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DS: Inequality, 700-800 Ranking

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

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

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New post 28 Aug 2008, 12:34
lexis wrote:
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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.

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

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New post 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

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

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

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New post 28 Aug 2008, 12:47
lexis wrote:
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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..
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Re: DS: Inequality, 700-800 Ranking [#permalink]

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New post 28 Aug 2008, 12:58
lexis wrote:
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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.

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

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New post 28 Aug 2008, 13:00
lexis wrote:
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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)

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

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New post 28 Aug 2008, 13:00
bhushangiri wrote:
x2suresh wrote:
lexis wrote:
Image


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.

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

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New post 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.

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

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New post 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.

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

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New post 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.

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

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New post 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

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

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

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New post 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.

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

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New post 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 :shock:

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.

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

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New post 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 :shock:



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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New post 28 Aug 2008, 15:15
lexis wrote:
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(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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New post 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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New post 29 Aug 2008, 08:23
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\(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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New post 29 Aug 2008, 08:32
lexis wrote:
Image
\(Green: x^3>y

Blue: sqrt{x}>y


Pink: x>y\)



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

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