December 15, 2018 December 15, 2018 10:00 PM PST 11:00 PM PST Get the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) December 16, 2018 December 16, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 51218

Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) xy<xy
[#permalink]
Show Tags
Updated on: 05 Jul 2013, 14:21
Question Stats:
32% (02:39) correct 68% (02:22) wrong based on 1001 sessions
HideShow timer Statistics
Originally posted by Bunuel on 12 Nov 2009, 18:09.
Last edited by Bunuel on 05 Jul 2013, 14:21, edited 1 time in total.
Added the OA




Manager
Joined: 11 Sep 2009
Posts: 129

Re: Hard inequality: Is xy < 0
[#permalink]
Show Tags
12 Nov 2009, 18:44
Statement 1: \(\frac{x^3*y^5}{x*y^2} < 0\)
\(x^2*y^3 < 0\)
We know that \(x^2\) must be positive, so therefore y < 0. Since we do not know whether x is positive or negative, however, this is insufficient.
Statement 2: \(xy < xy\)
I'll approach this the way I would as if I was writing it on the GMAT with time constraints, by picking numbers:
(1) Is it possible for xy > 0 AND the above criteria to be satisfied? Yes. (i.e. x = 2, y = 3) (2) Is it possible for xy < 0 AND the above criteria to be satisfied? Yes. (i.e. x = 2, y = 3)
Therefore, not sufficient.
Evaluating Both Statements
We know from Statement 1 that y < 0. Therefore we can revise the above approach for Statement 2 to specifics:
(1) Is it possible for y < 0, x < 0 (i.e. xy > 0) AND the above criteria to be satisfied? Yes. (i.e. x = 2, y = 3) (2) Is it possible for y < 0, x > 0 (i.e. xy < 0) AND the above criteria to be satisfied? Yes. (i.e. x = 2, y = 3)
Therefore, insufficient.
The answer is E.




VP
Joined: 05 Mar 2008
Posts: 1405

Re: Hard inequality: Is xy < 0
[#permalink]
Show Tags
12 Nov 2009, 18:50
Bunuel wrote: Is \(xy<0\)?
(1) \(\frac{x^3*y^5}{x*y^2}<0\)
(2) \(xy<xy\) I'm going with e on this one (1) \(\frac{x^3*y^5}{x*y^2}<0\) x^31 * y*52 < 0 x^2*y^3<0 y < 0 but x<0 or x > 0 so insufficient because it doesn't matter what x is the answer will be negative however plugging back into the original equation gives two different answers (2) \(xy<xy\)[/quote] if x = 1 and y = 2 1  2 < 1 + 2 if x>0 and y > 0 then the equations will be equal if x> 0 and y < 0 yes if x<0 and Y < 0 yes and both would make \(xy<0\) a different answer



VP
Joined: 05 Mar 2008
Posts: 1405

Re: Hard inequality: Is xy < 0
[#permalink]
Show Tags
13 Nov 2009, 20:30
good question..keep posting these types



Math Expert
Joined: 02 Sep 2009
Posts: 51218

Re: Hard inequality: Is xy < 0
[#permalink]
Show Tags
13 Nov 2009, 21:07



VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1033

Re: Hard inequality: Is xy < 0
[#permalink]
Show Tags
05 May 2011, 21:12
a gives x can > or < 0. y is <0. not sufficient.
b x>0,y>0 or x>0 and y< 0 both satisfies. Not sufficient.
a+b both x >0  x<0 and y <0. Not sufficient.
Thus E



Manager
Joined: 26 Dec 2011
Posts: 93

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) xy<xy
[#permalink]
Show Tags
07 Mar 2012, 09:05
For the second case, I tried to solve by squaring on both the sides. So you get, 2xy<2xy, which is ....xy>xy...only case possible is when one of them is negative... will it then not prove that xy<0. Please let me know where am I wrong?



Math Expert
Joined: 02 Sep 2009
Posts: 51218

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) xy<xy
[#permalink]
Show Tags
07 Mar 2012, 09:19
pavanpuneet wrote: For the second case, I tried to solve by squaring on both the sides. So you get, 2xy<2xy, which is ....xy>xy...only case possible is when one of them is negative... will it then not prove that xy<0. Please let me know where am I wrong? Welcome to GMAT Club. Below is an answer to your question. You can square both sides of an inequality if and only both sides are nonnegative. For xy<xy we know that xy is nonnegative, but xy can be negative, as well as positive (nonnegative), hence you cannot apply squaring. GENERAL RULE: A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are nonnegative (the same for taking an even root of both sides of an inequality).For example: \(2<4\) > we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) > we can square both sides and write: \(x^2<y^2\); But if either of side is negative then raising to even power doesn't always work. For example: \(1>2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are nonnegative. B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).For example: \(2<1\) > we can raise both sides to third power and write: \(2^3=8<1=1^3\) or \(5<1\) > \(5^2=125<1=1^3\); \(x<y\) > we can raise both sides to third power and write: \(x^3<y^3\). Hope it helps.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 26 Dec 2011
Posts: 93

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) xy<xy
[#permalink]
Show Tags
08 Mar 2012, 03:32
Thanks a lot for your prompt response! Its clear to me now!



