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

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Director
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If x#-y is (x-y)/(x+y)>1? (1) x>0 (2) y<0 [#permalink]  23 Jun 2006, 07:17
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Question Stats:

53% (02:20) correct 47% (01:07) wrong based on 58 sessions
If x#-y is (x-y)/(x+y)>1?

(1) x>0
(2) y<0

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-y-is-x-y-x-y-1-1-x-0-2-y-96090.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 26 Apr 2012, 20:24, edited 2 times in total.
Edited the question and added the OA. Topic is locked.
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VP
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Re: GMATprep DS [#permalink]  23 Jun 2006, 07:33
gmatmba wrote:
If x not equal to -y
is (x-y)/(x+y) >1?

(1) x > 0
(2) y < 0

E)

for solving this we need to establish if |x|>|y|
neither of the conditions provides an answer
consider:
a)x=2;y=-3
2+3/2-3=-5<1
b)x=3;y=-2
3+2/3-2=5>1
Senior Manager
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E

X+Y # 0

is x-y/x+y>1

Let's analyze different scenerio,
1) x+, y+
2) x-, y-
3) x+, y-
4) x-, y+

for (1) and (2) |x-y|<|x+y| => |x-y|/|x+y| < 1 it can be positive or negative depending on x>y or x<y

Look at example:
for (1)
(1, 2) =x-y/x+y = -1/3
(2, 1) =x-y/x+y = 1/3

for (2)
(-1, -2) =x-y/x+y = -1/3
(-2, -1) =x-y/x+y = 1/3

For (3) and (4) |x-y|>|x+y| => |x-y|/|x+y| > 1

if x+ and y-
(1,-2) =x-y/x+y = -3
(2,-1) =x-y/x+y = +3

if x- and y+
(-1,2) =x-y/x+y = -3
(-2,1) =x-y/x+y = 3

A) says x>0 does not tell us anything about Y insufficient
B) says y<0 does not tell us anything about X insufficient

togather x+ and y- scnerio 3, it can be positve or negative so insufficient!

Hence E
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KUDOS
This Q had me thoroughly confused.. but I finally got it.

Q : (x-y)/(x+y) > 1 i.e (x-y) > (x+y)

S1: x > 0
Let x = 4, y = -7 then 11 >-3 True.
Let x = 4, y = -1 then 3 > 5 False

S2: y < 0
Using the same values as above, different answers. ... Not sufficient.

S1 & S2:
Same set of values as above provide different answers
Therefore Not sufficient.

Director
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chiragr wrote:
E

X+Y # 0

is x-y/x+y>1

Let's analyze different scenerio,
1) x+, y+
2) x-, y-
3) x+, y-
4) x-, y+

for (1) and (2) |x-y|<|x+y| => |x-y|/|x+y| < 1 it can be positive or negative depending on x>y or x<y

Look at example:
for (1)
(1, 2) =x-y/x+y = -1/3
(2, 1) =x-y/x+y = 1/3

for (2)
(-1, -2) =x-y/x+y = -1/3
(-2, -1) =x-y/x+y = 1/3

For (3) and (4) |x-y|>|x+y| => |x-y|/|x+y| > 1

if x+ and y-
(1,-2) =x-y/x+y = -3
(2,-1) =x-y/x+y = +3

if x- and y+
(-1,2) =x-y/x+y = -3
(-2,1) =x-y/x+y = 3

A) says x>0 does not tell us anything about Y insufficient
B) says y<0 does not tell us anything about X insufficient

togather x+ and y- scnerio 3, it can be positve or negative so insufficient!

Hence E

Thats great explanation...thanks Chirag and hass_mba
How long did it take you guys to do it?
I couldn't do this on the test due to time constraints and got it wrong...
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Re: If x not equal to -y is (x-y)/(x+y) >1? (1) x > 0 (2) [#permalink]  26 Apr 2012, 19:26
if you reorganize the question it becomes x - y < x + y ==> 0 < 2y ==> 0 < y.... Why is this logic not possible?
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Re: If x not equal to -y is (x-y)/(x+y) >1? (1) x > 0 (2) [#permalink]  26 Apr 2012, 20:23
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Expert's post
calreg11 wrote:
if you reorganize the question it becomes x - y < x + y ==> 0 < 2y ==> 0 < y.... Why is this logic not possible?

Given: $$\frac{x-y}{x+y}>1$$. When you are then writing $$x-y>x+y$$, you are actually multiplying both sides of inequality by $$x+y$$: never multiply an inequality by variable (or expression with variable) unless you know the sign of variable (or expression with variable). Because if $$x+y>0$$ you should write $$x-y>x+y$$ BUT if $$x+y<0$$, you should write $$x-y<x+y$$ (flip the sign when multiplying by negative expression).

COMPLETE SOLUTION:

If $$x\neq{-y}$$ is $$\frac{x-y}{x+y}>1$$?

Is $$\frac{x-y}{x+y}>1$$? --> Is $$0>1-\frac{x-y}{x+y}$$? --> Is $$0>\frac{x+y-x+y}{x+y}$$? --> Is $$0>\frac{2y}{x+y}$$?

(1) $$x>0$$ --> Not sufficient.

(2) $$y<0$$ --> Not sufficient.

(1)+(2) $$x>0$$ and $$y<0$$ --> numerator (y) is negative, but we can not say whether the denominator {positive (x)+negative (y)} is positive or negative. Not sufficient.

In case of any question please post it here: if-x-y-is-x-y-x-y-1-1-x-0-2-y-96090.html
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Re: If x not equal to -y is (x-y)/(x+y) >1? (1) x > 0 (2)   [#permalink] 26 Apr 2012, 20:23
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