Author 
Message 
Senior Manager
Joined: 07 Jan 2008
Posts: 385

DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 12:13
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
HideShow timer Statistics
This topic is locked. If you want to discuss this question please repost it in the respective forum.



Manager
Joined: 15 Jul 2008
Posts: 205

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 12:34
lexis wrote: Is the answer A ? st 1) seems sufficient. sq.root(x) > y tells you know two things. first, x is +ve and second that y has to be smaller than the negative root of x. So you know x is positive and y is negatvie. therefore x>y indeed. st 2 does not say anything abt +ve ve aspects of x and y. so insufficient. Hence A.



Director
Joined: 27 Jun 2008
Posts: 527
WE 1: Investment Banking  6yrs

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 12:37
1) rt x>y Square both the sides, x>y^2  x>y AD
2) x^3>y Let x = +2 , +8>y, insuff
IMO
A



SVP
Joined: 30 Apr 2008
Posts: 1855
Location: Oklahoma City
Schools: Hard Knocks

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 12:41
Agree with A. Statement 1 becomes: x > y^2, so it's sufficient. Statement 2 has way to many possibilities. x could be fraction or negative so it's insufficient. Kind of an easy problem actually. lexis wrote:
_________________
 J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
GMAT Club Premium Membership  big benefits and savings



SVP
Joined: 07 Nov 2007
Posts: 1765
Location: New York

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 12:47
lexis wrote: 1) sqrt(x)>y from this it is clear that x must be +ve.. x=4 y=1 x>y x=1/4 y=1/3 x<y insuffcient 2) x^3>y x *x^2 >y (x^2 is always +ve) x=4 y=1 x>y x=1/2 y=1/4 x<y insuffcieint combine. x must be +ve and not <1 and >0 e.g x=1/4 y=1/3 (Statement 2 fails) so x>y (any x>1 such that sqrt(x)>y) C Any other short cuts..
_________________
Your attitude determines your altitude Smiling wins more friends than frowning



Director
Joined: 12 Jul 2008
Posts: 513
Schools: Wharton

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 12:58
lexis wrote: I get C (1) Insufficient This tells us x and y are positive but nothing else. For example: x = 1/4, y = 1/4 sqrt(x) = 1/2 y = 1/4 sqrt(x) > y OR x = 4, y =1 sqrt(x) = 2 > y = 1 (2) Insufficient x^3 > y This is obviously not sufficient. (1) and (2) Sufficient Split this into two (or three) cases: x < 1: Here, (2) tells us x > y. The nth power of any positive number less than 1 is smaller than that number. So if x^3 > y, then x > y. x > 1: Here, (1) tells us x > y. The nth root of any positive number greater than 1 is smaller than that number. So if sqrt(x) > y, then x > y. x = 1: Here, either (1) or (2) is enough. x = sqrt(x) = x^3 = 1 > x > y
Last edited by zonk on 28 Aug 2008, 13:01, edited 1 time in total.



Manager
Joined: 14 Jun 2008
Posts: 162

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 13:00
lexis wrote: 1) unsuff consider, \(x = \frac{1}{4}, y = \frac{1}{3}\) \(\sqrt{x}= \frac{1}{2}\) \(\sqrt{x} > y\) but \(x < y\) 2) insuff using fractions and negatives, but i think, from (1) we get \(x > y^2\) hence \(x is +ve\) this condition along with (2) makes teh condition true. i would say (C)



Director
Joined: 12 Jul 2008
Posts: 513
Schools: Wharton

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 13:00
bhushangiri wrote: x2suresh wrote: lexis wrote: 1) sqrt(x)>y from this it is clear that x must be +ve.. x=4 y=1 x>y x=1/4 y=1/3 x<y this is where i disagree with you. if x=1/4 then sq.root(x) can be 1/2 or 1/2. For satisfying the condition sq.root(x)>y, y < 1/2. That would automatically make x>y. That is why i think A should be the answer.insuffcient 2) x^3>y x *x^2 >y (x^2 is always +ve) x=4 y=1 x>y x=1/2 y=1/4 x<y insuffcieint combine. x must be +ve and not <1 and >0 e.g x=1/4 y=1/3 (Statement 2 fails) so x>y (any x>1 such that sqrt(x)>y) C Any other short cuts.. In GMAT universe, square roots are always positive.



Manager
Joined: 15 Jul 2008
Posts: 205

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 13:02
x2suresh wrote: You forgot the fraction values..
Not quite. I dont need fraction values here. we know x > 0. Taking your example, if x=1/4 then sqrt(x) = 1/2 or 1/2. To satisfy the statement 1 sqrt(x) > y, you need to account for all possible values of sqrt(x) and for that y has to be smaller than 1/2. So you have +ve x and ve y. Thus sufficient.



Manager
Joined: 15 Jul 2008
Posts: 205

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 13:03
zoinnk wrote: In GMAT universe, square roots are always positive.
No. In GMAT universe, squares are always positive. Square roots can be ve.



Director
Joined: 12 Jul 2008
Posts: 513
Schools: Wharton

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 13:05
bhushangiri wrote: zoinnk wrote: In GMAT universe, square roots are always positive.
No. In GMAT universe, squares are always positive. Square roots can be ve. No, that is incorrect. Square roots are always positive in GMAT universe. I've read this several times in OG and MGMAT guides.



