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If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0

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CEO
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Joined: 20 Mar 2014
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Re: If x ≠ -y, is (x-y)/(x+y)> 1 ? (1) x > 0 (2) y < 0  [#permalink]

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New post 11 Jan 2016, 13:03
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NoHalfMeasures wrote:
If x ≠ -y, is (x-y)/(x+y)> 1 ?
(1) x > 0
(2) y < 0


The question can be rewritten as,

(x-y)/(x+y)> 1 ---> (x-y)/(x+y)-1 >0 ---> -2y / (x+y)>0 ---> is y/(x+y) < 0?

Per statement 1, x>0. Taking (x,y) as (1,-4) and (1,-0.5) will give you 2 different answers for the question asked. Not sufficient.

Per statement 2, y<0. Again taking (x,y) as (1,-4) and (1,-0.5) will give you 2 different answers for the question asked. Not sufficient.

As the same 2 cases are applicable for the above 2 statements even after combining , you still get 2 different answers for the question asked.

Hence, E is the correct answer.
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Re: If x ≠ -y, is (x-y)/(x+y)> 1 ? (1) x > 0 (2) y < 0  [#permalink]

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New post 13 Jan 2016, 21:17
NoHalfMeasures wrote:
If x ≠ -y, is (x-y)/(x+y)> 1 ?
(1) x > 0
(2) y < 0


(x-y)/(x+y) >1 if only: (x-y) > (x+y) > 0 (condition 1) OR (x-y) < (x+y) < 0 (condition 2)

From (condition 1): (x-y) + (x+y) > (x+y) + (x+y) >0 ==> 2x>0 AND (x+y) - (x+y) > (x+y) - (x-y) ==> 0>2y
==> if we know x>0 AND y<0, we can make sure that (x-y)/(x+y)>1 ==> Answer choice is C

(it does not need to work further on condition 2)
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Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

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New post 09 Jan 2018, 19:24
gmatcracker2010 wrote:
If x ≠ -y is \(\frac{x-y}{x+y}>1\)?

(1) x > 0
(2) y < 0


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

(x-y)/(x+y) > 1
⇔ (x-y)(x+y) > ( x + y )^2, by multiplying both sides by (x+y)^2
⇔ x^2 - y^2 > x^2 + 2xy + y^2
⇔ 2y^2 + xy < 0
⇔ y(2y+x)<0

Since we have 2 variables (x and y) and 0 equations,C is most likely to be the answer. So, we should consider 1) & 2) first.

Conditions 1) & 2):
x = 3, y = -1 : Yes
x = -1, y = -1 : No

They are not sufficient.

Therefore, the answer is E.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

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New post 26 Mar 2019, 22:32
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Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0   [#permalink] 26 Mar 2019, 22:32

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