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if x is different from -y, in (x-y)/(x+y) greater than 1 ?

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if x is different from -y, in (x-y)/(x+y) greater than 1 ? [#permalink] New post 14 Apr 2011, 01:55
00:00
A
B
C
D
E

Difficulty:

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Question Stats:

63% (02:10) correct 37% (00:55) wrong based on 27 sessions
if x is different from -y, in (x-y)/(x+y) greater than 1 ?
1. x > 0
2. y< 0
[Reveal] Spoiler: OA
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Re: Did Sal had this question wrong ? [#permalink] New post 14 Apr 2011, 03:10
whichscore wrote:
if x is different from -y, in (x-y)/(x+y) greater than 1 ?
1. x > 0
2. y< 0


\frac{x-y}{x+y}>1
\frac{x-y}{x+y}-1>0
\frac{x-y-x-y}{x+y}>0
\frac{-2y}{x+y}>0
-\frac{y}{x+y}>0
\frac{y}{x+y}<0

Case I:
If y<0 or -y>0 -----A
Then, x+y>0
x>-y-----B
Combining A and B:
x>-y>0

Case II:
If y>0 or -y<0 -----C
Then, x+y<0
x<-y-----D
Combining C and D:
x<-y<0

1. x > 0
CaseI:
x>0
Is -y between 0 and x

If
x=5
y=-3
-y=-(-3)=3 will be between 0 and x(5) and the expression will be true.

If
x=5
y=-7
-y=-(-7)=7 will NOT be between 0 and x(5) and the expression will be false.
Not Sufficient.

2. y < 0
CaseI:
x>0
Is -y between 0 and x

If
x=5
y=-3; Here y<0
-y=-(-3)=3 will be between 0 and x(5) and the expression will be true.

If
x=5
y=-7; Here y<0
-y=-(-7)=7 will NOT be between 0 and x(5) and the expression will be false.
Not Sufficient.

Combing both; we use the same sample set:
x=5
y=-3
AND
x=5
y=-7
To prove its insufficiency.

Ans: "E"
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Re: Did Sal had this question wrong ? [#permalink] New post 14 Apr 2011, 04:11
I plugged numbers to get the answer :

A per (1)

x = 5, y = 5

then 0/10 < 1

x = 5, y = -1

6/4 > 1

So insufficient

As per (2)

x = 6, y = -2

8/4 > 1

x = 2 , y = -3

5/-1 < 1

So insufficient

As evident above, (1) and (2) together are also insufficient

Answer - E
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Re: Did Sal had this question wrong ? [#permalink] New post 14 Apr 2011, 17:18
Expert's post
whichscore wrote:
if x is different from -y, in (x-y)/(x+y) greater than 1 ?
1. x > 0
2. y< 0


Even though fluke has already provided the detailed solution, there are a couple of points I would like to reinforce here.

People often get confused when they read 'if x is different from -y'. Why do they give this information? They do that because you have (x+y) in the denominator and hence it cannot be 0. That is, x + y = 0 or x = -y is not valid. Hence, they are just clarifying that the fraction is indeed defined.

Also, we are used to having 0 on the right of an inequality. What do we do when we have a 1? We take the 1 to the left hand side and get a 0 on the right.
Is \frac{(x-y)}{(x+y)} > 1? (Don't forget it is a question, not given information)
Is \frac{(x-y)}{(x+y)} - 1 > 0?
which simplifies to: Is y/(x+y) < 0 ?

We know how to deal with this inequality! Note here that we have no information about x and y as yet (except that x is not equal to -y which is more of a technical issue rather than actual information)

1. x > 0
No information about y so not sufficient.
2. y< 0
No information about x so not sufficient.

Both together, we know that y is negative. We need the sign of (x+y) now. But (x+y) may be positive or negative depending on whether x or y has greater absolute value. Hence not sufficient.
Answer (E)
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Re: Did Sal had this question wrong ? [#permalink] New post 14 Apr 2011, 21:31
1) + 2) When x > 0 and y < 0, x- y will always be +ve but nothing can be concluded about (x + y). y can be very small -Infinity or y can be close to zero. Hence sign of x - y is unknown. Hence the division i.e x-y / (x + y) may or may not be greater than 1. I don't think any calculation is required in this problem.
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Re: Did Sal had this question wrong ? [#permalink] New post 17 Apr 2011, 04:55
fluke wrote:
whichscore wrote:
if x is different from -y, in (x-y)/(x+y) greater than 1 ?
1. x > 0
2. y< 0


\frac{x-y}{x+y}>1
\frac{x-y}{x+y}-1>0
\frac{x-y-x-y}{x+y}>0
\frac{-2y}{x+y}>0
-\frac{y}{x+y}>0
\frac{y}{x+y}<0

Case I:
If y<0 or -y>0 -----A
Then, x+y>0
x>-y-----B
Combining A and B:
x>-y>0

Case II:
If y>0 or -y<0 -----C
Then, x+y<0
x<-y-----D
Combining C and D:
x<-y<0

1. x > 0
CaseI:
x>0
Is -y between 0 and x

If
x=5
y=-3
-y=-(-3)=3 will be between 0 and x(5) and the expression will be true.

If
x=5
y=-7
-y=-(-7)=7 will NOT be between 0 and x(5) and the expression will be false.
Not Sufficient.

2. y < 0
CaseI:
x>0
Is -y between 0 and x

If
x=5
y=-3; Here y<0
-y=-(-3)=3 will be between 0 and x(5) and the expression will be true.

If
x=5
y=-7; Here y<0
-y=-(-7)=7 will NOT be between 0 and x(5) and the expression will be false.
Not Sufficient.

Combing both; we use the same sample set:
x=5
y=-3
AND
x=5
y=-7
To prove its insufficiency.

Ans: "E"


Dear Fluke
i had below approach
please correct me if i am wrong

the term simplifies to -\frac{y}{x+y}>0

Case I
Numerator and denominator both are positive
-y > 0
y < 0
and
x + y > 0
x > -y
but since -y>0 so x also also be >0

Case 2
Numerator and denominator both are negative

-y < 0
y > 0
and
x + y < 0
x < -y
but since -y<0 so x will also be < 0

now Clearly none of the statement alone is sufficient
and taking them together gives us only Case I
but not takes care about the case 2
so answer is E :) :)
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Re: Did Sal had this question wrong ? [#permalink] New post 19 May 2011, 22:45
by resolving -

x/ (x+y) > 1

a + b, x>0 y<0

x=2,y= -1 LHS > RHS

x=1,y =-2 LHS < RHS

E
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Re: Did Sal had this question wrong ?   [#permalink] 19 May 2011, 22:45
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