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Buy "All-In-One Standard ($149)", get free Daily quiz (2 mon). Coupon code : SPECIAL # Is |x - y| > |x + y|?  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Manager Joined: 14 Apr 2010 Posts: 176 Is |x - y| > |x + y|? [#permalink] ### Show Tags 02 Aug 2010, 23:16 1 10 00:00 Difficulty: 55% (hard) Question Stats: 63% (02:04) correct 37% (02:38) wrong based on 250 sessions ### HideShow timer Statistics Is |x - y| > |x + y|? (1) x^2 - y^2 = 9 (2) x - y = 2 ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 52935 Is |x - y| > |x + y|? [#permalink] ### Show Tags 04 Aug 2010, 00:34 4 3 bibha wrote: Is |x-y| > |x+y| i. x^2-y^2 = 9 ii. x-y=2 Please take the trouble of explaining one step at a time I'd start solving by analyzing the stem. First approach: Is $$|x-y|>|x+y|$$? In which cases $$|x-y|$$ will be more than $$|x+y|$$? If $$x$$ and $$y$$ have the same sign (both positive or both negative) then absolute value of their sum ($$|x+y|$$) will be more than absolute value of their difference ($$|x-y|$$), as in this case $$x$$ and $$y$$ will contribute to each other in $$|x+y|$$, for example $$|5+3|$$ or $$|-5-3|$$ and in $$|x-y|$$ they will not. If $$x$$ and $$y$$ have opposite signs then absolute value of their difference ($$|x-y|$$) will be more than absolute value of their sum ($$|x+y|$$), as in this case $$x$$ and $$y$$ will contribute to each other in $$|x-y|$$, for example $$|5-(-3)|$$ or $$|-5-3|$$ and in $$|x+y|$$ they will not. So basically the question is: do $$x$$ and $$y$$ have opposite signs? Second approach: Is $$|x-y|>|x+y|$$? As both sides of inequality are non-negative we can safely square them: is $$(x-y)^2>(x+y)^2$$? --> is $$x^2-2xy+y^2>x^2+2xy+y^2$$? --> is $$xy<0$$? Again the question becomes: do $$x$$ and $$y$$ have opposite signs? Next: (1) x^2-y^2 = 9 --> we cannot say from this whether $$x$$ and $$y$$ have opposite signs. Not sufficient. (2) x-y=2 --> we cannot say from this whether $$x$$ and $$y$$ have opposite signs. Not sufficient. (1)+(2) $$x+y=4.5$$ and $$x-y=2$$ --> directly tells us that $$|x-y|=2<|x+y|=4.5$$. Sufficient. (If you solve system of equations you'll see that $$x$$ and $$y$$ have the same sign: they are both positive). Answer: C. PLUGGING NUMBERS: On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another. (1) $$x^2-y^2=9$$ --> try $$x=5$$ and $$y=4$$ ($$x^2-y^2=25-16=9$$) --> $$|x-y|=1<|x+y|=9$$, so we have answer YES. Now let's try another set of numbers to get NO: $$x=5$$ and $$y=-4$$ --> $$|x-y|=9>|x+y|=1$$, so now we have answer NO. Thus this statement is not sufficient. We can do the same with statement (2) as well. For this question I don't recommend to use number plugging for (1)+(2), as it's quite straightforward algebraically that taken together 2 statements are sufficient. Hope it helps. _________________ ##### General Discussion Manager Status: Waiting to hear from University of Texas at Austin Joined: 24 May 2010 Posts: 68 Location: Changchun, China Schools: University of Texas at Austin, Michigan State Re: approach to these kind of qs?? [#permalink] ### Show Tags 03 Aug 2010, 04:17 1 bibha wrote: Is |x-y| > |x+y| i. x^2-y^2 = 9 ii. x-y=2 Please take the trouble of explaining one step at a time To me I try to address (ii) first because it looks easiest Pick x and y that meet the conditions given (x=5 y=3) |5-3| > |5+3| FALSE Try different x and y thinking we are using absolutes I picked two negatives (x=-3 y=-5) |-3-(-5)| > |-3+-5| 2 > 8 False So we see that ii alone is sufficient. The answer is B or D To investigate i I am lost at the algebra, so I just try to think about which squares are 9 apart 3^2=9 4^2=16 5^2=25 luckily on the GMAT you typically work with smaller squares So I pick X=5 and Y=4 because $$5^2-4^2=9$$ Since we are squaring keep in mind that X can be 5 or -5 While Y can be 4 or -4 The test in the given equation |x-y| > |x+y| (5,4) gives us 1 > 9 False (5,-4) gives us 9 > 1 True (-5,4) gives us 9 > 1 True (-5,-4) gives us 1 > 9 False So A alone doesn't work The correct answer is B. In proofreading my work I found a major error in my calculation with the absolutes. This changed my answer on this question. On the real GMAT I would have guessed probably E after 3-4 minutes of frustration! Director Status: Apply - Last Chance Affiliations: IIT, Purdue, PhD, TauBetaPi Joined: 17 Jul 2010 Posts: 611 Schools: Wharton, Sloan, Chicago, Haas WE 1: 8 years in Oil&Gas Re: approach to these kind of qs?? [#permalink] ### Show Tags 03 Aug 2010, 09:28 TallJTinChina wrote: bibha wrote: Is |x-y| > |x+y| i. x^2-y^2 = 9 ii. x-y=2 Please take the trouble