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

(1) x>0
(2) y<0
[Reveal] Spoiler: OA
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Re: gmat prep DS [#permalink]

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gmatcracker2010 wrote:
Please explain the attached problem.

My take on it:

is (X+y)/ (x-y) > 1

ie. x + y > x -y
ie. 0 > y

which is B but OA is E.


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 cannot say whether the denominator {positive (x)+negative (y)} is positive or negative. Not sufficient.

Answer: E.

The problem with your solution is that 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).

So again: given inequality can be simplified as follows: \(\frac{x-y}{x+y}>1\) --> \(0>1-\frac{x-y}{x+y}<0\) --> \(0>\frac{x+y-x+y}{x+y}\) --> \(0>\frac{2y}{x+y}\) --> we can drop 2 and finally we'll get: \(0>\frac{y}{x+y}\).

Now, numerator is negative (\(y<0\)), but we don't know about the denominator, as \(x>0\) and \(y<0\) can not help us to determine the sign of \(x+y\). So the answer is E.

Hope it helps.
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Re: GMATPrep DS question [#permalink]

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New post 01 Aug 2010, 15:03
Hi,

twi ways to approach:
1) Check it out by plugging in numbers:
a) x < y: makes the numerator negative, so doesn't work
b) x = y: creates a 0, so doesn't work
c) x > y: makes the denominator greater than the numerator, so doesn't work
d) y < 0: makes numerator greater than denominator, so WORKS

2) solve the equation:
(x-y)/(x+y)>1
=> x-y > x + y
=> x-x > y + y
=> 0 > 2 y
=> y must be negative

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Re: GMATPrep DS question [#permalink]

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New post 01 Aug 2010, 15:28
The OA is E. I thought it would be B. I think the catch is that x cannot be cancelled from both sides...

If x - y > x + y
Cancel x

y < 0
Should be B. OA says E.
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Re: GMATPrep DS question [#permalink]

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New post 01 Aug 2010, 15:37
\(\frac{x-y}{x+y}>1\)
Let us make cross multiplication with respect to the sign of x+y.
- Case 1:
x-y>x+y>0
or x+y>0 and y<0
- Case 2:
x-y<x+y<0
or x+y<0 and y>0

Both statement 1 and 2 together cannot say about the sign of x+y, thereofore correct answer is E.
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Re: GMATPrep DS question [#permalink]

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New post 01 Aug 2010, 15:56
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But Bogos, what is wrong in just cancelling x from both sides of
x-y > x+y?

You mentioned about w.r.t the sign of x+y, why is that important?
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Re: GMATPrep DS question [#permalink]

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New post 03 Aug 2010, 10:22
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bogos wrote:
You can refer to this review (3rd point):
inequalities-review-69016.html


Bogos
I read about inequalities and it says that one can subtract the same amout from both sides
so if x-y/x+y > 1 => x-y > x+y => y<0

where is the flaw?
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Re: GMATPrep - DS - Inequalities [#permalink]

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thirst4edu wrote:
Rephrasing the question
\(\frac{(x-y)}{(x+y)} > 1\)
\((x-y) > (x+y)\) , Since x <> -y, we can multiply (x+y) both sides
Adding -x both sides
-y > y
Adding y both sides
0 > y , which is y < 0, given by statement 2, so why Answer is E(both together not sufficient)? Am I missing something?


What you did is only true is (x+y)>0.
If you multiply both sides of an inequality with a number, the sign remains same if the number or expression is positive, whereas it flips if it is negative.


A counter example to show answer is indeed (e) is taking x=5 and y=-20 ... you'll get 25/-15 = -5/3 which is less than 1 ... on the other hand x=5 & y=-1 would get you 6/4 which is greater than 1.
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Re: gmat prep DS [#permalink]

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New post 22 Dec 2010, 07:59
Thanks Bunuel, that explains it perfectly
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Re: gmat prep DS [#permalink]

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New post 13 Mar 2011, 08:17
(x-y/x+y)>1
What is the problem if i try numbers
(1) x>0 no information about y clearly insuf.
(2) y <0 no information about x. insuf.

For C
x=2, y=-1
2+1/2-1>1
again
x=2, y=-3
2+3/2-3<1

so ans E.

Help if i wrong.
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if x is different from -y, in (x-y)/(x+y) greater than 1 ? [#permalink]

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if x is different from -y, in (x-y)/(x+y) greater than 1 ?
1. x > 0
2. y< 0
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Re: Did Sal had this question wrong ? [#permalink]

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New post 14 Apr 2011, 04:10
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]

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New post 14 Apr 2011, 05:11
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]

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New post 14 Apr 2011, 18:18
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]

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New post 14 Apr 2011, 22:31
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]

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New post 17 Apr 2011, 05:55
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]

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New post 19 May 2011, 23:45
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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If x is not equal to -y,is (x -y)/(x+y) >1? [#permalink]

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New post 15 Nov 2012, 18:55
Hi,

Request help with the following question.Thanks..


If x is not equal to -y,is (x -y)/(x+y) >1?

1.x > 0
2.y < 0

P.S. - While i can solve the question by testting numbers,i would be most grateful if somebody could help me understand the fraction theory being tested on this question.
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Re: If x is not equal to -y,is (x -y)/(x+y) >1? [#permalink]

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New post 15 Nov 2012, 19:14
shivanigs wrote:
Hi,

Request help with the following question.Thanks..


If x is not equal to -y,is (x -y)/(x+y) >1?

1.x > 0
2.y < 0

P.S. - While i can solve the question by testting numbers,i would be most grateful if somebody could help me understand the fraction theory being tested on this question.

Here you go:
First, lets rephrase the question-
\((x -y)/(x+y) >1\)
\((x -y)/(x+y) -1 >0\)
\(-2y/(x+y) >0\)
or
is \(y/(x+y) < 0\) ?

Statement 1:
x >0
This doesnt help us understanding the sign of numerator or denominator. Not sufficient.
Statement 2:
y<0
now we know, that numerator is postive, but we still dont know what is the sign for x+y. Not sufficient.

Combining 1 and 2. x>0, y<0
We want to know if
y/(x+y) < 0
Numerator is negative here, and denominator can be positive or negative depending upon absolute values of x and y. Not sufficien

Hence, Ans E it is!
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Re: If x is not equal to -y,is (x -y)/(x+y) >1? [#permalink]

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New post 15 Nov 2012, 19:21
shivanigs wrote:
Hi,

Request help with the following question.Thanks..


If x is not equal to -y,is (x -y)/(x+y) >1?

1.x > 0
2.y < 0

P.S. - While i can solve the question by testting numbers,i would be most grateful if somebody could help me understand the fraction theory being tested on this question.


Responding to a pm:

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

What do you do when you are given >1 ? It is hard to find implications of '>1'. It is much easier to handle '>0'

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

When will this be positive? In 2 cases:
1. When y is negative and (x+y) is positive i.e. x is positive and x has greater absolute value than y's absolute value.
2. When y is positive and (x+y) is negative i.e. x is negative and x has greater absolute value than y's absolute value.

Both statements together tell us that y is negative and x is positive but we don't know whether 'x's absolute value is greater than y's absolute value'.
Hence (E)
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Re: If x is not equal to -y,is (x -y)/(x+y) >1?   [#permalink] 15 Nov 2012, 19:21

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