Please help me with this question

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Please help me with this question [#permalink]

28 May 2011, 06:51

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59% (02:12) correct

40% (01:46) wrong

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I found this question in a practice test and am not able to comprehend the solution to the problem. The question is as follows:

Is x + y > 0 ?
(I) x² - y² > 1
(II) x/y + 1 > 0

I marked the answer as B using the following approach,
x/y + 1 > 0
x/y > -1
x > -y
x + y > 0

But the correct answer this E. Can someone please explain this? Thanks!

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Re: Please help me with this question [#permalink]

28 May 2011, 07:11

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Welcome to GMAT Club!

Here is a 10-sec solution:

x^2, y^2, x/y - are insensitive to changing simultaneously signs of x and y but x+y reverses its sign. So, it's E
In other words, let's say, if x=-5; y=-1 satisfies both conditions, x=5; y=1 will satisfy them too, but the answer will be different.
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Re: Please help me with this question [#permalink]

28 May 2011, 07:14

gujralvikas wrote:

x/y + 1 > 0
x/y > -1
x > -y
x + y > 0

x/y > -1 ---> x > -y (only if y>=0)! So you forgot to consider x < -y for y<0
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Re: Please help me with this question [#permalink]

29 May 2011, 09:40

Let x = 2, y = 1

and x = -2, y = -1

then both (1) and (2) are correct, but x+y > 0 or x+y < 0.

Answer - E
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Re: Please help me with this question [#permalink]

29 May 2011, 15:15

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gujralvikas wrote:

I am not solving the question since walker has already given you a very impressive logical solution and subhashghosh has shown the solution using 'plugging in numbers'.
But, let me add something here, "You multiply/divide an inequality by a number only if you know whether the number is negative or positive."
If the number is positive, fine. Just go ahead and do what you do in case of equations.
If the number is negative, there is no problem either but remember, you have to flip the inequality sign.

e.g.
Given: x < y
Multiply by 5: 5x < 5y
Multiply by -5: -5x > -5y
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Re: Please help me with this question [#permalink]

06 Jun 2011, 03:22

a+b

x = +|- 3 and y = +|- 2 gives different values for x+y > 0. Not sufficient.

E it is.
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Re: Please help me with this question [#permalink] 06 Jun 2011, 03:22

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