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If x y, is (x-y)/(x+y) > 1? 1.x>0 2.y<0

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If x y, is (x-y)/(x+y) > 1? 1.x>0 2.y<0 [#permalink]

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New post 06 Oct 2008, 18:10
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A
B
C
D
E

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If x≠y, is (x-y)/(x+y) > 1?

1.x>0
2.y<0

Kudos [?]: 127 [0], given: 7

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Kudos [?]: 127 [0], given: 7

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Re: DS - OG 11E Q139 [#permalink]

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New post 06 Oct 2008, 18:12
I picked up B - becuase i solved the equation, x cancels out and we are left with y<0....!!!

Where i am wrong? :?: :?: :!: :!: :evil: :evil: :evil:

Last edited by GODSPEED on 09 Oct 2008, 03:14, edited 1 time in total.

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Re: DS - OG 11E Q139 [#permalink]

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New post 07 Oct 2008, 00:10
GODSPEED wrote:
I picked up B - becuase i solved the equation, x cancels out and we are left with y<0....!!!

Where am i wrong? :?: :?: :!: :!: :evil: :evil: :evil:



In equality, this cannot be done unless we are absolutely sure that both denominator and nominator are positive. For example, -5/-2 > 1 and hence we cannot say that -5 > -2.

In order to know whether (x-y)/(x+y) > 1, we need to know that both numerator and denominator are of the same sign and that numerator is greater than denominator.

Stmt1 says that x is positive, but no clue about y. Hence, insufficient.

Stmt2 says y is negative but again no clue about x. Hence, insufficient.

Combining the two, (x-y) is positive. But, again not sure whether (x+y) is positive or negative. Hence, E.

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Re: DS - OG 11E Q139 [#permalink]

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New post 07 Oct 2008, 02:04
My two cents:

1 Insuff. becasue if y=0, the sentence is not correct.

2 Insuff. (x+|y|)/(x-|y|)>1? no if x<0

1+2
(x+|y|)/(x-|y|)>1 always becasue x>0

Therefore, C

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Re: DS - OG 11E Q139 [#permalink]

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New post 07 Oct 2008, 03:48
GODSPEED wrote:
If x≠y, is (x-y)/(x+y) > 1?

1.x>0
2.y<0


rephrase question

x^2 - y^2 / x+y > 0

from 1

x is +ve however nothing is mentioned about / y/ value...insuff

from 2

/x/ is not mentioned...insuff

both

E

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Re: DS - OG 11E Q139 [#permalink]

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New post 07 Oct 2008, 05:59
E. We must know if x>y.

1. x>0, so we have x-y/x+y. If y>x and y<0, x-y<0 and x-y/x+y < 0.
2. y<0, so we have x-(-y)/x+(-y), so x+y/x-y, if x<y x-y<0

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Re: DS - OG 11E Q139 [#permalink]

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New post 07 Oct 2008, 07:31
yezz wrote:
GODSPEED wrote:
If x≠y, is (x-y)/(x+y) > 1?

1.x>0
2.y<0


rephrase question

x^2 - y^2 / x+y > 0

from 1

x is +ve however nothing is mentioned about / y/ value...insuff

from 2

/x/ is not mentioned...insuff

both

E


Yezz.. Can you explain how you rephrased the equation ??
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Re: DS - OG 11E Q139 [#permalink]

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New post 07 Oct 2008, 16:18
amitdgr wrote:
yezz wrote:
GODSPEED wrote:
If x≠y, is (x-y)/(x+y) > 1?

1.x>0
2.y<0


Yezz.. Can you explain how you rephrased the equation ??


Apologies i did a mistake :

(x-y)/(x+y) > 1 = (x-y)/(x+y) - 1 = (x-y)/(x+y) - (x+y)/(x+y) = (x-y) - (x+y) / x+y = -2y/x+y
question becomes

is -2y/ x+y > 0 ( y +ve and x is -ve, /x/>/y/) or ( y -ve and x =ve , /x/>/y/)

from 1

x>o no mentioning of y ......insuff

from 2

y < 0 , no mentioning of x......insuff

both

x +ve , y -ve however no info about /x/, and that of y..........E

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Re: DS - OG 11E Q139 [#permalink]

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New post 09 Oct 2008, 03:15
OA is E

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Re: DS - OG 11E Q139 [#permalink]

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New post 09 Oct 2008, 03:44
scthakur wrote:
GODSPEED wrote:
I picked up B - becuase i solved the equation, x cancels out and we are left with y<0....!!!

Where am i wrong? :?: :?: :!: :!: :evil: :evil: :evil:



In equality, this cannot be done unless we are absolutely sure that both denominator and nominator are positive. For example, -5/-2 > 1 and hence we cannot say that -5 > -2.

In order to know whether (x-y)/(x+y) > 1, we need to know that both numerator and denominator are of the same sign and that numerator is greater than denominator.

Stmt1 says that x is positive, but no clue about y. Hence, insufficient.

Stmt2 says y is negative but again no clue about x. Hence, insufficient.

Combining the two, (x-y) is positive. But, again not sure whether (x+y) is positive or negative. Hence, E.


One thing i dont understand is when we solve (x-y)/(x+y)>1 then we get -2y>0 => y needs to be less than 0 and no dependency on x
can u explain this thing .can we reduce the given equation or use the original one!!!
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Re: DS - OG 11E Q139   [#permalink] 09 Oct 2008, 03:44
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