Manager
Status: Do till 740 :)
Joined: 13 Jun 2011
Posts: 87
Concentration: Strategy, General Management
GPA: 3.6
WE: Consulting (Computer Software)

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) xy<xy
[#permalink]
Show Tags
17 Nov 2012, 22:37
Hi bunuel,
) Holds true when: A. 0<x<y B. x<0<y C. y<x<0 D. y<0<x
pls let me know how this works
when you have xy>x y
if we take x=8 y=4
this will be always equal.xy can be never be greater than x y if both x and y have same signs Please let me know how you deduced it.



Math Expert
Joined: 02 Sep 2009
Posts: 51218

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) xy<xy
[#permalink]
Show Tags
18 Nov 2012, 04:08



Intern
Joined: 07 May 2011
Posts: 30

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) xy<xy
[#permalink]
Show Tags
19 Nov 2012, 18:42
I really like one of the ideas forwarded above.
data point 1 was fairly obvious. with data point 2, we need only to stare at it and the question xy<0 for a few seconds. the more you think about it the more it becomes obvious that you would need to pick numbers such that xy<0 can be proven and xy<0 can be debunked. you could do this by using x>0 and y<0 (to prove xy<0) and using x>0,y>0 (to debunk).



Intern
Joined: 06 Apr 2011
Posts: 13

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) xy<xy
[#permalink]
Show Tags
18 Feb 2013, 22:54
excellent question bunuel. after 3 mins of trying and testing numbers i frustatingly put C as an answer.... .. Thanks for explanation.



Retired Moderator
Joined: 05 Jul 2006
Posts: 1723

Re: Hard inequality: Is xy < 0
[#permalink]
Show Tags
13 May 2013, 11:13
Bunuel wrote: Yes, OA is E.
Number plugging as AKProdigy87 did is probably the best way here. But still if needed below is short overview of algebra of this question:
(1) y<0. Not sufficient.
(2) Holds true when: A. 0<x<y B. x<0<y C. y<x<0 D. y<0<x
Not sufficient.
(1)+(2) C. and D. options are left from (2) but still insufficient, as xy may or may not be negative.
Answer: E. Hi Bunuel, I want to double check the following with you ... the 4 cases u ve mentioned is detailed version of the following or not for the inequality to hold true 1) xy<0 2) /y/>/x/ and this is because the equality (=) in /yx/ > = /x/  /y/ holds true only in 2 cases /x/ = /y/ or /x/>/y/ and xy>0 am i right ?? plz let me know.. thanking you in advance.



Retired Moderator
Joined: 05 Jul 2006
Posts: 1723

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) xy<xy
[#permalink]
Show Tags
15 May 2013, 04:16
Bunuel wrote: shankar245 wrote: Hi bunuel,
) Holds true when: A. 0<x<y B. x<0<y C. y<x<0 D. y<0<x
pls let me know how this works
when you have xy>x y
if we take x=8 y=4
this will be always equal.xy can be never be greater than x y if both x and y have same signs Please let me know how you deduced it. No, the red part is not correct. Consider x=4 and y=8 OR x=4 and y=8. x=4 , y = 8 /48/ = 4 , /4/  /8/ = 4 x=4, y=8 /4(8)/ = 4 , /4//8/ = 48 = 4



Intern
Joined: 05 Feb 2013
Posts: 25
Location: Ukraine

Re: Hard inequality: Is xy < 0
[#permalink]
Show Tags
15 May 2013, 22:55
Bunuel wrote: Yes, OA is E.
Number plugging as AKProdigy87 did is probably the best way here. But still if needed below is short overview of algebra of this question:
(1) y<0. Not sufficient.
(2) Holds true when: A. 0<x<y B. x<0<y C. y<x<0 D. y<0<x
Not sufficient.
(1)+(2) C. and D. options are left from (2) but still insufficient, as xy may or may not be negative.
Answer: E. Hi Bunuel, Can you explain how you got the following set of inequality? : A. 0<x<y B. x<0<y C. y<x<0 D. y<0<x Thanks



