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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]  19 Jun 2010, 02:45
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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]  19 Jun 2010, 03:34
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gmatcracker2010 wrote:

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.

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]  01 Aug 2010, 14:03
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

Cheers
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Re: GMATPrep DS question [#permalink]  01 Aug 2010, 14:28
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]  01 Aug 2010, 14:37
$$\frac{x-y}{x+y}>1$$
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]  01 Aug 2010, 14:56
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]  03 Aug 2010, 09:22
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]  27 Oct 2010, 13:20
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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
-y > y
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]  22 Dec 2010, 06:59
Thanks Bunuel, that explains it perfectly
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Re: gmat prep DS [#permalink]  13 Mar 2011, 07:17
(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 not equal to -y,is (x -y)/(x+y) >1? [#permalink]  15 Nov 2012, 17:55
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]  15 Nov 2012, 18:14
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]  15 Nov 2012, 18:21
Expert's post
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: x ≠ -y, is (x-y)/(x+y) > 1 [#permalink]  16 May 2013, 12:48
gmacforjyoab wrote:
if x ≠ -y, is (x-y)/(x+y) > 1?

1) x> 0
2) y<0

The answer is [E]. I marked it [B], but as I began to write the solution, the obviousness of the solution hit me!

The expression, $$(x-y)/(x+y) > 1$$ when simplified gives us y < 0. Clearly the above will drive us directly to [B]. But when we consider this, in the case y < 0 and x > 0, and no relation is given between x and y, what happens if |x| < |y|. In such a case x+ y < 0. But clearly, x-y is greater than 0! Hence the solution doesn't stick! This is the reason why [B] can not be the answer! Similarly if x < 0 and y < 0, we cannot for sure determine the sign of the fraction.

Also, clearly the individual choice x > 0 does not help! so, neither of the choices lead to our answer. To actually solve the above we need a relation determined between x and y so as to determine the sign of (x-y)/(x+y). Wonderful question gmacforjyoab!

Hope my procedure is what you expected as an answer!

Regards,
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Re: x ≠ -y, is (x-y)/(x+y) > 1 [#permalink]  16 May 2013, 12:58
if x ≠ -y, is (x-y)/(x+y) > 1?

1) x> 0
2) y<0

is [(x-y)-(x+y)]/(x+y) > 0 , i.e. is -2y / x+y > 0 i.e. is 2y / x+y < 0? true if 1) y <0 , x>0 and /y/ > /x/ 2) Y>0 AND X<0 AND /X/>/Y/

from 1

x is +ve .... insuff

from 2

y -ve insuff

both

this looks like the 1st case described above but we dont know whether /y/ > /x/ ..... insuff

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Re: If x#-y is (x-y)/(x+y)>1? (1) x>0 (2) y<0 [#permalink]  03 Oct 2014, 12:45
having a hard time seeing how \frac{(x -y)}{(x+y)} -1 > 0 ?
becomes \frac{-2y}{(x+y)} > 0 ?

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Re: If x#-y is (x-y)/(x+y)>1? (1) x>0 (2) y<0 [#permalink]  03 Oct 2014, 12:47
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bsmith37 wrote:
having a hard time seeing how \frac{(x -y)}{(x+y)} -1 > 0 ?
becomes \frac{-2y}{(x+y)} > 0 ?

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Re: If x#-y is (x-y)/(x+y)>1? (1) x>0 (2) y<0 [#permalink]  03 Oct 2014, 14:13
Bunuel wrote:
bsmith37 wrote:
having a hard time seeing how \frac{(x -y)}{(x+y)} -1 > 0 ?
becomes \frac{-2y}{(x+y)} > 0 ?

yes, even in your explanation i just dont see the progression how the "1-" goes away and we change the numerator to "x+y-x+y" to arrive at 2y
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Re: If x#-y is (x-y)/(x+y)>1? (1) x>0 (2) y<0 [#permalink]  03 Oct 2014, 14:17
Expert's post
bsmith37 wrote:
Bunuel wrote:
bsmith37 wrote:
having a hard time seeing how \frac{(x -y)}{(x+y)} -1 > 0 ?
becomes \frac{-2y}{(x+y)} > 0 ?

yes, even in your explanation i just dont see the progression how the "1-" goes away and we change the numerator to "x+y-x+y" to arrive at 2y

You need to brush-up fundamentals:

$$1-\frac{x-y}{x+y}$$;

$$\frac{x+y}{x+y}-\frac{x-y}{x+y}$$;

$$\frac{(x+y)-(x-y)}{x+y}$$;

$$\frac{2y}{x+y}$$.
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Re: If x#-y is (x-y)/(x+y)>1? (1) x>0 (2) y<0 [#permalink]  03 Oct 2014, 14:19
yes perhaps i do. thanks
Re: If x#-y is (x-y)/(x+y)>1? (1) x>0 (2) y<0   [#permalink] 03 Oct 2014, 14:19
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