of explaining one step at a time To me I try to address (ii) first because it looks easiest Pick x and y that meet the conditions given (x=5 y=3) |5-3| > |5+3| FALSE Try different x and y thinking we are using absolutes I picked two negatives (x=-3 y=-5) |-3-(-5)| > |-3+-5| 2 > 8 False So we see that ii alone is sufficient. The answer is B or D To investigate i I am lost at the algebra, so I just try to think about which squares are 9 apart 3^2=9 4^2=16 5^2=25 luckily on the GMAT you typically work with smaller squares So I pick X=5 and Y=4 because $$5^2-4^2=9$$ Since we are squaring keep in mind that X can be 5 or -5 While Y can be 4 or -4 The test in the given equation |x-y| > |x+y| (5,4) gives us 1 > 9 False (5,-4) gives us 9 > 1 True (-5,4) gives us 9 > 1 True (-5,-4) gives us 1 > 9 False So A alone doesn't work The correct answer is B. In proofreading my work I found a major error in my calculation with the absolutes. This changed my answer on this question. On the real GMAT I would have guessed probably E after 3-4 minutes of frustration! In plugging in numbers approach - like here, (ii) works for 2 examples selected, but how do we universally say it is true? Is there not a more general approach? I think for speed we just select a few examples, but how do we guarantee it? I am always confused on this point. So I jump to prove it for good and I made a mistake here I saw x^2-y^2 = 9, took it to mean (x+y)(x-y)=9 and said x+y,x-y have to be both 3 or -3 and jumped at A. Wrong, does not say that they have to be integers... Please explain or point me to the strategy in dealing with these please! _________________ Consider kudos, they are good for health Intern Joined: 23 Nov 2009 Posts: 3 Re: approach to these kind of qs?? [#permalink] ### Show Tags 03 Aug 2010, 10:24 1 1 IMO C Stmt 1: After simplifying we get (x+y)(x-y)=9. So insuff. Stmt 2: x-y=2 now if x=5 and y=3 then, |5-3| > |5+3| does not hold true. but if x=1 and y=-1 then |1-(-1)| > |1-1| holds true. So stmt 2 is also insuff. combining 1 & 2, we have value of x-y=2 from stmt 2, so subsituting x-y=2 in (x-y)(x+y)=9, we get x+y=4.5. Now 2 < 4.5 ....so |x-y|<|x+y|.... please correct if I have mistaken anything here.... Director Status: Apply - Last Chance Affiliations: IIT, Purdue, PhD, TauBetaPi Joined: 17 Jul 2010 Posts: 611 Schools: Wharton, Sloan, Chicago, Haas WE 1: 8 years in Oil&Gas Re: approach to these kind of qs?? [#permalink] ### Show Tags 07 Aug 2010, 03:37 Bunuel wrote: bibha wrote: Is |x-y| > |x+y| i. x^2-y^2 = 9 ii. x-y=2 Please take the trouble of explaining one step at a time I'd start solving by analyzing the stem. First approach: Is $$|x-y|>|x+y|$$? In which cases $$|x-y|$$ will be more than $$|x+y|$$? If $$x$$ and $$y$$ have the same sign (both positive or both negative) then absolute value of their sum ($$|x+y|$$) will be more than absolute value of their difference ($$|x-y|$$), as in this case $$x$$ and $$y$$ will contribute to each other in $$|x+y|$$, for example $$|5+3|$$ or $$|-5-3|$$ and in $$|x-y|$$ they will not. If $$x$$ and $$y$$ have opposite signs then absolute value of their difference ($$|x-y|$$) will be more than absolute value of their sum ($$|x+y|$$), as in this case $$x$$ and $$y$$ will contribute to each other in $$|x-y|$$, for example $$|5-(-3)|$$ or $$|-5-3|$$ and in $$|x+y|$$ they will not. So basically the question is: do $$x$$ and $$y$$ have opposite signs? Second approach: Is $$|x-y|>|x+y|$$? As both sides of inequality are non-negative we can safely square them: is $$(x-y)^2>(x+y)^2$$? --> is $$x^2-2xy+y^2>x^2+2xy+y^2$$? --> is $$xy<0$$? Again the question becomes: do $$x$$ and $$y$$ have opposite signs? Next: (1) x^2-y^2 = 9 --> we can not say from this whether $$x$$ and $$y$$ have opposite signs. Not sufficient. (2) x-y=2 --> we can not say from this whether $$x$$ and $$y$$ have opposite signs. Not sufficient. (1)+(2) $$x+y=4.5$$ and $$x-y=2$$ --> directly tells us that $$|x-y|=2<|x+y|=4.5$$. Sufficient. (If you solve system of equations you'll see that $$x$$ and $$y$$ have the same sign: they are both positive). Answer: C. PLUGGING NUMBERS: On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another. (1) $$x^2-y^2=9$$ --> try $$x=5$$ and $$y=4$$ ($$x^2-y^2=25-16=9$$) --> $$|x-y|=1<|x+y|=9$$, so we have answer YES. Now let's try another set of numbers to get NO: $$x=5$$ and $$y=-4$$ --> $$|x-y|=9>|x+y|=1$$, so now we have answer NO. Thus this statement is not sufficient. We can do the same with statement (2) as well. For this question I don't recommend to use number plugging for (1)+(2), as it's quite straightforward algebraically that taken together 2 statements