Manager
Joined: 14 Jun 2008
Posts: 162

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 13:07
bhushangiri wrote: zoinnk wrote: In GMAT universe, square roots are always positive.
No. In GMAT universe, squares are always positive. Square roots can be ve. NEVER, square roots are +ve by definition. pg 114 OG 11



SVP
Joined: 07 Nov 2007
Posts: 1765
Location: New York

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 13:08
bhushangiri wrote: x2suresh wrote: You forgot the fraction values..
Not quite. I dont need fraction values here. we know x > 0. Taking your example, if x=1/4 then sqrt(x) = 1/2 or 1/2. To satisfy the statement 1 sqrt(x) > y, you need to account for all possible values of sqrt(x) and for that y has to be smaller than 1/2. So you have +ve x and ve y. Thus sufficient. Bushan, Let me put this way. Here given equation is sqrt(x)>yI pick value for sqrt(x)=1/2 and y=1/3 such that it satisfies sqrt(x)>y Now I will find the value for x = sqrt(x)*sqrt(x) =1/4 y=1/3 x<y
_________________
Your attitude determines your altitude Smiling wins more friends than frowning



Manager
Joined: 15 Jul 2008
Posts: 205

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 13:29
x2suresh wrote:
Here given equation is sqrt(x)>y I pick value for sqrt(x)=1/2 and y=1/3 such that it satisfies sqrt(x)>y If you pick sqrt(x)=1/2, then by default you have to consider 1/2. It is the charecteristic property of the square root function that it will have two values a +ve and a ve. You can't consider the +ve alone unless it is categoricaly mentioned. So if you have a situation where sqrt(x) > y, then y has to be less than the ve root. But here i guess the answer will be C. Zoink and Set009 are so sure about the GMAT rule for roots. Now I will find the value for x = sqrt(x)*sqrt(x) =1/4 y=1/3 x<y Zoink and Sset009.. thanx for bringing that up.
Last edited by bhushangiri on 28 Aug 2008, 13:35, edited 1 time in total.



SVP
Joined: 30 Apr 2008
Posts: 1855
Location: Oklahoma City
Schools: Hard Knocks

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 13:30
When someone pulls out the page number from the OG11, that's like quoting Jesus from the Bible. It's heresy to claim otherwise bhushangiri wrote: zoinnk wrote: In GMAT universe, square roots are always positive.
No. In GMAT universe, squares are always positive. Square roots can be ve.
_________________
 J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
GMAT Club Premium Membership  big benefits and savings



Manager
Joined: 15 Jul 2008
Posts: 205

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 13:34
jallenmorris wrote: When someone pulls out the page number from the OG11, that's like quoting Jesus from the Bible. It's heresy to claim otherwise i am not disputing that. Outside of GMAT universe however, i stand by my explanation. But that is for another forum another place and another time.



GMAT Instructor
Joined: 04 Jul 2006
Posts: 1254
Location: Madrid

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 15:15
lexis wrote: (1) If sqrt(x) > y, x > y if x is greater than or equal to 1, since for such values of x, x is greater than or equal to sqrt(x) . If 0< x < 1, sqrt(x) > x, so the relationship between x and y cannot be determined. (2) If x^3 > y, x > y if 0 < x < 1, since for such values of x , x > x^3. However, if x > 1, x < x^3, so the relationship between x and y is unclear. (T) For x not be to greater than y, (1) tells us that 0<x<1, but (2) tells us that x is greater than 1, a contradiction. Thus x must be greater than y.



GMAT Tutor
Joined: 24 Jun 2008
Posts: 1346

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
28 Aug 2008, 16:42
sset009 wrote: bhushangiri wrote: zoinnk wrote: In GMAT universe, square roots are always positive.
No. In GMAT universe, squares are always positive. Square roots can be ve. NEVER, square roots are +ve by definition. pg 114 OG 11 The OG says the exact opposite of that. If you read that page, it says "Every positive number n has two square roots, one positive and the other negative." It is only when dealing with the square root symbol (radicals) that you can never have a negative result: \(\sqrt{x} \geq 0\) for all x, because, by definition \(\sqrt{x}\) is equal to the nonnegative square root of x only. Note that radicals are not always positive; they can be zero.
_________________
GMAT Tutor in Toronto
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com



Senior Manager
Joined: 07 Jan 2008
Posts: 385

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
29 Aug 2008, 08:23
\(Green: x^3>y Blue: sqrt{x}>y Pink: x>y\)



SVP
Joined: 07 Nov 2007
Posts: 1765
Location: New York

Re: DS: Inequality, 700800 Ranking [#permalink]
Show Tags
29 Aug 2008, 08:32
lexis wrote: \(Green: x^3>y Blue: sqrt{x}>y Pink: x>y\) NICE FIGURE.. Good job.
_________________
Your attitude determines your altitude Smiling wins more friends than frowning




Re: DS: Inequality, 700800 Ranking
[#permalink]
29 Aug 2008, 08:32



Go to page
1 2
Next
[ 26 posts ]



