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If x not equal to y , is x-y/x+y > 1 ? 1. x > 0 2. y

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SVP
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If x not equal to y , is x-y/x+y > 1 ? 1. x > 0 2. y [#permalink] New post 09 Aug 2006, 12:57
If x not equal to y , is x-y/x+y > 1 ?
1. x > 0
2. y < 0

rephrasing the question is x-y/x+y > 1 ie : is x-y > x+y ps: ( x+ y can not be zero)

x-y-x-y > 0 ie: is y > zero

Clearly the answer is b

am i right ????
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 [#permalink] New post 09 Aug 2006, 13:11
Looking at each statement individually:

1) X>0 - the function can be greater or less then 1 depending on what Y is - Insuff

2) Y<0 - the function can be greater or less then 1 depending on what X is - Insuff

1&2) if X>0 and Y<0 function will always be greater then 1

C

Yezz,

You can't multiply x+y to both sides because if x+y is negative you have to change the inequality sign.
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Re: 700+ INIQUALITIES ( HELP NEEDED) [#permalink] New post 09 Aug 2006, 13:16
yezz wrote:

rephrasing the question is x-y/x+y > 1 ie : is x-y > x+y ps: ( x+ y can not be zero)

This is wrong. We don't know the sign of (x+y) so you can not multiply. If x+y is -ve then the sign of equality should be reversed.

Answer should be E.

Here is the explanation:

Question:Is (x-y)/(x+y) >1 ?

Lets draw a matrix
x-------y-------------answer to above question
pos---pos------------------always less than 1.
pos---neg------------------may be less or greater depending on values of x and y
neg---pos------------------may be less or greater depending on values of x and y
neg---neg----------------- always less than 1.

St1: x>0: Look at choices 1 and 2: INSUFF

St2: y<0: Look at choices 2 and 4: INSUFF

Combined: x> 0 and y <0: Look at choice 2.: INSUFF
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Senior Manager
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 [#permalink] New post 09 Aug 2006, 13:42
My bad

PS, you're right. If both X>0 and y<0, function could be greater or less then 1 depending on what values you use.

E it is.
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 [#permalink] New post 09 Aug 2006, 13:45
Thanks folks for your time to explain the issue to me .... you ve been of great help.

I have seen where my logic went wrong.

I clearly need more practise .

:lol:

Take care
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Re: 700+ INIQUALITIES ( HELP NEEDED) [#permalink] New post 09 Aug 2006, 13:53
yezz wrote:
If x not equal to y , is x-y/x+y > 1 ?
1. x > 0
2. y < 0

rephrasing the question is x-y/x+y > 1 ie : is x-y > x+y ps: (x+y cannot be zero)
x-y-x-y > 0 ie: is y > zero
Clearly the answer is b. am i right ????


first, the red part above is not correct. we cannot do that without knowing the valaues for (x-y) and (x+y). if they are +ves, we cannot do that. if (x-y) = 5 and (x+y) =7 and if we do so, 5>7 and this is not true.

using 1 and 2, we cannot say that (x-y)/(x+y) >1.

from i, x-y is +ve.
from ii, x+y can be -ve or +ve. if it is +ve, it is <(x-y) and the answer is yes. if it is -ve, it is again <(x-y) but (x-y)/(x+y) < 1 and the answer is no.

so it is E.
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Re: 700+ INIQUALITIES ( HELP NEEDED) [#permalink] New post 09 Aug 2006, 17:06
yezz wrote:
If x not equal to y , is x-y/x+y > 1 ?
1. x > 0
2. y < 0

rephrasing the question is x-y/x+y > 1 ie : is x-y > x+y ps: ( x+ y can not be zero)

x-y-x-y > 0 ie: is y > zero

Clearly the answer is b

am i right ????


x,y={ I,+,-,0,+-1/2, +-sqrt(2), ==}

(1) x=3,y= -2 Yes
x=3,y=1 No
BCE

(2) x=3,y= -2 Yes
x= 0, y=-2 No

CE

(1)&(2)
x=3,y= -2 Yes
x=1/2 , y= -1/4 Yes
x= 1/4, y= -1/2 No

Hence E

Heman

Last edited by heman on 14 Aug 2006, 18:53, edited 1 time in total.
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 [#permalink] New post 11 Aug 2006, 02:20
Clear E.

1) x>0

When x = 3 and 2
3-2/3+2 = 1/5 <1

Not Suff

2) y<0

When x = 2 and y = -3
2+3/2-3 = -5< 1
And
When x = 3 and y = -2
3+2/3-2 = 5 >1
Not suff.

Together.
Same as 2
Manager
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 [#permalink] New post 11 Aug 2006, 08:00
Quote:
Clear E.

1) x>0

When x = 3 and 2
3-2/3+2 = 1/5 <1

Not Suff

2) y<0

When x = 2 and y = -3
2+3/2-3 = -5< 1
And
When x = 3 and y = -2
3+2/3-2 = 5 >1
Not suff.

Together.
Same as 2



Well said. Easy E.
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 [#permalink] New post 12 Aug 2006, 20:02
so the approach here is just plugging numbers till you disprove one option after another -- is it?

I got the answer E but it took 3-4 mins.
  [#permalink] 12 Aug 2006, 20:02
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