are sufficient. Hope it helps. Excellent Bunuel!! Simply Excellent. When it is said that x-y=2 and x^2-y^2=9 why does it not jump at me that x+y must be 4.5!!! Got your point about plugging in numbers.. Can disprove but proving universally is difficult... _________________ Consider kudos, they are good for health Manager Joined: 24 Apr 2010 Posts: 55 Re: approach to these kind of qs?? [#permalink] ### Show Tags 07 Aug 2010, 08:09 well I eliminated A and B out as in both case x and y can be both and negative in either case.... which will obvoiusly give dual case... so D also out.... now here was big problem C and E.... I was not able to decide here i was trying to establish a case where x must be + or - same with y if we can do that we will have 1 another yes or no as answer any help here? Intern Joined: 01 Aug 2006 Posts: 33 Re: Is |x - y| > |x + y|? [#permalink] ### Show Tags 24 Jan 2014, 14:24 Squaring and rearranging, is xy <0? (x>0, y<0; or x<0, y>0) 1.x^2-y^2 = 9 -> x and y can take same/diff signs 2. x - y = 2 -> cannot be sure Combining, (x+y)(x-y) = 9; since x -y = 2, x +y =4.5 No need to solve because we get distinct values for x and y => we know the signs of x and y => we know the sign of xy. C. SVP Joined: 06 Sep 2013 Posts: 1696 Concentration: Finance Re: approach to these kind of qs?? [#permalink] ### Show Tags 30 Jan 2014, 16:20 Bunuel wrote: bibha wrote: Is |x-y| > |x+y| i. x^2-y^2 = 9 ii. x-y=2 Please take the trouble of explaining one step at a time I'd start solving by analyzing the stem. First approach: Is $$|x-y|>|x+y|$$? In which cases $$|x-y|$$ will be more than $$|x+y|$$? If $$x$$ and $$y$$ have the same sign (both positive or both negative) then absolute value of their sum ($$|x+y|$$) will be more than absolute value of their difference ($$|x-y|$$), as in this case $$x$$ and $$y$$ will contribute to each other in $$|x+y|$$, for example $$|5+3|$$ or $$|-5-3|$$ and in $$|x-y|$$ they will not. If $$x$$ and $$y$$ have opposite signs then absolute value of their difference ($$|x-y|$$) will be more than absolute value of their sum ($$|x+y|$$), as in this case $$x$$ and $$y$$ will contribute to each other in $$|x-y|$$, for example $$|5-(-3)|$$ or $$|-5-3|$$ and in $$|x+y|$$ they will not. So basically the question is: do $$x$$ and $$y$$ have opposite signs? Second approach: Is $$|x-y|>|x+y|$$? As both sides of inequality are non-negative we can safely square them: is $$(x-y)^2>(x+y)^2$$? --> is $$x^2-2xy+y^2>x^2+2xy+y^2$$? --> is $$xy<0$$? Again the question becomes: do $$x$$ and $$y$$ have opposite signs? Next: (1) x^2-y^2 = 9 --> we can not say from this whether $$x$$ and $$y$$ have opposite signs. Not sufficient. (2) x-y=2 --> we can not say from this whether $$x$$ and $$y$$ have opposite signs. Not sufficient. (1)+(2) $$x+y=4.5$$ and $$x-y=2$$ --> directly tells us that $$|x-y|=2<|x+y|=4.5$$. Sufficient. (If you solve system of equations you'll see that $$x$$ and $$y$$ have the same sign: they are both positive). Answer: C. PLUGGING NUMBERS: On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another. (1) $$x^2-y^2=9$$ --> try $$x=5$$ and $$y=4$$ ($$x^2-y^2=25-16=9$$) --> $$|x-y|=1<|x+y|=9$$, so we have answer YES. Now let's try another set of numbers to get NO: $$x=5$$ and $$y=-4$$ --> $$|x-y|=9>|x+y|=1$$, so now we have answer NO. Thus this statement is not sufficient. We can do the same with statement (2) as well. For this question I don't recommend to use number plugging for (1)+(2), as it's quite straightforward algebraically that taken together 2 statements are sufficient. Hope it helps. I like second approach best, squaring both sides is definetely something you can do here in order to manipulate the problem more easily getting rid of the absolute values in both sides and saving time instead of plugging numbers Just my 2cents Cheers J Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6962 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Is |x - y| > |x + y|? [#permalink] ### Show Tags 10 Nov 2015, 10:46 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Is |x - y| > |x + y|? (1) x^2 - y^2 = 9 (2) x - y = 2 If we modify the question, |x - y| > |x + y|?, or (|x - y|)^2 > (|x + y|)^2?, or (x - y)^2 > (x + y)^2?, or x^2-2xy+y^2>x^2+2xy+y^2?, or -2xy>2xy?, or 0>4xy?, or xy<0?. The conditions do not have xy<0, so there are 2 variables (X,y) and 2 equations are given from the 2 conditions, so there is high chance (C) will be our answer. Looking at the conditions together, x-y=2, x+y=9/2. This is sufficient and the answer becomes (C). For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Re: Is |x - y| > |x + y|?  [#permalink]