Retired Moderator
Joined: 05 Jul 2006
Posts: 1723

Re: Hard inequality: Is xy < 0
[#permalink]
Show Tags
17 May 2013, 15:45
Sergiy wrote: Bunuel wrote: Yes, OA is E.
Number plugging as AKProdigy87 did is probably the best way here. But still if needed below is short overview of algebra of this question:
(1) y<0. Not sufficient.
(2) Holds true when: A. 0<x<y B. x<0<y C. y<x<0 D. y<0<x
Not sufficient.
(1)+(2) C. and D. options are left from (2) but still insufficient, as xy may or may not be negative.
Answer: E. Hi Bunuel, Can you explain how you got the following set of inequality? : A. 0<x<y B. x<0<y C. y<x<0 D. y<0<x Thanks /xy/ >= /x//y/ is always true ( try it ) and the equality (equal sign) holds in 2 cases 1) when xy> 0 ( same sign) and /x/>/y/ ( both conditions together) 2) x=y=0 Now , the inequality holds true in any other case (/xy/ > /x//y/) i.e. xy<0 ( different signs) or /x/</y/ if u translate this last bit into inequalities it gives the above 4 scenarios that i reformatted below A. xy>0 but /y/> /x/ b. xy<0 , y>x c. xy<0 and /y/ > /x/ d. xy<0 , x>y hope this helps



Senior Manager
Joined: 13 May 2013
Posts: 425

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) xy<xy
[#permalink]
Show Tags
30 Jun 2013, 12:18
Is xy<0?
Is either x or y negative?
(1)(x^3*y^5)/(x*y^2)<0
(x^3  x^1)*(y^5  y^2) < 0 (x^2)*(y^3) < 0
(x^2) is positive regardless of whether x is positive or negative. If (x^2)*(y^3) < 0 then (y^3) must be negative and because a number raised to an odd power holds its sign, y must be negative. However, we do not know whether x is positive or negative meaning ab could be (a)*(b) = ab OR (a)*(b) = ab. INSUFFICIENT
(2) xy<xy
xy<xy 3  2 < 3  (2) = 32 < 1 ===> 1<1 Invalid 3  2 < 3(2) = 32 < 5 ===> 1<5 VALID (x, y) 2  3 < 23 = 1 < 1 ===> 1<1 VALID (x, y) 3  2 < 32 = 32 < 1 ===> 1<1 Invalid 3  2 < 32 = 32 < 5 ===> 1<5 VALID (x, y) INSUFFICIENT
1+2) xy<xy
2  3 < 2(3) 23 < 5 ===> 1<5 Sufficient (x = positive, y = negative) 2  3 < 2  (3) 1 < 1 ===> 1 < 1 Sufficient (x = negative, y = negative)
There are two valid statements where x and y are negative and where x is positive and y is negative meaning we don't know whether ab is negative or positive. INSUFFICIENT
(E)



Current Student
Joined: 31 Mar 2013
Posts: 66
Location: India
GPA: 3.02

Re: Hard inequality: Is xy < 0
[#permalink]
Show Tags
06 Nov 2013, 10:41
Bunuel wrote: Yes, OA is E.
Number plugging as AKProdigy87 did is probably the best way here. But still if needed below is short overview of algebra of this question:
(1) y<0. Not sufficient.
(2) Holds true when: A. 0<x<y B. x<0<y C. y<x<0 D. y<0<x
Not sufficient.
(1)+(2) C. and D. options are left from (2) but still insufficient, as xy may or may not be negative.
Answer: E. Bunuel, I'm finding this confusing. Can you help me out. Your post on this thread isxyxy123108.html says that for statement 2 to be true both x and y need to have the same sign. Now, this question if you look at statement 2, it says that x and y have opposite signs. Hence, it is never possible for xy<xy to be true. So statement 2 seems sufficient. Where is my error in reasoning? Thank you.



Director
Joined: 23 Jan 2013
Posts: 568

Re: Hard inequality: Is xy < 0
[#permalink]
Show Tags
06 Nov 2013, 23:13
emailmkarthik wrote: Bunuel wrote: Yes, OA is E.
Number plugging as AKProdigy87 did is probably the best way here. But still if needed below is short overview of algebra of this question:
(1) y<0. Not sufficient.
(2) Holds true when: A. 0<x<y B. x<0<y C. y<x<0 D. y<0<x
Not sufficient.
(1)+(2) C. and D. options are left from (2) but still insufficient, as xy may or may not be negative.
Answer: E. Bunuel, I'm finding this confusing. Can you help me out. Your post on this thread isxyxy123108.html says that for statement 2 to be true both x and y need to have the same sign. Now, this question if you look at statement 2, it says that x and y have opposite signs. Hence, it is never possible for xy<xy to be true. So statement 2 seems sufficient. Where is my error in reasoning? Thank you. I agree. Inequality rule saying that xy<xy is true only when x and y are having different signs exists. Does not it? If so, B is correct




Re: Hard inequality: Is xy < 0 &nbs
[#permalink]
06 Nov 2013, 23:13



Go to page
1 2
Next
[ 28 posts ]



