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# Is x + y > 0 ?

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Intern
Joined: 30 Sep 2010
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Is x + y > 0 ? [#permalink]

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26 Dec 2010, 08:31
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55% (hard)

Question Stats:

59% (01:18) correct 41% (01:08) wrong based on 152 sessions

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Is x + y > 0 ?

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

Kudos [?]: 157 [0], given: 1

Veritas Prep GMAT Instructor
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26 Dec 2010, 22:55
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Expert's post
surendar26 wrote:
Is x + y > 0 ?
(I) x² - y² > 1
(II) x/y + 1 > 0

(I) x² - y² > 1
(x + y)(x - y) is positive. So either both are positive or both are negative. Also, absolute value of x is greater than absolute value of y.
e.g. x = 3, y = 2, then (x + y) = 5 and (x+y)(x - y) = 5
x = -4, y = -2, then (x + y) = -6 and (x + y)(x - y) = 12
(x + y) can be positive or negative. Not sufficient.

(II) x/y + 1 > 0
(x+y)/y > 0
So either both are positive or both are negative.
e.g. y positive. y = 4, x = 3, then (x+y) = 7 and (x + y)/y = 7/4
y negative. y = -4, x = 3, then (x+y) = -1 and (x + y)/y = (-1)/(-4) = 1/4
So x + y can be positive or negative. Not sufficient.

Taking both together,
(x+y), (x -y) and y, all have the same signs. The same examples as shown for statement I above satisfy this condition.
e.g. y positive. x = 3, y = 2, then (x + y) = 5, (x - y) = 1
y negative. x = -4, y = -2, then (x + y) = -6, (x - y) = -2
(x + y) can be positive or negative. Not sufficient.