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22 Jul 2017, 02:54
bibha wrote:
Is |x - y| > |x + y|?

(1) x^2 - y^2 = 9
(2) x - y = 2

Let us try to solve it step by step.

1. It says (x+y)(x-y)=9.
It means either both of them are positive or negative.
In that case, we can not say anything about their magnitudes.
Hence Insufficient.

2. It says x-y=2

If x=5, y=3, then x-y<x+y

If x=1, y=-2, x-y>x+y

If x=-5, y=-7, x-y<x+y(absolute values)

Hence Insufficient.

1+2, we have to take x+y and x-y of same signs.
it discards the usage of X=1 and y=-2.

Hence sufficient
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Re: Is |x - y| > |x + y|?  [#permalink]

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26 Jul 2017, 11:10
bibha wrote:
Is |x - y| > |x + y|?

(1) x^2 - y^2 = 9
(2) x - y = 2

|x-y| > |x+y| ? This can only happen if they are opposite signs, but looking the statements, plugging in number seems to be the fastest approach.

1) $$x^2$$-$$y^2$$ = 9
=> (x+y)(x-y)=9
x=5, y=4 ; x+y=9 and x-y=1 => No
x=5, y= -4 ; x=y=1 and x-y=9 => Yes

Insufficient.

2) x-y = 2
x+y = ?
Insufficient.

1+2)
(x+y)(x-y) = 9
x-y = 2
=> x+y = 4.5
Sufficient.

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Re: Is |x - y| > |x + y|?  [#permalink]

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01 Oct 2018, 20:43
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Re: Is |x - y| > |x + y|?   [#permalink] 01 Oct 2018, 20:43
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