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# If x not equals to -y, is (x-y)/(x+y)>1? 1. x >0 2.

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Manager
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If x not equals to -y, is (x-y)/(x+y)>1? 1. x >0 2. [#permalink]

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12 Jan 2009, 03:06
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If x not equals to -y, is (x-y)/(x+y)>1?
1. x >0
2. y< 0

Kudos [?]: 225 [0], given: 4

Intern
Joined: 16 Jun 2008
Posts: 17

Kudos [?]: 2 [0], given: 0

Location: Kiev, Ukraine

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12 Jan 2009, 06:21
(x-y)/(x+y)>1 means that (x-y)>(x+y)

We solve this equation and get 2y<0, so y<0. Second assumption gives us exactly this condition, so second assumption is enough to answer that (x-y)/(x+y)>1.

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Manager
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Kudos [?]: 225 [0], given: 4

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13 Jan 2009, 14:39
OA is E.
Can anyone suggest how to tackle this one?

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13 Jan 2009, 20:48
sandipchowdhury wrote:
If x not equals to -y, is (x-y)/(x+y)>1?
1. x >0
2. y< 0

I would suggest pluging in:

1: x > 0 but y? could be -ve 0 or +ve.

i. If y = 0, (x-y)/(x+y) = 1.
ii. If y > x > 0, (x-y)/(x+y) < 0.
iii. If x > y > 0, (x-y)/(x+y) < 1.
iv. If y < 0 and lyl < x, (x-y)/(x+y) > 1.
iv. If y < 0 and lyl > x, (x-y)/(x+y) < 0.

2: y < 0 but x ? could be -ve 0 or +ve. same as in 1.

i. If x = 0, (x-y)/(x+y) = -1.
ii. If x > 0 and x > lyl, (x-y)/(x+y) > 1.
iii. If x > 0 and x < lyl, (x-y)/(x+y) < 0.
iv. If 0 < x < y, (x-y)/(x+y) > 0 but <1.
v. If 0 < y < x, (x-y)/(x+y) > 0 but <1.

1&2:

i. If lyl < x, (x-y)/(x+y) > 1.
ii. If lyl > x, (x-y)/(x+y) < 0.

So -- E.

PS: Appreciate you for pointing out typos, if any.
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Kudos [?]: 841 [0], given: 19

Manager
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Posts: 94

Kudos [?]: 225 [0], given: 4

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13 Jan 2009, 22:54
is there a shorter way to solve this?
Can we simplify the equation before pluging in?

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16 Jan 2009, 11:10
Thanks cowalko, who generously pointed a mistake.
I knew I was missing something while doing this. My intuition!

GT

cowalko wrote:
Subject: DS-Inequality GMATprep

GMAT TIGER wrote:
sandipchowdhury wrote:
If x not equals to -y, is (x-y)/(x+y)>1?
1. x >0
2. y< 0

I would suggest pluging in:

1: x > 0 but y? could be -ve 0 or +ve.

i. If y = 0, (x-y)/(x+y) = 1.
ii. If y > x > 0, (x-y)/(x+y) < 0.
iii. If x > y > 0, (x-y)/(x+y) < 1.
iv. If y < 0 and lyl < x, (x-y)/(x+y) > 1.
iv. If y < 0 and lyl > x, (x-y)/(x+y) < 0.

2: y < 0 but x ? could be -ve 0 or +ve. same as in 1.

i. If x = 0, (x-y)/(x+y) = -1.
ii. If x > 0 and x > lyl, (x-y)/(x+y) > 1.
iii. If x > 0 and x < lyl, (x-y)/(x+y) < 0.
iv. If 0 < x < y, (x-y)/(x+y) > 0 but <1.
v. If 0 < y < x, (x-y)/(x+y) > 0 but <1.

1&2:

i. If lyl < x, (x-y)/(x+y) > 1.
ii. If lyl > x, (x-y)/(x+y) < 0.

So -- E.

PS: Appreciate you for pointing out typos, if any.

Just one note not affecting the solution: Neither y nor x can be 0, because the problem says that x does not equal -y. X only equals -Y when x and y are 0

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Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

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GMAT Tutor
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16 Jan 2009, 14:47
nlutsenko wrote:
(x-y)/(x+y)>1 means that (x-y)>(x+y)

We solve this equation and get 2y<0, so y<0. Second assumption gives us exactly this condition, so second assumption is enough to answer that (x-y)/(x+y)>1.

You can't simplify as you've done: you multiplied both sides of the inequality by x+y, which you can't do, because you don't know if x+y is negative or positive. If x+y is negative, you'd need to reverse the inequality.

sandipchowdhury wrote:
is there a shorter way to solve this?
Can we simplify the equation before pluging in?

Yes, the question can be answered quickly. Even using both statements, on the left side of the inequality, the numerator is certainly positive, but the denominator could be positive and could be negative. So the left side of the inequality could be positive or negative. Since it can be negative, and it can also clearly be larger than 1 (choose x = 2, y = -1), the statements together are insufficient. So E.
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Manager
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16 Jan 2009, 22:20
Thanks Ian. Good thinking.

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Re: DS-Inequality GMATprep   [#permalink] 16 Jan 2009, 22:20
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