negative to make the assumption true.

Checkpoints : All positive , negative and neutral values of x and y
Let x=-5, y=7. x+y is positive, x-y is negative. Does not meet the condition
Let x=5, y=7. x+y is positive and x-y is negative. Does not meet the condition.
Let x=7, y=-5. x+y is positive, x-y is positive. Satisfys the condition.
Let x=7, y=5. x+y is positive, x-y is positive. Satisfys the condition.
Inference, x has to be greater than y to satisfy option B.

Hence, option B is sufficient to answer the question.
Choice B is correct.
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Location: India
Re: Is x > y ? (1) x + y > 0 (2) x^2 – y^2 > 0 [#permalink]

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16 Jan 2018, 00:10
Quantic wrote:
Bunuel wrote:
Is x > y ?

(1) x + y > 0
(2) x^2 – y^2 > 0

Check Points : Positive , negative and neutral test cases.
Assumption : Lets assume condition 1 to be true . We have x+y > 0

Negative test case : x=-5, y=7. x+y>0 Yes
Is X > y . Answer No

Positive test case : x=7, y=5. x+y>0 Yes

We do not have a definite answer to this answer choice. Hence A is not sufficient.

Assumption : Lets assume condition 2 to be true . We have x^2 – y^2 > 0. Breaking it down leads to (x+y)(x-y) > 0
=>if ( x+y ) is positive, (x-y) has to be positive
=> if (x+y) is negative, (x-y) has to be negative to make the assumption true.

Checkpoints : All positive , negative and neutral values of x and y
Let x=-5, y=7. x+y is positive, x-y is negative. Does not meet the condition
Let x=5, y=7. x+y is positive and x-y is negative. Does not meet the condition.
Let x=7, y=-5. x+y is positive, x-y is positive. Satisfys the condition.
Let x=7, y=5. x+y is positive, x-y is positive. Satisfys the condition.
Inference, x has to be greater than y to satisfy option B.

Hence, option B is sufficient to answer the question.
Choice B is correct.

Hi

Good approach, but I think this is slightly tricky here.

In case of second condition, we should also take a case where x= -7, y= 5. x+y is negative, and x-y is also negative. But here x is NOT greater than y. So x does not have to be greater than y to meet second statement's condition.

I think answer should be C (after combining both statements).
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Joined: 03 Jun 2015
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WE: Information Technology (Computer Software)
Re: Is x > y ? (1) x + y > 0 (2) x^2 – y^2 > 0 [#permalink]

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16 Jan 2018, 00:18
amanvermagmat wrote:
Quantic wrote:
Bunuel wrote:
Is x > y ?

(1) x + y > 0
(2) x^2 – y^2 > 0

Check Points : Positive , negative and neutral test cases.
Assumption : Lets assume condition 1 to be true . We have x+y > 0

Negative test case : x=-5, y=7. x+y>0 Yes
Is X > y . Answer No

Positive test case : x=7, y=5. x+y>0 Yes

We do not have a definite answer to this answer choice. Hence A is not sufficient.

Assumption : Lets assume condition 2 to be true . We have x^2 – y^2 > 0. Breaking it down leads to (x+y)(x-y) > 0
=>if ( x+y ) is positive, (x-y) has to be positive
=> if (x+y) is negative, (x-y) has to be negative to make the assumption true.

Checkpoints : All positive , negative and neutral values of x and y
Let x=-5, y=7. x+y is positive, x-y is negative. Does not meet the condition
Let x=5, y=7. x+y is positive and x-y is negative. Does not meet the condition.
Let x=7, y=-5. x+y is positive, x-y is positive. Satisfys the condition.
Let x=7, y=5. x+y is positive, x-y is positive. Satisfys the condition.
Inference, x has to be greater than y to satisfy option B.

Hence, option B is sufficient to answer the question.
Choice B is correct.

Hi

Good approach, but I think this is slightly tricky here.

In case of second condition, we should also take a case where x= -7, y= 5. x+y is negative, and x-y is also negative. But here x is NOT greater than y. So x does not have to be greater than y to meet second statement's condition.

I think answer should be C (after combining both statements).

:D
I missed this test case.
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Re: Is x > y ? (1) x + y > 0 (2) x^2 – y^2 > 0 [#permalink]

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16 Jan 2018, 05:49
Bunuel wrote:
Is x > y ?

(1) x + y > 0
(2) x^2 – y^2 > 0

We know that it is an easy (C) because the moment we see x^2 - y^2 in stmnt 2, it makes us think of (x + y)*(x - y).
Stmnt 1 says that (x + y) is positive which means (x - y) > 0 too. This implies x > y.

We just need to ensure that no statement alone is sufficient.
Stmnt 1 is simple enough. Either x or y could be greater say x = 2, y = 3 or x = 3, y = 2. Not sufficient
Stmnt 2 alone just implies that absolute value of x is greater than absolute value of y so x = 3, y = 2 is possible, also x = -3, y = 2 is possible. Not sufficient

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Re: Is x > y ? (1) x + y > 0 (2) x^2 – y^2 > 0   [#permalink] 16 Jan 2018, 05:49
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