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17361 [1], given: 232 Manager Joined: 30 Aug 2010 Posts: 91 Kudos [?]: 191 [0], given: 27 Location: Bangalore, India Re: Inequalities [#permalink] ### Show Tags 31 Jan 2011, 05:46 surendar26 wrote: Is x + y > 0 ? (I) x² - y² > 1 (II) x/y + 1 > 0 Let me solve it the way GMAT expects us to do. qtn: Is X+Y positive? stmnt1: $$x^2 - y^2 > 1$$ ==> $$(x+1)(x-y) > 1$$ ==> $$(x+y)(x-y)$$ shud surely be > 0 as it is > 1 (> 1, instead of > 0, is given just to make the statement more indirect/confusing) ==> $$(x+y)(x-y)$$ is positive ==> $$(x+y)$$ and $$(x-y)$$ both, at the same time, are positive or nagative...not suff. stmnt2 $$\frac{x}{y} + 1$$ > 0 ==> $$(x+y)/y$$ > 0 ==> $$(x+y)/y$$ is positve again $$(x+y)$$ and $$y$$both, at the same time, are positive or nagative...not suff. stmnts1 and 2 together: $$X+Y$$ cab be positive or negative...NOT suff. Answer "E". Regards, Murali. Kudos? Kudos [?]: 191 [0], given: 27 Math Expert Joined: 02 Sep 2009 Posts: 41891 Kudos [?]: 128912 [3], given: 12183 Re: Check this one out - Algebra [#permalink] ### Show Tags 04 Feb 2011, 12:27 3 This post received KUDOS Expert's post Merging similar topics. mariyea wrote: Is x + y > 0 ? (I) x² - y² > 1 (II) x/y + 1 > 0 Simple logic would probably be the best way to deal with this question (without much calculation, algebra and number plugging). Is x + y > 0 ? Question asks whether the sum of $$x$$ and $$y$$ is positive. (1) x² - y² > 1 --> if $$x$$ is some big enough positive number and $$y$$ is some small enough positive number (for example $$x=2$$ and $$y=1$$) then the answer will obviously be YES as the sum of two positive values is positive BUT if you consider the same values but with the minus sign ($$x=-2$$ and $$y=-1$$) then again the answer will obviously be NO as the sum of two negative values is negative. Not sufficient. (2) x/y + 1 > 0 --> exact same approach for this statement: if both $$x$$ and $$y$$ are positive (which satisfies the given statement as x/y+1=positive/positive+positive) then the answer will be YES BUT if both $$x$$ and $$y$$ are negative (which also satisfies the given statement as x/y+1=negative/negative+positive=positive+positive) then the answer will be NO. Not sufficient. (1)+(2) Two positive values and two negative values from (1), also satisfy (2), so we still have two answers. Not sufficient. Answer: E. _________________ Kudos [?]: 128912 [3], given: 12183 Senior Manager Joined: 30 Nov 2010 Posts: 257 Kudos [?]: 117 [0], given: 66 Schools: UC Berkley, UCLA Re: Check this one out - Algebra [#permalink] ### Show Tags 05 Feb 2011, 10:46 Bunuel wrote: Merging similar topics. mariyea wrote: Is x + y > 0 ? (I) x² - y² > 1 (II) x/y + 1 > 0 Simple logic would probably be the best way to deal with this question (without much calculation, algebra and number plugging). Is x + y > 0 ? Question asks whether the sum of $$x$$ and $$y$$ is positive. (1) x² - y² > 1 --> if $$x$$ is some big enough positive number and $$y$$ is some small enough positive number (for example $$x=2$$ and $$y=1$$) then the answer will obviously be YES as the sum of two positive values is positive BUT if you consider the same values but with the minus sign ($$x=-2$$ and $$y=-1$$) then again the answer will obviously be NO as the sum of two negative values is negative. Not sufficient. (2) x/y + 1 > 0 --> exact same approach for this statement: if both $$x$$ and $$y$$ are positive (which satisfies the given statement as x/y+1=positive/positive+positive) then the answer will be YES BUT if both $$x$$ and $$y$$ are negative (which also satisfies the given statement as x/y+1=negative/negative+positive=positive+positive) then the answer will be NO. Not sufficient. (1)+(2) Two positive values and two negative values from (1), also satisfy (2), so we still have two answers. Not sufficient. Answer: E. Yeah I chose E too. I wanted to see how you would approach this q. I took the statements apart to find the q simpler to solve... Thank you Bunuel! _________________ Thank you for your kudoses Everyone!!! "It always seems impossible until its done." -Nelson Mandela Kudos [?]: 117 [0], given: 66 Intern Joined: 17 Apr 2011 Posts: 4 Kudos [?]: [0], given: 0 Is x + y > 0 ? [#permalink] ### Show Tags 28 May 2011, 06:51 1 This post was BOOKMARKED Is x + y > 0 ? (1) x^2 - y^2 > 1 (2) x/y + 1 > 0 [Reveal] Spoiler: 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! Last edited by Bunuel on 10 Jun 2014, 14:03, edited 1 time in total. Renamed the topic and edited the question. Kudos [?]: [0], given: 0 CEO Joined: 17 Nov 2007 Posts: 3584 Kudos [?]: 4584 [2], given: 360 Concentration: Entrepreneurship, Other Schools: Chicago (Booth) - Class of 2011 GMAT 1: 750 Q50 V40 Re: Please help me with this question [#permalink] ### Show Tags 28 May 2011, 07:11 2 This post received KUDOS Expert's post 2 This post was BOOKMARKED 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. _________________ HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame Kudos [?]: 4584 [2], given: 360 CEO Joined: 17 Nov 2007 Posts: 3584 Kudos [?]: 4584 [0], given: 360 Concentration: Entrepreneurship, Other Schools: Chicago (Booth) - Class of 2011 GMAT 1: 750 Q50 V40 Re: Please help me with this question [#permalink] ### Show Tags 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 _________________ HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame Kudos [?]: 4584 [0], given: 360 SVP Joined: 16 Nov 2010 Posts: 1597 Kudos [?]: 592 [0], given: 36 Location: United States (IN) Concentration: Strategy, Technology Re: Please help me with this question [#permalink] ### Show Tags 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 _________________ Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant) GMAT Club Premium Membership - big benefits and savings Kudos [?]: 592 [0], given: 36 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7674 Kudos [?]: 17361 [2], given: 232 Location: Pune, India Re: Please help me with this question [#permalink] ### Show Tags 29 May 2011, 15:15 2 This post received KUDOS Expert's post gujralvikas wrote: 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! 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 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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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: Is x + y > 0 ? [#permalink]

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13 Mar 2016, 07:40
Here Combining both the statements => x+y can be +ve as values can be both positive
and the sum can be negative as when both will be positive
so E
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Re: Is x + y > 0? [#permalink]

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01 Nov 2016, 21:12
shubhraghosh wrote:
Here is an interesting question I came across today. It's a good question in my opinion for all the tricks DS questions are known to present

Is x+y > 0?

(1) x^2 - y^2 > 1
(2) (x/y) + 1 > 0

Will post the OA shortly, but would be great to see some analysis here.

Cheers!

Ans: B
for the first statement
(1) if you plug in the numbers -4 for x and -1 for y, the statement will hold and x + y < 0 but if you plug in any two +ve numbers x + y > 0. Not Sufficient
(2) (x/y) > -1
Multiply both sides by y
x > -y
x + y > 0
Sufficient
Ans: B

I would really like to see a method for the first statement and not just plug in numbers, I had to plug in 3 numbers before I could get here which took a little more than 1 min.

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Re: Is x + y > 0?   [#permalink] 01 Nov 2016, 21:12
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