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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] New post 23 Jun 2006, 07:17
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E

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Question Stats:

59% (02:21) correct 41% (01:13) wrong based on 50 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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Re: GMATprep DS [#permalink] New post 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
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 [#permalink] New post 23 Jun 2006, 07:46
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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 [#permalink] New post 23 Jun 2006, 08:25
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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

Different answers ... not sufficient.

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.

Answer : E
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 [#permalink] New post 23 Jun 2006, 12:20
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... :x
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Re: If x not equal to -y is (x-y)/(x+y) >1? (1) x > 0 (2) [#permalink] New post 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] New post 26 Apr 2012, 20:23
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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.

Answer: E.

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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