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# DS: Inequality, 700-800 Ranking

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28 Aug 2008, 12:13
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Re: DS: Inequality, 700-800 Ranking [#permalink]

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28 Aug 2008, 12:34
lexis wrote:

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.

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Re: DS: Inequality, 700-800 Ranking [#permalink]

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28 Aug 2008, 12:37
1) rt x>y
Square both the sides,

2) x^3>y
Let x = -+2 , -+8>y, insuff

IMO

A

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Re: DS: Inequality, 700-800 Ranking [#permalink]

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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:

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**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Kudos [?]: 607 [0], given: 32 SVP Joined: 07 Nov 2007 Posts: 1795 Kudos [?]: 1037 [0], given: 5 Location: New York Re: DS: Inequality, 700-800 Ranking [#permalink] ### Show Tags 28 Aug 2008, 12:47 lexis wrote: 1) sqrt(x)>y from this it is clear that x must be +ve.. x=4 y=1 x>y x=1/4 y=1/3 x<y insuffcient 2) x^3>y x *x^2 >y (x^2 is always +ve) x=4 y=1 x>y x=-1/2 y=-1/4 x<y insuffcieint combine. x must be +ve and not <1 and >0 e.g x=1/4 y=1/3 (Statement 2 fails) so x>y (any x>1 such that sqrt(x)>y) C Any other short cuts.. _________________ Your attitude determines your altitude Smiling wins more friends than frowning Kudos [?]: 1037 [0], given: 5 Director Joined: 12 Jul 2008 Posts: 514 Kudos [?]: 162 [0], given: 0 Schools: Wharton Re: DS: Inequality, 700-800 Ranking [#permalink] ### Show Tags 28 Aug 2008, 12:58 lexis wrote: I get C (1) Insufficient This tells us x and y are positive but nothing else. For example: x = 1/4, y = 1/4 sqrt(x) = 1/2 y = 1/4 sqrt(x) > y OR x = 4, y =1 sqrt(x) = 2 > y = 1 (2) Insufficient x^3 > y This is obviously not sufficient. (1) and (2) Sufficient Split this into two (or three) cases: x < 1: Here, (2) tells us x > y. The nth power of any positive number less than 1 is smaller than that number. So if x^3 > y, then x > y. x > 1: Here, (1) tells us x > y. The nth root of any positive number greater than 1 is smaller than that number. So if sqrt(x) > y, then x > y. x = 1: Here, either (1) or (2) is enough. x = sqrt(x) = x^3 = 1 --> x > y Last edited by zonk on 28 Aug 2008, 13:01, edited 1 time in total. Kudos [?]: 162 [0], given: 0 Manager Joined: 14 Jun 2008 Posts: 162 Kudos [?]: 36 [0], given: 0 Re: DS: Inequality, 700-800 Ranking [#permalink] ### Show Tags 28 Aug 2008, 13:00 lexis wrote: 1) unsuff consider, $$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) Kudos [?]: 36 [0], given: 0 Director Joined: 12 Jul 2008 Posts: 514 Kudos [?]: 162 [0], given: 0 Schools: Wharton Re: DS: Inequality, 700-800 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. Kudos [?]: 162 [0], given: 0 Manager Joined: 15 Jul 2008 Posts: 206 Kudos [?]: 66 [0], given: 0 Re: DS: Inequality, 700-800 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. Kudos [?]: 66 [0], given: 0 Manager Joined: 15 Jul 2008 Posts: 206 Kudos [?]: 66 [0], given: 0 Re: DS: Inequality, 700-800 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. Kudos [?]: 66 [0], given: 0 Director Joined: 12 Jul 2008 Posts: 514 Kudos [?]: 162 [0], given: 0 Schools: Wharton Re: DS: Inequality, 700-800 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. Kudos [?]: 162 [0], given: 0 Manager Joined: 14 Jun 2008 Posts: 162 Kudos [?]: 36 [0], given: 0 Re: DS: Inequality, 700-800 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 Kudos [?]: 36 [0], given: 0 SVP Joined: 07 Nov 2007 Posts: 1795 Kudos [?]: 1037 [0], given: 5 Location: New York Re: DS: Inequality, 700-800 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)>y I 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 Kudos [?]: 1037 [0], given: 5 Manager Joined: 15 Jul 2008 Posts: 206 Kudos [?]: 66 [0], given: 0 Re: DS: Inequality, 700-800 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. Kudos [?]: 66 [0], given: 0 SVP Joined: 30 Apr 2008 Posts: 1870 Kudos [?]: 607 [0], given: 32 Location: Oklahoma City Schools: Hard Knocks Re: DS: Inequality, 700-800 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$$.

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Re: DS: Inequality, 700-800 Ranking [#permalink]

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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.

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Re: DS: Inequality, 700-800 Ranking [#permalink]

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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.

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Re: DS: Inequality, 700-800 Ranking [#permalink]

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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 non-negative square root of x only.

Note that radicals are not always positive; they can be zero.
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Re: DS: Inequality, 700-800 Ranking [#permalink]

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29 Aug 2008, 08:23

$$Green: x^3>y Blue: sqrt{x}>y Pink: x>y$$

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Re: DS: Inequality, 700-800 Ranking [#permalink]

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29 Aug 2008, 08:32
lexis wrote:

$$Green: x^3>y Blue: sqrt{x}>y Pink: x>y$$

NICE FIGURE.. Good job.
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Re: DS: Inequality, 700-800 Ranking   [#permalink] 29 Aug 2008, 08:32

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