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

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

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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 09 Oct 2013, 09:00, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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Re: (x-y)/ (x+y) < 1 [#permalink]

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New post 24 Sep 2009, 20:51
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I think it is B.

x-y < x+y => y>0

So stmt 2 gives the answer
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Re: (x-y)/ (x+y) < 1 [#permalink]

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New post 24 Sep 2009, 23:26
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seems like an E to me

Stament 1 does not allow for 1 answer:

X = 5, Y may be positive, say 2, then 5-2 / 5+2 < 1, while if Y= -2, then 5-(-2) /5-2 > 1

Same for statement 2.

Both combined allow division by 0: X=1, Y=-1, then 1-(-1) /1-1, which also does not give a clear answer.

even if you take X=2, Y=-1, then get 2-(-1) / 2-1 > 1, while if X=1, Y=-3, then 1-(-3) / 1-3 < 1

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Re: (x-y)/ (x+y) < 1 [#permalink]

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New post 25 Sep 2009, 07:53
(x-y)/ (x+y) < 1 reduces to y>0? so my ans = B

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Re: (x-y)/ (x+y) < 1 [#permalink]

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New post 25 Sep 2009, 13:51
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1) insuff.
4 cases.
a)y>0 and y<x answer is yes
b)y>0 and y>x answer is yes
c)y<0 and absolute value y>x answer is yes
d)y<0 and absolute value y<x answer is no
2)insuff
4 cases
a and b is like the above where is x>0 and answer is yes
c)x<0 and y<x answer is yes
d)x<0 and y>x answer is no

together
x>0 and y<0
insuff
depends on the comparison between the absolute value of y and x.

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Re: (x-y)/ (x+y) < 1 [#permalink]

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New post 27 Sep 2009, 04:00
I would go for E unless statement 2 states Y is < -1, pls post OA

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Re: (x-y)/ (x+y) < 1 [#permalink]

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New post 27 Sep 2009, 09:39
(x-y)/(x+y)=(x+y-2y)/(x+y)=1-2y/(x+y) <1 ?

1, x>0, can't find the sign for y/(x+y), insuff
2, y<0, -2y>0 but we don't know whether x+y post or neg insuff

both 1&2, x>0 & y<0 but we don't know x+y>0 or not, insuff

E for me

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Re: (x-y)/ (x+y) < 1 [#permalink]

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New post 02 Oct 2009, 04:58
Sometimes its better to put values and check :-D

X=2,
Y=-3

This will the condition true

However,

X=3,
Y=-2

This will make it false.

E :-D

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Re: (x-y)/ (x+y) < 1 [#permalink]

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New post 23 Oct 2009, 16:30
Putting numbers is actually a cool strategy.......but it is time consuming....any shortcut way to solve this problem?

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Re: Is (x-y)/ (x+y) < 1 1) x > 0 2) y < 0 [#permalink]

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New post 09 Oct 2013, 08:53
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Re: If x#-y is (x-y)/(x+y)>1? [#permalink]

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New post 09 Oct 2013, 09:01
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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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-y-is-x-y-x-y-1-1-x-0-2-y-96090.html
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Re: If x#-y is (x-y)/(x+y)>1? [#permalink]

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New post 12 Oct 2015, 19:25
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If x#-y is (x-y)/(x+y)>1?   [#permalink] 12 Oct 2015, 19:25
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